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abcminiuser
PostPosted: Feb 16, 2010 - 04:04 AM
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For an updated version of this tutorial in PDF format, please see this page of my website.

Newbie's Guide to AVR Interrupts
By Dean Camera, 2010

Hi all. This is a small guide to teach those new to the embedded world about the wonders of processor interrupts. Feedback appreciated.


Part 1: What are Interrupts?

One of the most fundamental and useful principles of modern embedded processors are interrupts. There seems to be a lot of confusion about interrupts; what are they, and how do we use them? In short, an interrupt is a way for an external (or, sometimes, internal) event to pause the current processor's activity, so that it can complete a brief task before resuming execution where it left off. Consider the following:

Let's say we are at home, writing an excellent tutorial on how a principle of modern embedded processors works. We are very interested in this topic, so we are devoting all our concentration to our keyboard. However, half way though, the phone rings. Despite not being by the phone waiting for a call, we are able to stop what we are doing, take the call, and go back where we left off once we have hung up.

This is how a processor interrupt works. We can set up the processor so that it is looking for a specific external event (like a pin going low, or a timer overflowing) to become true, while it goes on and performs other tasks. When these events occur, we stop the current task, handle the event, and resume back where we left off. This gives us a great deal of flexibility; rather than having to actively poll our events to see if they have happened, we can instead just ignore them completely and trust our interrupt routines to process them instead.

What we are doing is called asynchronous processing - that is, we are processing the interrupt events outside the regular "execution thread" of the main program. It's important to note that we are not doing two things at once (although it may appear that way) - the regular program is stopped while an interrupt routine runs.

When correctly set up, we can link a specific interrupt source to a specific handler routine, called an Interrupt Service Routine, or ISR for short.

Part 2: What can cause interrupts?

There are two main sources of interrupts:

Arrow Hardware Interrupts, which occur in response to a changing external event such as a pin going low, or a timer reaching a preset value
Arrow Software Interrupts, which occur in response to a command issued in software

The 8-bit AVRs lack software interrupts, which are usually used for special operating system tasks like switching between user and kernel space, or for handling exceptions. Because of this, we'll only be looking at hardware interrupts.

Each AVR model contains a different set of interrupt sources. We can find out which interrupts our chosen model has by looking in the "Interrupts" chapter of the AVR's datasheet. This chapter contains a table, similar to the following fragment from the AT90USB1287 datasheet:

Code:
Vector No. | Address | Source | Interrupt Definition
2          | $0002   | INT0   | External Interrupt Request 0
3          | $0004   | INT1   | External Interrupt Request 1
4          | $0006   | INT2   | External Interrupt Request 2


This contains a complete list of all the possible interrupts in our AVR, giving each interrupt's Vector table address, Source Interrupt Name, and Source Interrupt Description. Take a look at your AVR's list - see the variety of interrupt sources, from all sorts of on-chip peripherals of the AVR.

Hardware interrupts can be useful to process sparingly occurring events, such as buttons presses or alarm inputs. They can also be useful for situations where a low, fixed latency is required, for example when processing inputs from a rotary encoder.

Part 3: So how to we define an ISR?

How we do interrupts varies between programming languages, but at the lowest level, we are simply making a general AVR function somewhere in the AVR's FLASH memory space, and "linking" it to a specific interrupt source by placing a JMP or RJMP instruction to this function at the address specified in the table we just looked at. At the start of the AVR's FLASH memory space lies the Interrupt Vector Table, which is simply a set of hardwired addresses which the AVR's processor will jump to when each of the interrupts fire. By placing a jump instruction at each interrupt's address in the table, we can make the AVR processor jump to our ISR which lies elsewhere.

For an ISR to be called, we need three conditions to be true:

Arrow Firstly, the AVR's global Interrupts Enable bit (I) must be set in the MCU control register SREG. This allows the AVR's core to process interrupts via ISRs when set, and prevents them from running when cleared. It defaults to being cleared on power up, so we need to set it.

Arrow Secondly, the individual interrupt source's enable bit must be set. Each interrupt source has a separate interrupt enable bit in the related peripheral's control registers, which turns on the ISR for that interrupt. This must also be set, so that when the interrupt event occurs the processor runs the associated ISR.

Arrow Thirdly, The condition for the interrupt must be met - for example, for the USART Receive Complete (USART RX) interrupt, a character must have been received.

When all three conditions are met, the AVR will fire our ISR each time the interrupt event occurs.

Again, the method used to define an ISR differs between langages and compilers, so I'll just put the AVR-GCC and AVR ASM versions here.

Code:
#include <avr/interrupt.h>

ISR({Vector Source}_vect)
{
   // ISR code to execute here
}


Code:
.org 0x{Vector Address}
jmp MyISRHandler


MyISRHandler:
  ; ISR code to execute here
  reti


Note that in the case of the assembly version, we need to add a "reti" instruction at the end of our interrupt instead of the usual "ret" instruction to return to the main program's execution; this special instruction has the dual function of exiting the ISR, and automatically re-enabling the Global Interrupt Enable bit. This happens inside the C version too when the function returns, we just don't see it normally.

This raises the next point; by default, interrupts are themselves not interruptable. When an interrupt fires, the AVR CPU will automatically disable the Global Interrupt Enable bit, to prevent the ISR from being itself interrupted. This is to prevent stack overflows from too many interrupts occurring at once and to prevent the ISRs from running too long, as most uses of interrupts are to have minimal latency when processing an event. It's perfectly possible to set the Global Interrupt Enable bit again as part of the ISR so that nested interrupts can occur, but this is highly not recommended as it is dangerous.

Part 4: Enabling a Specific Interrupt?

If you simply add in an ISR to your existing program, you will find that it appears to do nothing when you try to fire it. This is because while we have defined an ISR, we haven't enabled the interrupt source! As mentioned in part 3, we need to meet all three criteria for the ISR to fire.

Firstly, we need to set the I bit in the SREG register. This is the Global Interrupt Enable bit, without which the AVR will simply ignore any and all interrupts. In assembly, we have two special single-cycle instructions for dealing with the I bit:

sei, which SEts the I flag
cli, which CLears the I flag

While in C, we have to use special macros from our libc library's header. In the case of AVR-GCC and its avr-libc library, we just use the sei() and cli() macro equivalents defined in <avr/interrupt.h>:

Code:
sei ; Enable Global Interrupts


Code:
sei(); // Enable Global Interrupts



Next, we also need to enable a specific interrupt source, to satisfy the second condition for firing an ISR. The way to do this varies greatly between interrupt sources, but always involves setting a specific flag in one of the peripheral registers. Let's set an interrupt on the USART Receive Complete (USART RX) interrupt. According to the datasheet, we want to set the RXCIE bit in the UCSRB register:

Code:
in r16, UCSRB
ori r16, (1 << RXCIE)
out UCSRB, r16


Code:
UCSRB |= (1 << RXCIE);



That should be enough to make the AVR's execution jump to the appropriate ISR when the interrupt occurs.

