what functionality does comma hold here,why it doesn't evoke compile time error?

Question

code snippet:

#include <stdio.h>
int computeSum(int number1, int number2) {
    int sum;
    sum = number1 + number2;
    return sum;

}
int main()
{
      int sum = 0,k=0;

      k,sum= computeSum(4,2);
      printf("%d %d",sum,k);

     return 0;
}

What is the functionality of comma ( , ) between k and sum,why it doesn't evoke any error?what is its significance?

It's just syntactically wrong. What are you trying to achieve? — Denis Sablukov, Jan 13 '19 at 11:48
@DenisSablukov Why do you think it is wrong? I'd think it's just the comma operator separating two expressions. The left expression is just the `k`, the right one is the assignment (only) to sum. (In particular, nothing is assigned to `k`). — Peter - Reinstate Monica, Jan 13 '19 at 11:50
@DenisSablukov "The left operand of a comma operator is evaluated as a void expression; there is a sequence point between its evaluation and that of the right operand. Then the right operand is evaluated; the result has its type and value." (from the n1570 draft, 6.5.17) — Peter - Reinstate Monica, Jan 13 '19 at 11:54

bruno · Answer 1 · 2019-01-13T12:05:51.427

2

k,sum= computeSum(4,2); just get the value of k then assign sum

perhaps you wanted to do k = sum = computeSum(4,2); ?

an assignment is an expression returning a value, not a statement like an if ...

edited Jan 13 '19 at 12:05

answered Jan 13 '19 at 11:51

bruno

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You sure it's syntactically invalid? – Peter - Reinstate Monica Jan 13 '19 at 11:52
2

It is very much valid syntax. It just doesn't do anything useful here. – StoryTeller - Unslander Monica Jan 13 '19 at 11:52
yes yes, I wanted to say this is not what he wanted to do, my bad English, sorry – bruno Jan 13 '19 at 11:53
Even though I upvoted now, you may want to reconsider "an assignment is an expression, not a statement": An expression can be a statement, aptly named "expression statement" (n1570 6.8.3). – Peter - Reinstate Monica Jan 13 '19 at 11:58
@PeterA.Schneider I wanted to say `a = if ...` is not allowed, yes an expression is also a statement by first an expression ... how to say that better ? – bruno Jan 13 '19 at 12:04

score 2 · Accepted Answer · answered Jan 13 '19 at 11:56

Precisely stated, the meaning of the comma operator in the general expression

e1 , e2

is evaluate the subexpression e1, then evaluate e2; the value of the expression is the value of e2.

When compiling your program with gcc compiler, it gives warning message:

warning: left-hand operand of comma expression has no effect
k,sum= computeSum(4,2);
^

Reason is, the expression

k,sum= computeSum(4,2);

is evaluate the subexpression k and then sum= computeSum(4,2);. The subexpression k result is unused.

score 2 · Answer 3 · answered Jan 13 '19 at 11:56

The one comma-operator you use, between the expression k and the assignment-expression, just makes a single expression out of two expressions, and orders them.

Of course, the first expression has no effect, thus the compiler warns you (if you ask for it):

prog.c: In function 'main':
prog.c:12:8: warning: left-hand operand of comma expression has no effect [-Wunused-value]
   12 |       k,sum= computeSum(4,2);
      |        ^

what functionality does comma hold here,why it doesn't evoke compile time error?

3 Answers3