unexpected output with comma operator

Question

I wrote a program

#include<stdio.h>
int main()
{
      int x=3;
       if((x)==1,2,4,5,6)
               printf("number found in the list\n");
       else
               printf("Number not found \n");

      return 0;
}

I was expecting the output to be "Number not found " but it is "Number found in the list " why it is so

See [What does the comma operator `,` do in C?](http://stackoverflow.com/a/18444099/1708801), it has the lowest precedence. — Shafik Yaghmour, Oct 06 '15 at 16:56
C does not support list types. The "comma" is called for a good reason **operator**, not just being a syntax element. — too honest for this site, Oct 06 '15 at 17:06

cadaniluk · Answer 1 · 2015-10-06T17:07:59.380

5

The == operator has higher precedence than ,, therefore the if-clause evaluates to

if (((x) == 1),2,4,5,6)

which is always true since the last "element" of the comma operator "counts" (6).

From the C11 standard:

The left operand of a comma operator is evaluated as a void expression [...]. Then the right operand is evaluated; the result has its type and value.

edited Oct 06 '15 at 17:07

answered Oct 06 '15 at 16:56

cadaniluk

15,027
2
39
67

score 2 · Accepted Answer · answered Oct 06 '15 at 17:14

You're expecting (x) == 1,2,4,5,6 to be evaluated as "x is equivalent to one of 1, 2, 4, 5, or 6", but that's not how the comma operator works. This is actually evaluated as (x)==1, then 2, then 4, then 5, and then finally 6. Since 6 is not 0, the condition evaluates to true and the first branch is taken.

You'll either need to write

if ( x == 1 || x == 2 || x == 4 || x == 5 || x == 6 )

or, you'll need to use a loop:

int list[] = {1,2,4,5,6};
size_t listSize = sizeof list / sizeof *list;
int found = 0;

for ( size_t i = 0; !found && i < listSize; i++ )
  found = (x == list[i]);

if ( found )
{
  printf( "Number found in list\n" );
}
...

unexpected output with comma operator

2 Answers2