Part 5: Things you Need to Know

There are a few things to keep in mind when using interrupts in your program:

1) When an interrupt is executing, your main program isn't. That means that if your application isn't robust and your receive many interrupts in a very short space of time, your main program's execution will slow to a halt. Keep this in mind when you decided to attach a 4MHz clock input to an external interrupt pin.

2) Data shared between the ISR and your main program must be both volatile and global in scope in the C language. Without the volatile keyword, the compiler may optimize out accesses to a variable you update in an ISR, as the C language itself has no concept of different execution threads. Take the following example:

Code:
#include <avr/interrupt.h>

int MyValue;

ISR(SomeVector_vect)
{
   MyValue++;
}

int main(void)
{
   SetupInterrupts();

   while (MyValue == 0); // Wait for interrupt

   TurnOnLED();
}


There is a good chance the program will freeze forever in the while loop, as (according to the optimizer) the "MyValue" global's value never changes in the loop. By making the global variable volatile, you are declaring to the compiler that the value may change due to circumstances it is not aware of, forcing it to generate code to reload the value each time.

3) There's another subtle bug in the above code sample; it's possible at the machine code level for an interrupt to occur half way between the fetching of the two bytes of the integer variable MyValue, causing corruption if the variable's value is altered inside the ISR. This can be avoided by making the fetching of the variable's value in the main program code "atomic", by disabling the global interrupt enable bit while the fetching is taking place. In avr-libc this can be done in the following manner:

Code:
#include <avr/interrupt.h>
#include <util/atomic.h>

volatile int MyValue;

ISR(SomeVector_vect)
{
   MyValue++;
}

int main(void)
{
   SetupInterrupts();

   int MyValue_Local;

   do
   {
      ATOMIC_BLOCK(ATOMIC_RESTORESTATE)
      {
         MyValue_Local = MyValue;
      }
   } while (MyValue_Local == 0) // Wait for interrupt

   TurnOnLED();
}


Your compiler may vary.

4) Each interrupt source has exactly one flag to indicate that the event occurred. If two or more of the same interrupt event occurs while you are in the middle of processing an ISR, you will only receive one ISR call for the event when the current ISR terminates, and the remaining interrupt events will be lost.

5) The interrupt source flag is usually cleared when the ISR fires. This isn't always the case (some interrupt flags are cleared when a particular register is read, such as the USART receive complete flag) however this is more of an exception to the rule. The upshot of this is that if you receive another interrupt to the source you are currently processing while in its ISR, you'll get another call to the ISR once the current one terminates. Again, you are still limited to a single "event occurred" flag for each interrupt, so you can miss interrupts if many happen at once before you have a chance to process them.

Part 6: Putting it all together

Rather than waste my efforts by repeating what I've already done before here, I will instead direct readers to my previous tutorial on Interrupt Driven USART Communications for a short, practical example of how to use interrupts in an application.

For an updated version of this tutorial in PDF format, please see this page of my website.

- Dean Twisted Evil

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Last edited by abcminiuser on Feb 04, 2012 - 02:18 PM; edited 4 times in total
 
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siddhant3s
PostPosted: Feb 16, 2010 - 08:48 AM
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Awesome. Another great tutorial from the man.
One addition to what I know about this:
Many times, the source name of the interrupt contains spaces ( Eg:TIMER1 COMPA. In this case, while writing the ISR, just replace all the spaces with underscores.
As a rule of thumb, ``just replace all the spaces with underscore and add a _vect"
 
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clawson
PostPosted: Feb 16, 2010 - 11:04 AM
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siddhant3s,

I wouldn't guess at interrupt vector routine names as you suggest - far better is to look them up in the compiler's manual. As Dean mentions GCC above the manual page in question is:

http://www.nongnu.org/avr-libc/user-man ... rupts.html

specifically the "Choosing the vector: Interrupt vector names" section.

Cliff

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Kartman
PostPosted: Feb 18, 2010 - 05:50 AM
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Dean, your sample code has a shared int that isn't accessed atomically! Debateable as to the impact on the example given since only 0 is tested.
 
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abcminiuser
PostPosted: Feb 18, 2010 - 05:57 AM
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Good point Kartman, I've updated the text.

This isn't one of my finest works, I wrote it while I wasn't in the mood to write, so it's not particularly gripping. I'll re-write it at some point in the future -- or someone else can make their own version using the information here as a guide.

- Dean Twisted Evil

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stu_san
PostPosted: Feb 18, 2010 - 06:03 PM
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Good job, Dean! This encompasses all of the items (plus a few more) that I had set aside for my tutorial on the same subject. Even if you consider it not one of your finest works, it is a good and needed tutorial.

If I might suggest an item or two:

- Make all of your full program examples include a "forever" loop at the end of main, as well as a "return 0;". Many newbies quite literally grab your examples out of the tutorial. Mistakes in the example programs will always show up on the board.

- In your copious spare time Wink would you add a section on when an interrupt makes sense and when it does not? The classic example of this is external switch handling, with debounce and all that. It may be too much for here, but perhaps something to the effect of, "processors are fast. very fast. humans are mind-numbingly slow compared to processors. Anything that needs to interact with a human almost never needs a hardware interrupt. Instead, simple polling of the pin is preferable."

Again, thanks for your work! I (and many of us) know how little extra time one has when attending a university.

Stu

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mcuaust
PostPosted: Feb 21, 2010 - 09:32 AM
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Thanks for the tutorial Dean.
If you could post it in your standard PDF format that'd be appreciated too.
 
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pak_soft
PostPosted: Feb 21, 2010 - 12:00 PM
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while (MyValue_Local == 0); // <----

It may be typo error.
Thank Dean for your tutorial, can you post some examples about "atomic" usage?
 
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clawson
PostPosted: Feb 21, 2010 - 01:18 PM
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Quote:

If you could post it in your standard PDF format that'd be appreciated too

Here's a PDF of the above but you could have done this just as easiy as I did. Just highlighted Dean's post, Ctrl-C, switch to MS Word, Ctrl-V into a new document then Print... and use a PDF writer (I use CutePDF).

