Type conversion - unsigned to signed int/char

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本文详细解释了C语言中不同类型变量进行比较时的隐式类型转换规则，包括整数提升和常规算术转换等概念，并通过具体示例阐述了这些规则如何影响程序的输出结果。

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I tried the to execute the below program:

#include <stdio.h>

int main() {
    signed char a = -5;
    unsigned char b = -5;
    int c = -5;
    unsigned int d = -5;

    if (a == b)
        printf("\r\n char is SAME!!!");
    else
        printf("\r\n char is DIFF!!!");

    if (c == d)
        printf("\r\n int is SAME!!!");
    else
        printf("\r\n int is DIFF!!!");

    return 0;
}

For this program, I am getting the output:

char is DIFF!!! int is SAME!!!

Why are we getting different outputs for both?
Should the output be as below ?

char is SAME!!! int is SAME!!!

A1:

This is because of the various implicit type conversion rules in C. There are two of them that a C programmer must know: the usual arithmetic conversions and the integer promotions (the latter are part of the former).

In the char case you have the types (unsigned char) == (signed char). These are both small integer types. Other such small integer types are bool and short. The integer promotion rules state that whenever a small integer type is an operand of an operation, its type will get promoted to int, which is signed. This will happen no matter if the type was signed or unsigned.

In the case of the signed char, the sign will be preserved and it will be promoted to an intcontaining the value -5. In the case of the unsigned char, it contains a value which is 251 (0xFB ). It will be promoted to an int containing that same value. You end up with

if( (int)-5 == (int)251 )

In the integer case you have the types (unsigned int) == (signed int) . They are not small integer types, so the integer promotions do not apply. Instead, they are balanced by the usual arithmetic conversions , which state that if two operands have the same "rank" (size) but different signedness, the signed operand is converted to the same type as the unsigned one. You end up with

if( (unsigned int)-5 == (unsigned int)-5)

A2:

Cool question!

The int comparison works, because both ints contain exactly the same bits, so they are essentially the same. But what about the chars?

Ah, C implicitly promotes chars to ints on various occasions. This is one of them. Your code saysif(a==b), but what the compiler actually turns that to is:

if((int)a==(int)b)

(int)a is -5, but (int)b is 251. Those are definitely not the same.

EDIT: As @Carbonic-Acid pointed out, (int)b is 251 only if a char is 8 bits long. If int is 32 bits long, (int)b is -32764.

REDIT: There's a whole bunch of comments discussing the nature of the answer if a byte is not 8 bits long. The only difference in this case is that (int)b is not 251 but a different positive number, which isn't -5. This is not really relevant to the question which is still very cool.

From: http://stackoverflow.com/questions/17312545/type-conversion-unsigned-to-signed-int-char