关于C中sizeof运算符的困惑

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.

Answer 4

the size of one char is 1byte, a char can hold values up to 127(unsigned up to 255). when you say something like (a + b) a temporary variable is created and used to add a to b, because a and b can hold only 127, they are promoted by the compiler to be an int, just to be sure.

which is logical because if a = 100 and b = 100, the user would like to see 200 when he adds them and not 73 (which is the result of an overflow).