Cliff

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mcuaust
PostPosted: Feb 21, 2010 - 11:28 PM
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Thanks for doing that Cliff, and for the tip.
 
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hooverphonique
PostPosted: Feb 25, 2010 - 10:09 PM
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nice tut...

since this is labeled as an "AVR" tutorial and not "just" an "ATMega" tutorial, I have a few points:

On the ATMega, a "cli" and "ret" instruction works just as well as the usual "reti" to exit an interrupt.
On the XMega's however, it is extremely important to always use "reti", otherwise the mcu will stay at the current interrupt level (yes, the XMega has 3 nesting interrupt levels, in case you didn't know Smile

this is especially important for OS's that use an ISR for task switching by switching the stack, because the new stack in effect when exiting the ISR may not have the required number of "reti"s to back out of the interrupted state.
 
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Koshchi
PostPosted: Feb 25, 2010 - 10:26 PM
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Quote:
On the ATMega, a "cli" and "ret" instruction works just as well as the usual "reti" to exit an interrupt.

But they are not completely equivalent. If you do that then there is a possibility that the ISR will be entered again without any opcodes of the main thread being run. This is not likely to make a differance, but you should be aware of it. It is best to simply use RETI.

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hooverphonique
PostPosted: Feb 25, 2010 - 10:40 PM
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Koshchi wrote:
Quote:
On the ATMega, a "sei" and "ret" instruction works just as well as the usual "reti" to exit an interrupt.

But they are not completely equivalent. If you do that then there is a possibility that the ISR will be entered again without any opcodes of the main thread being run. This is not likely to make a differance, but you should be aware of it. It is best to simply use RETI.


you are correct, sir.. it may prevent one instruction from the main thread to be executed, before an isr could fire again..
FreeRTOS exhibits this behaviour in certain scenarios because the code path generates this exact instruction sequence sometimes (which is partly why it doesn't work properly on xmegas).

I meant "sei" earlier, btw, not "cli"..
 
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boaz_o
PostPosted: Apr 05, 2010 - 07:47 PM
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hi,
can you explain the difference between PCINT and INT0 ?

thanks for the effort writing this tut..

regards

boaz
 
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clawson
PostPosted: Apr 05, 2010 - 07:54 PM
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Quote:

can you explain the difference between PCINT and INT0 ?

PCINTs can only detect changes in state and they are grouped in 8's so each ISR is for 8 pins and you then have to work out in the ISR which pin it was that changed and triggered the ISR.

INTn's have an ISR each and can be configured for all of high-level, low-level, rising-edge and falling-edge.

So you get more functionality with INTn's but there's less of them.

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ayeosq
PostPosted: May 03, 2010 - 06:22 AM
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Code:

void SetupInterrupts(void)
{
    sei();//Enable Global Interrupts
    EIMSK=0x01;
    EICRA=0x03;
}
ISR(INT0_vect)
{
     PORTC=0xFE;
}

I am pretty new to this interrupt thing, first time doing avr programming
Am I doing it right?
I am working on a ATmega88...
 
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ayeosq
PostPosted: May 03, 2010 - 08:52 AM
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Ok, my bad, push the wrong input!
Cheers!
 
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ayeosq
PostPosted: May 03, 2010 - 09:02 AM
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Now I need to time the INTO is pushed...
Any one can help?
 
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qli029
PostPosted: Jul 21, 2010 - 01:26 AM
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very useful,
 
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jay.chhatpar
PostPosted: Jul 21, 2010 - 10:22 PM
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Hi! I am using a Rotary Encoder.Can someone please help me with the code. I have encoder sensor connected at PORTD pin2 INT0, Motor Connected at PORTB pin0,1.The encoder sensor gives pulses when motor rotates. I need to stop the motor after pulse count=30.
 
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clawson
PostPosted: Jul 22, 2010 - 10:25 AM
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Do NOT hijack threads - your question has nothing to do with discussion of this tutorial article. Try posting this question as a new thread in AVR Forum.

Moderator

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jay.chhatpar
PostPosted: Jul 24, 2010 - 07:14 AM
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It was the first time on forum didn't know i was hijacking the thread. I am sorry
 
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dpaulsen
PostPosted: Aug 25, 2010 - 07:15 PM
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I wish this tutorial was around when I first started. Would have saved me a lot of grief in the beginning. Kudos.

I have one suggestion.
abcminiuser wrote:
4) Each interrupt source has exactly one flag to indicate that the event occurred. If two such interrupts occur while you are in the middle of processing another interrupt's ISR, you will only receive one ISR call when the current ISR terminates. This means that if you have lots of interrupts occurring in a short space of time, you can miss interrupts.


When I first read this I thought you were saying that if two different interrupts occur then you will only service one of them, which I knew was wrong. Then when I read it a few times I realized you meant that if the same interrupt occurs in rapid succession. I think this could be made a little more clear for newbies.

Just a suggestion, take it or leave it.

-Daniel
 
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JohanEkdahl
PostPosted: Aug 25, 2010 - 08:57 PM
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Dean (abcminiuser) has several excellent qualities:

1. He writes top-notch tutorials.
2. He get a lot of praise for them, but that does not go to his head.
3. He does not leave his tutorials without attention. He reacts to reasonable suggestions for change.

Your suggestion seems reasonable. I'm sure that Dean will take you suggestion. Thank you for making AVRfreaks better!

Also, Dean, isn't it so that it does not have to be another interrupt that creates the blocking. Eg a for slow pin change ISR and the pin changing so fast it does so three times during the execution of the ISR, the first change will be serviced immediately, the second will be "noted by hardware" and the ISR will be executed again as soon as the first instance exits (or an SEI...), and the third one will be lost. Correct?
 
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abcminiuser
PostPosted: Aug 25, 2010 - 11:41 PM
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Quote:

When I first read this I thought you were saying that if two different interrupts occur then you will only service one of them, which I knew was wrong. Then when I read it a few times I realized you meant that if the same interrupt occurs in rapid succession. I think this could be made a little more clear for newbies.


I've altered the paragraph to make my intention clearer. This really wasn't one of my better tutorials, but any information on a subject is better than none, so I posted it anyway under the guise of "waste not, want not".

Quote:

Also, Dean, isn't it so that it does not have to be another interrupt that creates the blocking. Eg a for slow pin change ISR and the pin changing so fast it does so three times during the execution of the ISR, the first change will be serviced immediately, the second will be "noted by hardware" and the ISR will be executed again as soon as the first instance exits (or an SEI...), and the third one will be lost. Correct?


