sizeof()

[mehak]

why is the value of val 4?

#include<stdio.h>
int f(int);
int main(){
    int i=3,val;
    val=sizeof (f(i)+ +f(i=1)+ +f(i-1));
    printf("%d %d",val,i);
    return 0;
}
int f(int num){
        return num*5;
}

[LendraDwi]

Because sizeof() is use to calculate the data type

[Chervil]

Function f() returns an integer. Hence the result of evaluating the expression
f(i)+ +f(i=1)+ +f(i-1) is an integer too.

In this case sizeof() is giving the size of an integer which is typically 4 bytes, though it can vary across systems.

[dhayden]

Note that sizeof() does not evaluate its argument, it just determines its size. So in the example, f() never actually gets called and i never changes from it's initial value of 3.

[Chervil]

Well the value given by sizeof is known at compile time, so the expression isn't actually evaluated. But its type is deduced. For example:

val = sizeof (f(i)+ +f(i=1)+ +f(i-1) + 1.0);

gives 8 rather than 4 on my system, as the type of the expression is different.