Is there an easy way to turn on/off just one bit on a port, such as PORTD, without affecting any of the other bits?
turning on/off one bit on a port
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PORTD |= 0x01;
this will enable the first bit. If you know the name of the thing to be enabled, you can also do something like:
PORTD |= (1<<RXD);
where RXD is a mask for that bit.
edit: assuming you're using C, that is...
Thanks. Yes I'm using C.
What if I want to disable just a certain bit (i.e. PORTD starts at 0xFF but I want to selectively turn off bit 2).
same way,
PORTD = (0<<PDx); where x is your pin number
Same concept. Use
PORTD &= 0xFB;
or you can use an XOR to toggle the bit, ie:
PORTD ^= 0b0100;
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in the good old days you would have just written cbi(PORTD, 1); or sbi(PORTD,1); but it has been removed...(still in
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cbi and sbi for ever!!!
thats what i do
any comments? good or bad all are welcome
i agree with daqq
"cbi and sbi for ever!!!"
//***************************************
//* MACRO'S *
//***************************************
#define SBI(port,bit) asm("sbi %0, %1" : : "I" (_SFR_IO_ADDR(port)), "I" (bit))
#define CBI(port,bit) asm("cbi %0, %1" : : "I" (_SFR_IO_ADDR(port)), "I" (bit))
#define Sp(wrd,bit) CBI(wrd,bit) //set bit in port
#define Rp(wrd,bit) SBI(wrd,bit)// reset bit in port
#define Tp(wrd,bit) wrd ^= (1<<bit)//togle bit in port
OZ1 wrote:
same way,PORTD = (0<<PDx); where x is your pin number
Absolutely not.
This will turn off all bits in PORTD, as 0 shifted to anything will always be 0. This means that the above is equivalent to
PORTD = 0
regardless of the chosen PDx. To turn off a specific bit you have to AND the current port value with a mask where every bit is set except the bit you want to clear, thus:
PORTD &= ~(1<<PDx);
.
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Zalfrin wrote:
PORTD |= (1<<RXD);where RXD is a mask for that bit.
Not quite:
RXD is the number of the bit you want to set;
(1<<RXD) is the mask for that bit
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EW wrote a great bit manipulation a while ago, and I've shoved it into the Tutorials section where it belongs. See this thread.
- Dean :twisted:
Quote:
"cbi and sbi for ever!!!"
Give me a correction if I'm wrong, but I think the compilers (at least GCC and ICC) are using the cbi/sbi instructions when the code is like "PORTD |= 0x01" !?
Jensi
Yes, jensi, but I think that daqq is referring to the cbi and sbi macros that where available in older versions of avr-libc.
Hi all ,
i have just started programming on 8515 using the GCC , and immediately run into this here small problem i.e. easy port bit manipulation in C .
I come from using C compilers which have 'bit ' datatype using which I could go like :
bit sdi PORTA.1
//and in the code ..
sdi=1;
..
.. sdi=0;
Writing SPI code using the bit bangigng method or simply turning just one led on / off is much simpler this way.. is there any such syntax I could use with GCC or is it going to be & and | all the way.. ?? :(
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You're going it have to use a bitfield (do a forum search for examples) if you want to be able to use bits. Create a bitmask to a global flag register, then #define an alias to a bit so you can then use that alias like your old bit datatype.
I strongly suggest you just learn the "proper" C bitmask manipulations - there's a post in the Tutorials section ("Programming 101") which explains how to set, toggle and clear bits in a register/variable.
- Dean :twisted:
#define clear_bit(a,z) (a &= ~_BV(z))
#define set_bit(a,z) (a |= _BV(z))
clear_bit(PORTB, PB7);
works for me :)
old thread, but...
I'm not sure that gcc allows this, but when I was using the motorola/freescale software stationary for an embedded systems class, they did something like this:
struct PORT {
unsigned char bit7 : 1,
unsigned char bit6 : 1,
...
unsigned char bit0 : 1;
}
syntax is not quite right, but you get the idea. There were also unions inside the struct that allowed you to access the whole port if you so desired. Then,you would use something like this:
struct PORT PORTD @ 0x220 //address of PORTD
PORTD.bit0 = 1;
As I said, probably will not work in gcc, but it was a pretty handy way of accessing ports and registers and such.
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That's the gist of what you do with GCC, and it's called a bitfield. In GCC you have to define an alias name via #define rather than creating a struct at a fixed IO address - a search of the forums for "bitfield" will yield good results.
- Dean :twisted:
AVR program that toggles all bits of PORTB and mointer bit #2 of PORTC if it is On , turn ON bit 7 of PORTC else turn OFF bit 7 of PORTC
AVR program that toggles all bits of PORTB and mointer bit #2 of PORTC if it is On , turn ON bit 7 of PORTC else turn OFF bit 7 of PORTC
Like Zabrin says
To set a bit to one
PORTB |= (1<<PinNumber)
To reset a bit to zero
PORTB &= ~(1<<PinNumber)
To toggle a bit
PORTB ^= (1<<PinNumber)
In PinNumber you can put a number, varible or constand.
thank you , also please help me with that i am zero experience
AVR program that gets the status of bit5 of PORTB & send it to bit 7 of PORTC
AVR program that gets the status of bit5 of PORTB & send it to bit 7 of PORTC
statusB5 = PINB & (1<<5)
if (statusB5) {PORTC |= (1<<7)}
else {PORTC &= ~(1<<7)}
I add an atmel solution it uses a library I have personally so it's more complex but you can check it if you want feel free to use the library only GPIO and Timers have been tested
Last Edited: Tue. Oct 9, 2018 - 05:38 PM
please can you help me also to find a link ti download Atmel studio 7
warm regards
mohamad najem wrote:
please can you help me also to find a link ti download Atmel studio 7
mohamad najem wrote:
please can you help me also to find a link ti download Atmel studio 7
warm regards
Were you a sleep in class the day they showed how to google?
http://www.microchip.com/mplab/a...
Jim
AVR program using assembly language to monitor PORT B, if its high, send 55 hex to PORT D, else send AB hex to PORT D. how can i make it??
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Elio sebaaly wrote:
PORTB is EIGHT bits - which one did you have in mind?
to monitor PORT B, if its high
Anyway the general technique is to set B as input. D as output, then read PINB to get the input state, determine whether the condition is met (not clear what condition you are testing for - see above) then write 55/AB to D based on this.
