Baffling interview question I found - post-increment operator

**Absurd** · 09-26-2015

Hi.
I found this interview question online, and it really confuses me...

Consider the following code snippet:

Code:

int main(int argc, char* argv[]) {

    int x=97;
    int y=sizeof(x++);

    printf("x is %d\n", x);

    return 0;
}

What is the output?
1. x is 97
2. x is 98
3. x is 99
4. run-time error

My guess was 2, but when I compiled and ran the code, I was amazed to find out that this produce "x is 97"!
What's going on here?

*EDIT

Funny thing is when I change the above to this:

Code:

int main(int argc, char* argv[]) {

    int x=97;
    int y=x++;

    printf("x is %d\n", x);

    return 0;
}

It produces "x is 98" as expected...

**laserlight** · 09-26-2015

Unless its operand is a variable length array, sizeof does not evaluate its operand.

**Absurd** · 09-26-2015

So sizeof cancels out the post-increment operator?

**Absurd** · 09-26-2015

And what about this?

Code:

int main(int argc, char* argv[]) {

    int x=97;
    int y=97;
    int z=sizeof(x+y);

    printf("z is %d\n", z);

    return 0;
}

This produce "z is 4".
But the only way to determine that is to first evaluate what x+y is, no?

**Ktulu** · 09-26-2015

As laserlight stated, sizeof does not evaluate its operand in this case. Meaning sizeof(x+y) will give the same result as sizeof(x*y*x*y*x*y) and sizeof(x).

**laserlight** · 09-26-2015

Originally Posted by Absurd

So sizeof cancels out the post-increment operator?

Not really: it just does not evaluate it. Consider this idiom that you may have seen me recommend:

Code:

int *p = malloc(sizeof(*p) * n);

If sizeof evaluated its operand, then sizeof(*p) should be a mistake: p is being initialised at that point, so what could it point to? Indeed, this was my objection when I first saw the use of this idiom, until realising that my objection was invalid because sizeof does not evaluate its operand.

Originally Posted by Absurd

But the only way to determine that is to first evaluate what x+y is, no?

No, the one only needs to determine what are the types of x and y, and hence what is the type of (x + y). No evaluation is needed.

**phantomotap** · 09-26-2015

Unless its operand is a variable length array, sizeof does not evaluate its operand.

O_o

I think the statement could use some code.

Code:

#include <stdio.h>

void DoSomething1()
{
    printf("%zu\n", sizeof(char[printf("Hello, World!\n")]));
}

void DoSomething2()
{
    printf("%zu\n", sizeof(printf("Hello, World!\n")));
}

int main()
{
    DoSomething1();
    DoSomething2();
    return(0);
}

Soma

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