Correct as ever Johan - any blocking ISR thread can potentially cause you to miss events, if two or more of the same event occurs while in an ISR. This is (for the benefit of others) due to each interrupt source having only a single flag to denote when the ISR event precondition has occurred.


One thing I forgot to mention - but which isn't really all that useful to most people - is this:

Quote:
and the ISR will be executed again as soon as the first instance exits (or an SEI...)


Isn't quite 100% right, assuming you interpret "immediately" to mean "the next clock cycle". On the AVR processors, it is guaranteed that at least one clock cycle of the main execution thread will run after a RETI before the next ISR can execute. If your app is tight enough that running a single cycle of the main thread between interrupts makes any difference at all however, you really need to rethink your choice of clock and (possibly) processor.

- Dean Twisted Evil

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JohanEkdahl
PostPosted: Aug 26, 2010 - 08:13 AM
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Quote:
Isn't quite 100% right, assuming you interpret "immediately" to mean "the next clock cycle".

No, didn't intend to mean "next cycle". I wrestled a bit with finding different wording, but nothing came out nice. "Right away", "with no delay other than that of the hardware in a non-interrupt-blocking state" etc..

"Without waiting being blocked" might have done it.
 
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clawson
PostPosted: Aug 26, 2010 - 09:16 AM
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Quote:

it is guaranteed that at least one clock cycle of the main execution thread will run after a RETI before the next ISR can execute

OK can I be pedantic then? If you have a non-blocking ISR then when a nested ISR performs RETI it will be one opcode of that previous ISR and not mainline that will be executed before the next IF flag is serviced.

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mcuaust
PostPosted: Aug 26, 2010 - 10:56 AM
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Dean, Johan, etc.,
your willingness to help others, and your preparedness to work together, makes AVR Freaks such a resource.

Well done! (from a novice in the field).
 
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nobbyv
PostPosted: Sep 10, 2010 - 05:18 PM
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abcminiuser wrote:
[
5) The interrupt source flag is usually cleared when the ISR fires. This isn't always the case (some interrupt flags are cleared when a particular register is read, such as the USART receive complete flag) however this is more of an exception to the rule. The upshot of this is that if you receive another interrupt to the source you are currently processing while in its ISR, you'll get another call to the ISR once the current one terminates.
- Dean Twisted Evil


Just want to make sure I understand this point: so I have one interrupt set up. It fires and I enter my ISR and process some code. The same interrupt then fires again while processing the code in my ISR. The quote above means that I will complete my ISR, which I am already in, then start my ISR over again? Is this correct? Thanks
 
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Koshchi
PostPosted: Sep 10, 2010 - 05:46 PM
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Quote:
Is this correct?
Yes. Keep in mind though, if the same interrupt fires more than once while an ISR is being processed, then all but one of those triggerings will be lost.

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clawson
PostPosted: Sep 10, 2010 - 05:50 PM
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Also note that ONE opcode of the interrupted code will be called between each successive ISR entry. If your main (non interrupt code) is doing anything but an empty loop it probably won't do you any favours for the design to only allow one opcode of it to execute between each ISR. Usually in embedded design you move the "heavy work" out of the ISR and put it in the main code if it's not timing critical. Then just do the timing critical stuff in the ISR and then set a flag to let main handle the rest when it gets a chance.

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mcuaust
PostPosted: Sep 11, 2010 - 04:33 AM
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clawson wrote:
Also note that ONE opcode of the interrupted code will be called between each successive ISR entry.

Please explain what this means.
 
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Koshchi
PostPosted: Sep 11, 2010 - 05:02 AM
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It means exactly what it says: when an ISR finishes, at least one opcode of the code that was interrupted will be run before any other ISR is serviced.

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mcuaust
PostPosted: Sep 11, 2010 - 05:43 AM
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Thanks Koshchi, I understand it now.
 
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nobbyv
PostPosted: Sep 13, 2010 - 03:43 PM
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Koshchi wrote:
Quote:
Is this correct?
Yes. Keep in mind though, if the same interrupt fires more than once while an ISR is being processed, then all but one of those triggerings will be lost.


Understood. Thanks for the replies.
 
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macmad1984
PostPosted: Sep 28, 2010 - 06:05 PM
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Very good, thank you. This cleared up a lot I could not be bothered to read in the datasheets. Yes, I know, I'm lazy...

As a recommendation, more about the flow of the program would be helpful. For example, what happens to the PC during an interrupt? Does it get pushed to the stack like during a call? I gather from what you said about stack overflows that this may be the case. Sorry if I somehow missed this bit, and I don't mean to sound negative.

In short:
You have been very helpful, thanks a bunch for this great tutorial!Very Happy
 
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clawson
PostPosted: Sep 28, 2010 - 06:35 PM
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Quote:

As a recommendation, more about the flow of the program would be helpful. For example, what happens to the PC during an interrupt? Does it get pushed to the stack like during a call?

Yes. I know you don't like reading manuals (not a great way to start out as a programmer as about 15-20% of the job is reading data/manuals) but you can see this yourself by simulating a simple timer interrupt program in AVR Studio and breakpointing at the entry to the ISR. You will see the interrupted execution address at the base of the stack.

For example build this for mega328P and simulate in Studio Simulator 2:
Code:
#include <avr/io.h>
#include <avr/interrupt.h>

volatile uint8_t count;

int main(void)
{
   // build and simulate (V2) this for mega328P
   TIMSK0 |= (1<<TOIE0);
   TCCR0B |= (1<<CS00);
   sei();
   while(1) {
      PORTB = count;
      PORTD = ~count;
   }
}

ISR(TIMER0_OVF_vect) {
   count++;
}

The code builds to:
Code:
00000090 <main>:
volatile uint8_t count;

int main(void)
{
   // build and simulate (V2) this for mega328P
   TIMSK0 |= (1<<TOIE0);
  90:   80 91 6e 00    lds   r24, 0x006E
  94:   81 60          ori   r24, 0x01   ; 1
  96:   80 93 6e 00    sts   0x006E, r24
   TCCR0B |= (1<<CS00);
  9a:   85 b5          in   r24, 0x25   ; 37
  9c:   81 60          ori   r24, 0x01   ; 1
  9e:   85 bd          out   0x25, r24   ; 37
   sei();
  a0:   78 94          sei
   while(1) {
      PORTB = count;
  a2:   80 91 00 01    lds   r24, 0x0100
  a6:   85 b9          out   0x05, r24   ; 5
      PORTD = ~count;
  a8:   80 91 00 01    lds   r24, 0x0100
  ac:   80 95          com   r24
  ae:   8b b9          out   0x0b, r24   ; 11
  b0:   f8 cf          rjmp   .-16        ; 0xa2 <main+0x12>

