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

sizeof is a compile-time operator, so at the time of compilation sizeof and its operand get replaced by the result value. The operand is not evaluated (except when it is a variable length array) at all; only the type of the result matters.

short func(short x) {  // this function never gets called !!
   printf("%d", x);    // this print never happens
   return x;
}

int main() {
   printf("%d", sizeof(func(3))); // all that matters to sizeof is the 
                                  // return type of the function.
   return 0;
}

Output:

2

as short occupies 2 bytes on my machine.

Changing the return type of the function to double:

double func(short x) {
// rest all same

will give 8 as output.