Why is the return value of the fun function 8 instead of 7?

Question

With reference to Comma-Separated return arguments in C function [duplicate] ,

x=x+2,x+1;

will be evaluated as

x=x+2; 

However, in case of the following code

#include<stdlib.h>
#include<stdio.h>

int fun(int x)
{
    return (x=x+2,x+1); //[A]
}

int main()
{
   int x=5;
   x=fun(x);
   printf("%d",x); // Output is 8
}

Shouldn't line [A],be evaluated as

x=x+2;

giving x = 7

Answer 1

The statement return (x = x + 2, x + 1); is equivalent to:

x = x + 2; // x == 7
return x + 1; // returns 8

Answer 2

When writing return (x=x+2,x+1), the first expression is evaluated first so x=x+2 is evaluated, causing x to equal 7 as a side effect. Then the second expression is evaluated and returned, hence the function returns x+1 hence returns 8.

If you had written return (x+2,x+1);, the result would have been 6 because the first expression x+2 doesn't have any side effect.

Answer 3

Both parts in the return are evaluated respectively and the result of the last instruction is returned:

At first we have:

x = x + 2 // 7

Now x is updated to 7 before the second evaluation which gives:

x + 1 // 7 + 1 = 8

and finally return 8.

For better understanding consider the case of intermediate variable as follows:

return (y = x + 2, y + 1);

Answer 4

The QA you conveniently linked states

The comma operator evaluates a series of expressions. The value of the comma group is the value of the last element in the list.

so the value of

x+2,x+1;

is x+1 and there are no side effects.

Sample code:

#include<stdio.h>
int main(int argc, char * argv){
    int x;
    x = 0;
    x = (x+2, x+1);
    printf("%d\n", x);
    return 0;
}

results in 1 when run.

However, when you do

return (x=x+2, x+1)

you do have a side effect: x is incremented by two first, then x is incremented by 1 and the result is returned.