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

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.

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);

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

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

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.

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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.

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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.