compilation error in c statements

Question

I am trying to run this code.this code runs successfully.But when i change variables to constant numbers i get compilation error.

My code that works fine:

 int a =5,b=6,c=7;
 int d;
 d = a,b,c;
 printf("%d",d);

as i run the code its output is 5. but when i run this segment of code:

 d = 2,6,7;
 printf("%d",d);

i get compilation error.I tried i on other compiler also. But the error still exists. What i am doing wrong.

`int d; d = 2,6,7;` -- what do you think this piece of code *should* do? — DevSolar, Feb 06 '15 at 12:42
@UnicoRahul: It would be helpful if you'd **post** the compiler's error message... — DevSolar, Feb 06 '15 at 12:52
@UnicoRahul clearly, the comma operator does things which are completely unrelated to the things that you think it does. "c process this as a queue" is completely nonsensical, I do not know where you got this impression from. Just hit the books and learn exactly what comma does in C. — Mike Nakis, Feb 06 '15 at 12:53
Also, you cannot be coming to StackOverflow and saying things like "i get compilation error". You have to tell exactly what compilation error you get, and precisely on which line you get it. — Mike Nakis, Feb 06 '15 at 12:55
Is the error "error: expected identifier or '(' before numeric constant"? — Spikatrix, Feb 06 '15 at 12:55

lqhcpsgbl · Accepted Answer · 2015-02-06T12:58:31.817

Your first code use the variables and assignment d = a and b and c just as expression there, so run the code:

int main(int argc, char const *argv[])
{
 int a =5,b=5,c=7;
 int d;
 d = a,b,c+1;
 printf("%d",d);
 return 0;
}

You get 5, b and c+1 just valued and put them there useless.But if you run this code which includes comma expression:

int main(int argc, char const *argv[])
{
 int a =5,b=5,c=7;
 int d;
 d = (a,b,c+1);
 printf("%d",d);
 return 0;
}

You get 8 as the last one valued expression. You can use the number play as an expression with ():

int main(int argc, char const *argv[])
{
 int a =5,b=5,c=7;
 int d;
 d = (0,3,1);
 printf("%d",d);
 return 0;
}

get the last number or valued data.

It works for me the code below:

int main(int argc, char const *argv[])
{
 int a =5,b=5,c=7;
 int d;
 d = 0,3+1,1-1;
 printf("%d",d);
 return 0;
}

it output is 0, but if you don't with (), it meaningless by this way, why not just use d = 0;

1 Answers1