Confusion about sizeof operator in C

Question

I'm confused about sizeof operator in C.

#include <stdio.h>

int main(void)
{
    char num1=1, num2=2, result;
    result = num1 + num2;
    printf("Size of result: %d \n",sizeof result);
    printf("Size of result: %d \n",sizeof(num1+num2));
}

The results are 1 and 4 respectively. Why does this happen?

Answer 1

TL;DR answer:

• sizeof result is same as sizeof(char).
• sizeof(num1+ num2) is same as sizeof (int) ^why?

In your case, they produce 1 (guaranteed by standard) and 4 (may vary), respectively.

That said, sizeof produces a result of type size_t, so you should %zu format specifier to print the value.

---

Why:

First, for the addition operator +, quoting C11, chapter §6.5.6

If both operands have arithmetic type, the usual arithmetic conversions are performed on them.

Regarding usual arithmetic conversions, §6.3.1.8/p1

[....] Otherwise, the integer promotions are performed on both operands.[...]

and then from §6.3.1.1,/p2,

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.

So, sizeof(num1+num2) is the same as sizeof(int).

Answer 2

result is of char type, therefore sizeof is giving 1 while num1+num2 promoted to int type and therefore it gives 4 (size of int).

Note that when an arithmetic operation is performed on a type smaller than that of int and all of it's value can be represented by int then result will be promoted to int type.

Answer 3

num1 + num2 is becoming integer and hence the output is 4 whereas result is char which outputs 1.

You can refer this article Integer Promotion:

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions.