Why does sizeof(x++) not increment x?

Question

Here is the code compiled in dev c++ windows:

#include 

int main() {
    int x = 5;
    printf(\"%d and \", sizeof(x++)); // note 1
    print         


        

Answer 1

As the operand of sizeof operator is not evaluated, you can do this:

int f(); //no definition, which means we cannot call it

int main(void) {
        printf("%d", sizeof(f()) );  //no linker error
        return 0;
}

Online demo : http://ideone.com/S8e2Y

That is, you don't need define the function f if it is used in sizeof only. This technique is mostly used in C++ template metaprogramming, as even in C++, the operand of sizeof is not evaluated.

Why does this work? It works because the sizeof operator doesn't operate on value, instead it operates on type of the expression. So when you write sizeof(f()), it operates on the type of the expression f(), and which is nothing but the return type of the function f. The return type is always same, no matter what value the function would return if it actually executes.

In C++, you can even this:

struct A
{
  A(); //no definition, which means we cannot create instance!
  int f(); //no definition, which means we cannot call it
};

int main() {
        std::cout << sizeof(A().f())<< std::endl;
        return 0;
}

Yet it looks like, in sizeof, I'm first creating an instance of A, by writing A(), and then calling the function f on the instance, by writing A().f(), but no such thing happens.

Demo : http://ideone.com/egPMi

Here is another topic which explains some other interesting properties of sizeof:

• sizeof taking two arguments