000000b2 <__vector_16>:
   }
}

ISR(TIMER0_OVF_vect) {
  b2:   1f 92          push   r1
  b4:   0f 92          push   r0
  b6:   0f b6          in   r0, 0x3f   ; 63
  b8:   0f 92          push   r0
  ba:   11 24          eor   r1, r1
  bc:   8f 93          push   r24
   count++;
  be:   80 91 00 01    lds   r24, 0x0100
  c2:   8f 5f          subi   r24, 0xFF   ; 255
  c4:   80 93 00 01    sts   0x0100, r24
}
  c8:   8f 91          pop   r24
  ca:   0f 90          pop   r0
  cc:   0f be          out   0x3f, r0   ; 63
  ce:   0f 90          pop   r0
  d0:   1f 90          pop   r1
  d2:   18 95          reti

Put a breakpoint on the "push r1" at the start of the ISR.

Run the code. When it hits the breakpoint you will see that the SP value is 0x08FB. If you now open a memory window set the droplist to "Data" and then look at around 0x8F? You will see that the memory at 0x8FC contains 0x00 0x56. Looking at the code of the while(1) loop in main:
Code:
11:          sei();
+00000050:   9478        SEI                      Global Interrupt Enable
13:             PORTB = count;
+00000051:   91800100    LDS       R24,0x0100     Load direct from data space
+00000053:   B985        OUT       0x05,R24       Out to I/O location
14:             PORTD = ~count;
+00000054:   91800100    LDS       R24,0x0100     Load direct from data space
+00000056:   9580        LAT       R24            Load and Toggle
+00000057:   B98B        OUT       0x0B,R24       Out to I/O location
+00000058:   CFF8        RJMP      PC-0x0007      Relative jump

you'll see that 0x0056 is the LAT opcode. The system jumped to the vector after finishing the LDS opcode at location 0x0054

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badbyte
PostPosted: Nov 06, 2010 - 09:03 PM
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Hi People,
Im new in using the avr, wanted to try out the code with an interrupt example.
When reaching the line "MCUCR|=(1<<ISC01)|(1<<ISC00);" in debug mode, it instantly jumps to the ISR.
Any idea why?
Here is the code:
#include <avr/io.h>
#include <avr/signal.h>
#include <avr/interrupt.h>

int main(void)
{
DDRD=0x00;
DDRD=0xff;
// Enable INT0 External Interrupt
GICR |= (1<<INT0);

// Enable Interrupts
sei();
// Falling-Edge Triggered INT0
MCUCR|=(1<<ISC01)|(1<<ISC00);

while (1)
{

}
}

// External Interrupt 0 ISR
ISR(INT0_vect)
{
PORTB++;
}
 
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JohanEkdahl
PostPosted: Nov 06, 2010 - 09:16 PM
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Hey, badbyte!

We need to double check the bit numbers and stuff against the data sheet, but you're not telling what AVR model you're having.

Still, I have this from one AVR data sheet:
Quote:
When changing the ISCn1/ISCn0 bits, the interrupt must be disabled by clearing its Interrupt Enable bit in the EIMSK Register. Otherwise an interrupt can occur when the bits are change


I have this text right below the table describing the ISCnn bits. Yours too, perhaps?
 
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badbyte
PostPosted: Nov 06, 2010 - 09:18 PM
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Ah right. I'm using the Atmega32.
 
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JohanEkdahl
PostPosted: Nov 06, 2010 - 09:24 PM
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Quote:

I'm using the Atmega32.

And did the External Interrupt section in the data sheet have anything to say about the order in which SFR register bits should be manipulated?
 
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badbyte
PostPosted: Nov 06, 2010 - 10:17 PM
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Ummm, scratch that question I just solved it.
 
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gtpats
PostPosted: Dec 01, 2010 - 05:32 AM
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hi,

Thank you for the tutorial, gives me few questions.

1.I am trying to know how much time (micro secs) can the control be inside the ISR function. Does it have a timeout, or can we make it stay as long as we want - [The main program halts here]

2-as far as i know , 100 cycles is max, each cycle is 1/8000000 secs ; 0.125 uSec [for an 8MHZ, atmega128]
But when programming in C we do not know how many cycles we are using, is there a way to find the No.of cycles in ISR through any way?
3- Interrupt Latent time is min 4cycles
4- from data sheet "A return from an interrupt handling routine takes four clock cycles. During these 4clock cycles,the Program Counter (two bytes) is popped back from the Stack, the Stack Pointer is incremented by two, and the I-bit in SREG is set."

Can someone confirm 1. pls

- gt
 
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clawson
PostPosted: Dec 01, 2010 - 12:59 PM
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1) You said it yourself, main() (and it's called functions and other interrupts) are prevented from executing while an ISR is being serviced. As such ISRs should always be designed to operate for the absolute minimum time possible. Usually this need be nothing more than collecting a data byte from the interrupting device or setting some output state. If more is to be done set a volatile variable "flag" in the ISR then, back in main() check this flag from time to time and if set do the rest of the none time critical work triggered by the ISR.

2) Where on earth did you get this 100 cycle figure from? Sure (as in 1 above) it's a guideline to keep ISRs short and if you could make them 100 cycles or less that'd be great - but nothing forces this - they could be 532 million 581 thousand 295 cycles if you really want (though it's poor design).

3) There's certainly an overhead for jumping via the vector but, because of the minimum work usually performed in an ISR it's tricky to make them much less than about 30 cycles. Often this makes main loop polling a faster solution than an interrupt.

Cliff

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ereihani
PostPosted: Dec 21, 2010 - 08:14 PM
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Thanks Dean.
I was confused in my program for the volatile definition of my variable.
 
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hobadirl
PostPosted: Feb 04, 2011 - 01:54 PM
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Nice work thank you.........
 
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arian88
PostPosted: Jul 14, 2011 - 02:43 PM
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Thank you for so good tutorial, I read it.. but have one question..
what about if instead of UART Rx you want to check status of one pin..
Feks Port C, Pin7 (or others pin), to give event for that to check this pin is low or high and do something...

How can you do this trough Interrupts? Lets say we use Atmega8..

