whiteman7777 
can any one please give a full description if how this line works:
while ((SPSR & (1<<SPIF))==0);
does it shift the number "00000001" with respect of the amount of SPIF bit?
how it ANDs SPSR (8 bits while only 3 are used) with the rest of argumant? and how it generates only one bit as out put??
this whole matter really confuses me up :(
MBedder Read a generic C book instead of asking about any next operator which is new to you.
whiteman7777 each C guidance has given a different explain about these operators, i am really confused.
if you know any online texts containing thorough and true description avaiable for this please put the link here.
MBedder each C guidance has given a different explain about these operators
whiteman7777 ok i read the related part of the ebook here:
http://publications.gbdirect.co....
these i already knew, but yet again i have not got any thing clears up the working of that line:
while ((SPSR & (1<<SPIF))==0);
could you be nice enough to just explain this line?
jgmdesign 
could you be nice enough to just explain this line?
Good Luck, MBedder is not known to just answer a question, nor be nice either :)
I would answer it, but I don't really understand 'c'. Sorry
Jim
whiteman7777 Quote:could you be nice enough to just explain this line?Good Luck, MBedder is not known to just answer a question, nor be nice either :)
I would answer it, but I don't really understand 'c'. Sorry
Jim
----------------
yes i already have got that! no C book has ever explained such arguments.
abcminiuser 
while ((SPSR & (1<<SPIF))==0);
SPSR is a register - a hardware "variable" that maps to a portion of the AVR's on-chip SPI peripheral. When read, it will return 8 bits of data with the meanings given in the datasheet.
The "&" operator is a bit-wise AND. There's a tutorial in the Tutorials forums that explains bitwise operators, but the relevant portion of it here is that "X & Y" will return a set of bits which are set in both "X" and "Y", with all other bits being cleared to 0. This bitwise ANDing operation is commonly referred to as a "bitmask" operation.
The (1<<SPIF) section creates a "mask" (8-bit value in this case) with the SPIF bit set and all others cleared. The "<<" left shift operator moves the left operand value the right operand number of bits left. The SPIF constant is a bit number as given in the datasheet. By shifting the constant "1" to the left SPIF number of times, you end up with a binary value with the a single bit set at the position indicated by the value of SPIF. For example, if SPIF was equal to the value of 2, you would end up with the binary value of 0b00000100.
When you put these two concepts together, you end up with the code masking the current value of the SPSR register with a mask of the SPIF bit, resulting in the value 0 if the SPIF bit was not set in the SPSR register, or the SPIF mask if it was set. The "== 0" checks this and converts this into a binary 0 or 1, depending on the value of the SPIF bit.
- Dean :twisted:
clawson 
whiteman7777 ok, i think i have understood this, the number 00000001 is shifted left as long as its 1 matchs the SPIF's position in SPSR, and then gets ANDed with the SPIF itself (equivallantly positioned bit on SPSR to the position of the bit 1), then the AND is only comparing the bit 1 with the bit SPIF and as long as the output is 0 the program does not pass to the next line, it might be a codevision trick possibly because the SPSR register could not be addressed directly (?????), well then if the compiler knows the variable SPFR then why it has not been writen as:
while (SPIF==0);
?????????
this matter is turning to be somehow mind blowing :(
clawson You clearly haven't read the link I gave. Report back when you have.
Bottom line: because of SBR, SBI, CBI, CBR, SBIC, SBIS Atmel chose to define control bits as bit number rather than bit mask. The 1<< construct converts a bit number to a bit mask (it's far more difficult to convert a bit mask back into a bit number!). This conversion is done at compile/assembly time and leads to no runtime code overhead.
MBedder Looks like today is a real idiots' parade on AVR Freaks :(
whiteman7777 alright, that says nothing new, the bit 1 is shifted to the position of the bit number and the out put is a bit mask, now how is that two bytes are being ANDed and the output is a bit which is needed to decide?
clawson
now how is that two bytes are being ANDed and the output is a bit which is needed to decide?
whiteman7777 i believe you dont get what i mean, consider this:
0x03&0x06
what is the out put? yes it is
0x02
yes as you might have noticed it is still a byte, now the while sequence needs a single bit (0 or 1) to decide. not a byte.
well after al perhaps i am forgetting some concepts, you know i had no tutors and also have been away from the deal for a long time, i was studying meckanics because i am interested in making industrial robotic arms!
MBedder Hopefully none of us will ever have a close contact with any of your robots :lol:
clawson
0x03&0x06
what is the out put? yes it is
0x02
if ((0x03 & 0x06) == 0) {
// do this when tested bit is not set
}
if (0x03 & 0x06) {
// do this when tested bit is set
}
if (!(0x03 & 0x06)) {
// do this when tested bit is not set
}
if ((0x03 & 0x06) == 0x02) {
// do this when tested bit is set
}
What would not be valid is:
if ((0x03 & 0x06) == 1) {
// do this when tested bit is set
}
BTW 0x06 as a bit mask in this is a bad example. That is testing two bits. Usually that bit mask would be one of 0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40 or 0x80 so only a single bit is tested.
whiteman7777 the point is that a byte cant be only 0 or 1, its a byte, what do you mean of "the tested bit"??
see if i write
while (0x03&0x06);
it is like to write:
while (0x02==1);
which does not mean any thing!
and its exactly what happens here:
while ((SPSR & (1<<SPIF))==0);
:(
MBedder Idiot parade is approaching the main street :lol:
whiteman7777 ok i understood the mater, it is a bad thing to stay away from study and work!
thank you so much :)
clawson
which does not mean any thing!
while(true);
or
while(false);
false == 0
any other value is "true". So:
while (0x03 & 0x06);
is the same as:
while(true);
If that 0x03 were just 0x01 then it would be:
while(0x01 & 0x06);
which is the same as
while(false);
So now consider:
while ((SPSR & (1<<SPIF))==0);
where SPIF=7. So either bit 7 in SPSR is set or it isn't. If the bit is set the condition says:
while ((0b1??????? & (0x80))==0);
which is:
while ((0x80)==0);
which is false so execution continues after this while() condition. On the other hand if SPIF is not set in SPSR then it says:
while ((0b0??????? & (0x80))==0);
which is
while ((0x00)==0);
which is true so it sticks in this while() condition until the bit does become set.
whiteman7777 also i believe that russian guy has no information about the embedded programing and is here just to make some obsessive comments, it is a shame. i could not help but to say this.
whiteman7777 thanks for your time and your help clawson.
MBedder i believe that russian guy has no information about the embedded programing and is here just to make some obsessive comments, it is a shame.
To add a bit of clarity here, with respect to the subject line....
This operation is NOT unique to the SPSR. This construct is used with MANY registers, including PIN registers, USART status register, timer status registers, and more. Thus, it is important to NOT get locked in to thinking that this is unique to working with the SPSR.
Second important thing is that (1<<SPIF) is >>JUST<< a number! It is determined at the time the program is compiled rather than as the program executes. Thus, it is equivalent to
if (SPSR & a_number == 0)
The construct (1<<bitref) is simply a way to improve readability. Why? Few readers will even remember what bit position SPIF represents. They WILL remember its function and how it operates, but not its bit position, Further, the bit position might not be the same in all AVRs. So, this also improves portability.
Jim