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?

@jwqwerty: Not logically wrong per se, but useless, as you're obtaining size of a temporary object that exists only for the duration of determining its size. In day-to-day usage, you'd be looking for `sizeof(num1)+sizeof(num2)` if you e.g. want to allocate space to serialize the data. — SF., Jul 14 '16 at 08:15
@jwqwerty `sizeof` is used to get the size of data type like `sizeof (int), sizeof (my_struct)` or size of specific variable/object `sizeof my_var, sizeof abc`. Maybe you want to write `sizeof num1 + sizeof num2`? — i486, Jul 14 '16 at 08:58

Sourav Ghosh · Answer 1 · 2016-07-14T08:16:06.133

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).

haccks · Accepted Answer · 2016-07-14T07:53:31.297

4

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.

edited Jul 14 '16 at 07:53

answered Jul 14 '16 at 07:47

haccks

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Why is num1+num2 promoted to int type? – Jin Jul 14 '16 at 07:48
@jwqwerty; Because of arithmetic operation. – haccks Jul 14 '16 at 07:50
@ haccks3 Then I guess in the line "result = num1 + num2", num1+num2 should also be promoted to int because it involves arithemetic operation? – Jin Jul 14 '16 at 07:52
@jwqwerty; It's the result of `num1 + num2` that is promoted to `int`. Assigning this result to a `char` type will not change the type of that variable. – haccks Jul 14 '16 at 07:55
@ haccks3 Oh I get it! Thanks!! – Jin Jul 14 '16 at 07:58
1

@jwqwerty: The result of the operation may not fit in the same type as the original, so type promotion is applied. Take `unsigned char a=200,b=200` - if you add them, you get 400, which won't fit. If you still insist on writing this to `unsigned char result=a+b;` result will be 144, because the variable gets truncated, data is lost because the container won't fit it. But during the addition, the 'temporary' container is expanded. – SF. Jul 14 '16 at 08:19

score 3 · Answer 3 · answered Jul 14 '16 at 07:51

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.

score 0 · Answer 4 · answered Jul 14 '16 at 13:34

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).

Confusion about sizeof operator in C

4 Answers4