Why (char) + (char) = (int) in C?

Question

Is it for the same reason as char + char = int? Why? ?

I got different results on this source code by different compilers

#include <stdio.h>
int main() {
    char a = 100, b = 100;
    printf("%d\n", a + b);
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
}

`scanf("%d%d", &a, &b);` invokes *undefined behavior*. This will be actually dangerous because typically `char` is smaller than `int`. — MikeCAT, Aug 01 '20 at 16:58
Also note that whether `char` is signed is implementation-defined. What different results did you get? — MikeCAT, Aug 01 '20 at 16:59
@MikeCAT ok I see, then could you tell me why sizeof(a+b) is equal to 4? — Antithesis, Aug 01 '20 at 17:00
Re "*why sizeof(a+b) is equal to 4?*", [Integer promotion](https://www.geeksforgeeks.org/integer-promotions-in-c/) — ikegami, Aug 01 '20 at 17:02
@MikeCAT I put a=100 and b=100. the second a+b = 200 in VS, and 100 in Code Blocks & Dev-Cpp — Antithesis, Aug 01 '20 at 17:03
[c - Implicit type promotion rules - Stack Overflow](https://stackoverflow.com/questions/46073295/implicit-type-promotion-rules) — MikeCAT, Aug 01 '20 at 17:03
Use `%hhd` to read `char` via `scanf()`. `%d` is for reading `int`. — MikeCAT, Aug 01 '20 at 17:08
@MikeCAT, TonyTannous, ikegami Thank you! That was because of the integer promotion — Antithesis, Aug 01 '20 at 17:15

score 1 · Accepted Answer · answered Aug 01 '20 at 17:15

You get different results because scanf("%d%d", &a, &b) is incorrect. For each %d, scanf expects the address of an int object, but you provided the addresses of char objects. This results in (dangerous) undefined behaviour.

For char objects, use the following:

scanf("%hhd%hhd", &a, &b)    // In a environment with signed chars
  -or-
scanf("%hhu%hhu", &a, &b)    // In a environment with unsigned chars

Why (char) + (char) = (int) in C?

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