Code:
int main(void){
   setOutput(&DDRC, 0);
   setLow(&PORTC, 0);

   //enable global Interrups   
   sei();
   //give status, give sign for any logical
   setLow(&MCUCR, ISC01);
   setHigh(&MCUCR, ISC00);
   //enable INT0, PortD.2 (Atmega8 use PIND2 as INT0)
   setHigh(&GIMSK, INT0);


}


ISR(PCINT_value)
{
      setHigh(&PORTC, 0);
   _delay_ms(500);
   setLow(&PORTC, 0);
   _delay_ms(500);
}


I have write this code, and I know about something here is wrong..
1- But I dont understand how can I connect the port and pin to correctly Interrupt?
2- Can I choose wich of PCINT for wich port I want? Or each PCINT is for specific Pin?

NB! setHigh() and setLow() is just function i write it for myself for turn High and Low, and they works Wink )
 
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clawson
PostPosted: Jul 14, 2011 - 02:49 PM
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Quote:

Port C, Pin7

Quote:

Atmega8

For that you would need divine intervention from God herself! The mega8 only has PC0..PC5 - there is no PC7.

BTW the mega8 is one of the ancient AVRs so only has external/pin change interrupts on two pins (INT0/INT1, physical pins 4 and 5, also PD2, PD3). Use a modern chip like a mega88 if you want interrupts from all the IO.

(oh and your post is off-topic for the tutorial forum - it's not for diagnosing problems - it's for discussing the tutorial article in the first post of each thread).

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arian88
PostPosted: Jul 14, 2011 - 03:23 PM
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Ok, thanks anyway.. Maybe I should post new Thread and dont continue here, have still problem to understand how this tutorial can works with PD2 or PD3 to check they are high or low...

Again, thanks for this good tutorial
 
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mihaescuflorin90
PostPosted: Oct 29, 2011 - 08:16 AM
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Excellent tutorial,Dean buddy!
Congratulations!
 
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Nadwaga
PostPosted: Oct 10, 2012 - 03:36 PM
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Awesome tutorial that helped me a lot, thanks for sharing!

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behzadh47
PostPosted: Dec 07, 2012 - 12:53 PM
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Knowing that each address of program memory is 16 bits wide and PC is less than 16 bits wide, why the address of each interrupt vector differs from the previous one by two?
For instance:
the INT1 vector is located at address $0004 and the INT2 vector is located at address $ 0006.
 
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webpower
PostPosted: May 02, 2013 - 12:24 AM
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Thanks for posting this. Very useful.
 
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soundararajan
PostPosted: Aug 12, 2013 - 01:24 PM
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Thanks for this post Smile.

One minor comment.

As i understand AVR8 and AVR32 uses a different interrupt handling mechanism (correct me if i am wrong), and if that is the case may i request you to add a small note that this is for AVR8 (again, this i assume Smile ?
 
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clawson
PostPosted: Aug 12, 2013 - 05:03 PM
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Quote:

i request you to add a small note that this is for AVR8

Take a step back to the first page. This Tutorial Forum is part of the 8 bit section. You can see that in the link to here:

AVR Freaks Forum Index » AVR (8-bit) Technical Forums » AVR Tutorials

ALL the articles here are about AVR8. AVR32 stuff is separate. As far as I know they haven't chosen to setup a tutorial forum.

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Hillridge
PostPosted: Aug 15, 2013 - 06:06 PM
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I have a question about using:
ATOMIC_BLOCK(ATOMIC_FORCEON){}

Say I have timer2 set up to overflow once per second and the following code:
Code:

volatile unsigned long RealTimeSeconds;      
ISR(TIMER2_OVF_vect)
{
   RealTimeSeconds++;
}

Similarly to the tutorial, I want to access RealTimeSeconds from my main code, and want to be sure the ISR doesn't fire mid-read (however unlikely), but I want to access it from multiple routines.

Rather than doing this at each point I want RealTimeSeconds:
Code:

ATOMIC_BLOCK(ATOMIC_FORCEON){
   blah = RealTimeSeconds;
}


Can I do this?

Code:

unsigned long getSeconds()
{
   ATOMIC_BLOCK(ATOMIC_FORCEON){
      return RealTimeSeconds;
   }
}

blah = getSeconds();


Also, if the overflow occurs while the atomic code is executing, will the ISR be skipped entirely (effectively losing a second of time keeping) or just held off until the atomic block is done?
 
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abcminiuser
PostPosted: Aug 15, 2013 - 06:08 PM
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Yes, that's fine - although slightly slower since you then have the overhead of the function call (unless the compiler inlines it).

If the flag is set while interrupts are disabled it will then execute the ISR when the block ends - but if multiple occurrences of the same event cause the flag to be set, the ISR will only be run.

- Dean Twisted Evil

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Hillridge
PostPosted: Aug 15, 2013 - 06:25 PM
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Thanks, that's what I was assuming. I may just do it the original way though since wrapping the return statement generates this harmless but annoying warning:
control reaches end of non-void function [-Wreturn-type]
 
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abcminiuser
PostPosted: Aug 15, 2013 - 06:28 PM
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Better option:

Code:
unsigned long getSeconds()
{
   unsigned long temp;

   ATOMIC_BLOCK(ATOMIC_FORCEON){
      temp = RealTimeSeconds;
   }

   return temp;
}


Betterest option: use the above, but switch to the standard types from C99.

- Dean Twisted Evil

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clawson
PostPosted: Aug 15, 2013 - 06:29 PM
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Quote:

wrapping the return statement generates this harmless but annoying warning:
control reaches end of non-void function [-Wreturn-type]

try:
Code:
unsigned long getSeconds()
{
   unsigned long retval;
   ATOMIC_BLOCK(ATOMIC_FORCEON){
      retval = RealTimeSeconds;
   }
   return retval;
}

I bet you find the compiler generates exactly the same code but the warning is suppressed. Consider making the function "static inline" in a header file too so you don't have the CALL/RET overhead as long as it's not called in hundreds of places.

EDIT SNAP! (great minds think alike - we both posted at 6:28!)

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Hillridge
PostPosted: Aug 15, 2013 - 06:41 PM
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Thanks. Would the C99 Data type be an "unsigned long int" rather than just an "unsigned long"?

I'm favoring code size, so I'll leave it as is. I tested it as a static inline and got a 50 byte increase.
 
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abcminiuser
PostPosted: Aug 15, 2013 - 06:44 PM
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The exact-width types from <stdint.h> - they make your code easier to grok and somewhat more portable, since you can be sure that a "uint32_t" is an unsigned 32-bit integer regardless of platform.

