CODEWITHSUNDEEP
c
Tandura
Tandura

Reputation: 908

Functionality of sizeof

Why dose the folowing code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    printf("Hello world!\n");
    int x;
    long l;
    double d;
    printf("\n%d",sizeof(x+l+d));
    return 0;
}

prints on the console 8? I originaly thought that will convert x and l to double and display 24. Why 8?

Upvotes: 0

Views: 51

Answers (2)

Dabbler
Dabbler

Reputation: 9873

You are only passing one argument to sizeof, and that single argument's type is double. If you wrote sizeof(x) + sizeof(l) + sizeof(d), that would be something different (although still not 24, because not each argument is double).

Upvotes: 1

abligh
abligh

Reputation: 25199

sizeof returns the number of bytes used to store its argument. In this case its argument is x+l+d. x is an int, l a long, and d a double. When you add integer types to doubles, the result is promoted to form a double. So what you have written is the equivalent of sizeof(double). A double takes 8 bytes to store, so you are seeing 8 as the result.

Upvotes: 3

Related Questions