See:
https://en.wikibooks.org/wiki/C_Program ... e/stdint.h

- Dean Twisted Evil

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Koshchi
PostPosted: Aug 15, 2013 - 10:33 PM
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Quote:
Thanks. Would the C99 Data type be an "unsigned long int" rather than just an "unsigned long"?
"unsigend long" and "unsigned long int" are identical. In the first the "int" is implied.

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zu_adha
PostPosted: Oct 03, 2013 - 01:12 PM
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i still don't get this part, please explain more, what does this mean..
Quote:

3) There's another subtle bug in the above code sample; it's possible at the machine code level for an interrupt to occur half way between the fetching of the two bytes of the integer variable MyValue, causing corruption if the variable's value is altered inside the ISR. This can be avoided by making the fetching of the variable's value in the main program code "atomic", by disabling the global interrupt enable bit while the fetching is taking place. In avr-libc this can be done in the following manner:


Code:
#include <avr/interrupt.h>
#include <util/atomic.h>

volatile int MyValue;

ISR(SomeVector_vect)
{
   MyValue++;
}

int main(void)
{
   SetupInterrupts();

   int MyValue_Local;

   do
   {
      ATOMIC_BLOCK(ATOMIC_RESTORESTATE)
      {
         MyValue_Local = MyValue;
      }
   } while (MyValue_Local == 0) // Wait for interrupt

   TurnOnLED();
}
 
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clawson
PostPosted: Oct 03, 2013 - 02:28 PM
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Quote:

i still don't get this part, please explain more,

Consider the code:
Code:
#include <avr/interrupt.h>

volatile int MyValue;

ISR(SomeVector_vect)
{
   MyValue++;
}

int main(void)
{
   SetupInterrupts();

   while(1) {
     if (MyValue == 0x100) {
        TurnOnLED();
     }
   }
}

C guarantees that MyValue starts at 0 so you'd have thought that after 256 (0x100) interrupts the LED would be turned on - right?

So now consider the case when MyValue is holding 0x00FF. The code that does the "if (MyValue == 0x100) {" is going to look something very like:
Code:
.L9:
//==>       if (MyValue == 0x100) {
   lds r24,MyValue
   lds r25,MyValue+1
   cp r24,__zero_reg__
   sbci r25,1
   brne .L9

So the compiler collects the value in MyValue into registers R25:R24 then makes a comparison of this with 0x100.

Now consider that it has done the first of the two LDS there:
Code:
   lds r24,MyValue

At this moment R24 contains 0xFF. Then the interrupt code occurs and it heads off to the ISR():
Code:
//==>    MyValue++;
   lds r24,MyValue
   lds r25,MyValue+1
   adiw r24,1
   sts MyValue+1,r25
   sts MyValue,r24

That takes 0x00FF and increments it to 0x0100 and stores it back. So the "MyValue" location now holds 0x00 and MyValue+1 holds 0x01. It then returns from the interrupt to the second LDS in:
Code:
//==>       if (MyValue == 0x100) {
   lds r24,MyValue
   lds r25,MyValue+1
   cp r24,__zero_reg__
   sbci r25,1

Already this R24 holds 0xFF and now it loads the value from MyValue+1. That contains 0x01. So in total R25:R24 hold 0x01FF and not the 0x0100 we're hoping to see. So the if() test is false and the LED is not turned on. Now the interrupt fires again and this time 0x0100 in MyValue is incremented to 0x0101. On return from ISR() it now does the if() test in main() again. Only this time it succeeds in correctly reading R25:R24 as 0x0101. That does not == 0x100 either so again the LED is not lit.

Because of "mis-read" of MyValue the actual 0x0100 value was never seen so the if() test was never true and the LED never turned on (or at least not until MyValue has incremented and wrapped).

That is why you need atomic protection. In avr-gcc you do that with ATOMIC_BLOCK() which effectively says "just while I'm doing this one thing turn the interrupts off, you can turn them on when I've finished". So the code inside the ATOMIC_BLOCK that takes a copy of MyValue into MyValue_Local is guaranteed not to be interrupted and therefore cannot "mis-read" the variable by being interrupted part way through. If MyValue had 0x00FF it will still have that until at least the end of the ATOMIC_BLOCK and once it has 0x100 it will again have it for the entire reading process.

That is why you need atomic access protection when reading variables that are more than a byte wide. If MyValue had been char/uint8_t yo wouldn't have needed to worry as LSD/LD or whatever that would be used to read a single byte cannot be interrupted mid-way through.

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zu_adha
PostPosted: Oct 03, 2013 - 04:40 PM
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so this is the mean of atomic?

Quote:
"just while I'm doing this one thing turn the interrupts off, you can turn them on when I've finished".


so can i use that for usart instead using interrupt for sending a bit?
 
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clawson
PostPosted: Oct 03, 2013 - 06:36 PM
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Quote:

so this is the mean of atomic?

Well atomic means "uninterruptible" (effectively). (it dates from a time when atoms could not be divided - now we know they can be - but on the whole CPU opcodes cannot be, like the old view of atoms.
Quote:

so can i use that for usart instead using interrupt for sending a bit?

I haven't got the first idea what that question means or how it relates in any way to the question of atomicity. Are you talking about implementing "soft UART" by bit-banging an IO pin in an interrupt? If so I guess it's true that the byte that is being sent would not want to be updated after only 5 or the 8 bits had been transmitted or similar.

Maybe try your question in your native language?
(Mungkin mencoba pertanyaan Anda dalam bahasa asli Anda?)

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zu_adha
PostPosted: Oct 04, 2013 - 05:35 AM
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i mean, i have a code that using timer interrupt, but i dont used usart interrupt, so in some periode of time usart has stop/error and i know it maybe because when the uC sent the data my code interrupt that for the timer, so if i used atomic is that same like i use interrupt for usart?

can i use this
Code:

ATOMIC_BLOCK(ATOMIC_RESTORESTATE)
      {
void kirim(char *x)
{
   while(*x>0)
   {
      while (!( UCSRA & (1 << UDRE )));
      UDR = *x++;

   }
}

instead of interrupt uart?

and one more question. i forgot to put
#include <avr/interrupt.h>
but i put
#include <util/atomic.h>
only and i used interrupt
Code:
SEI()
but it compiled with no error.
i never see what inside atomic.h but why i can use interrupt when i forget to put #include <avr/interrupt.h> instead #include <util/atomic.h>

and help me here too Smile
http://www.avrfreaks.net/index.php?name=PNphpBB2&file=viewtopic&p=1103673#1103673

ps:
Quote:
Maybe try your question in your native language?
(Mungkin mencoba pertanyaan Anda dalam bahasa asli Anda?)

wow you can speak indonesian? or used google translate.. Smile what does it means? can i used my native language, did you understand that?
 
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clawson
PostPosted: Oct 04, 2013 - 09:47 AM
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Quote:

instead of interrupt uart?

That depends what you are trying to achieve. The contents of your ATOMIC_BLOCK there are possibly going to take thousands and thousands of CPU cycles (it block on UDRE and outputs potentially multiple bytes) and the whole thing wil occur without interrupts enabled (that's the purpose opf ATOMIC_BLOCK). So your question is can you really afford to have the interrupts disabled for that length? During that time your timer interrupt will not be serviced.

As with anything to do with interrupts whether it be the ISR()s themselves or ATOMIC_BLOCK's in the main code each should aim to complete in just a few tens or hundreds of CPU cycles. This way no process can "hog" the CPU. Doing a while(UDRE) loop while interrupts aer disabled goes against this.

If you are saying you have a timer interrupt and code that inputs/outputs to the UART that is non-interrupt and that sometimes the UART code is intermittent it suggests that something in your timer interrupt handler is taking too long to complete.

In the limit a timer ISR() should be doing little more than setting a flag to say that some period has expired. The main() code can then service this flag and do any "long work" necessary. The downside of this design is that the main code may not "see" the flag for some while so the triggered process can be "late". If there things you need to do that really are time critical then do put that code into the ISR() but always consider how long they are going to delay the return from interrupt by. If a long time think of another way to implement it.
Quote:

wow you can speak indonesian? or used google translate.. Smile what does it means? can i used my native language, did you understand that?

I just used Google translate (I am a moderator so could see you were from Indonesia). My suggestion to you is that if you have difficulty with asking questions or describing your project then just write it in Indonesian then either post it here "as is" or use Google Translate yourself. If you don't I will anyway. You may find it easier to describe your problem if you just use your natural language.

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binaryavr
PostPosted: Nov 17, 2013 - 12:23 AM
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Posts: 136


Which would you say is correct or more correct?

The I flag in the status register:
1) Globally activates all interrupts.
2) Permits/allows the AVR to process interrupts.

If I understand correctly, the global interrupt flag allows us to activate separate interrupts. It does not activate them, just allows us to go and activate each interrupt ourselves, in their respective control registers? If it is disabled, and we try to enable a separate interrupt we will be unable to do so? So I'm pretty sure 2) is correct. I just want make sure I correctly understood what I read in the manual.
 
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clawson
PostPosted: Nov 17, 2013 - 12:45 PM
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Quote:

If I understand correctly, the global interrupt flag allows us to activate separate interrupts.

Where on earth did you get that idea?

Surely it's simple. The I flag is simply the global on off switch for the entire interrupting system. It most definitely does not allow the control of individual interrupt sources, you do that with individual control bits like UDRIE, TXCIE, TWIE, SPIE, etc.

While not always true it is very often the case that the names of the bits that control individual interrupts end in the letters "IE" which stand for Interrupt Enable.

To use interrupts you therefore enable one or more ??IE bits. For any that you do set you provide ISR code and then, when you are ready to turn the whole system on and start receiving calls to the ISR(s) you execute SEI which sets the I bit in SREG.

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JohanEkdahl
PostPosted: Nov 17, 2013 - 01:27 PM
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Adding to Cliffs post:

The global interrupt flag controls if any interrupt will "fire", as in "an ISR will be called".

The individual interrupt enable bits on the individual peripherals controls the individual interrupts.

The individual interrupt enable bits are all "in parallel" with each other but in series with the global interrupt enable bit. So to get e.g. thr UART receive interrupt ISR executed both the individual interrupt enable bit for that needs to be set, and the global interrupt enable bit needs to be set.

Do note that the status bit in the UART status register that indicates an UART receive interrupt will be set even if the global interrupt enable bit is cleared.

In essence, the global interrupt enable bit controls if ISRs will be called or not.

If an interrupt condition arises while the global interrupt bit is cleared, but the individual interrupt enable bit in the peripheral is set - then when the global interrupt bit is set again the interrupt will actually fire (as in "the ISR will be called").

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binaryavr
PostPosted: Nov 17, 2013 - 10:42 PM
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As far as software interrupts go, AVR does not support software interrupts of any kind?
 
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clawson
PostPosted: Nov 17, 2013 - 10:55 PM
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Yes it does, read the datasheet (you can trigger an ext int by writing the port). I've never totally understood of what practical use that is.

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binaryavr
PostPosted: Dec 29, 2013 - 03:25 AM
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When using the mega16 and interrupts, is it necessary to set DDRD(PD2 and D3) as input and activate the pullup resistors?

Code:

LDI TMP, 0x00
OUT DDRD, TMP
LDI TMP, 0xFF
OUT PORTD, TMP


No there is no need to explain MCUCR, GICR, and SREG. I'm pretty sure DDRD does not need to be set(or that it is even possible) if we enable the pins as interrupt pins. I'm not sure about PORT.
 
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clawson
PostPosted: Dec 29, 2013 - 11:58 AM
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Pins used for interrupting need to be input but that's the default state anyway. As to whether you want pullup's or not depends on the external electronics. Is there an open connection that might float? If there is do you want it tied high or low in the quiescent state?

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binaryavr
PostPosted: Dec 31, 2013 - 09:40 AM
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Is it possible to have INT0 do one thing when you press down a button and do something else once the button is released? For example: You press down INT0 a LED turns on, you release INT0, the LED turns off. So, it does one thing during the falling edge and another during rising edge. Or would I have to use INT0 to turn the LED on and INT1 to turn the LED off.
 
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JohanEkdahl
PostPosted: Dec 31, 2013 - 01:56 PM
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Quote:

Is it possible to have INT0 do one thing when you press down a button and do something else once the button is released?

It is seldom that "driving" an interrupt with a switch ios a good idea. Search this site and Google for "button bounce", "contact bounce", "button de-bouncing" and similar for A LOT of earlier discussions.

In short: When a button makes or breaks contact it will not produce one single clean rising or falling edge. Instead it will do several on-off-on-off... transitions during a few milliseconds, creating "false button presses".

No, don't ask. Do the searches, and read up first. It has been covered more times than we can count.

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