unsigned char在函数调用中自动提升为int，为什么？ [英] Unsigned char automatically promoted to int in function call, why?

问题描述

为什么调用函数时 unsigned char 自动提升为 int ？在下面的例子中有一个 f（int）和 f（char）编译器将 unsigned char 参数强制转换为 char 并调用 f（char），因为它们具有相同的位数。它调用 f（int），即使这意味着将参数提升为具有更多位的类型。任何指向规则被定义的指针？标准或编译器/平台特定？

Why is an unsigned char automatically promoted to an int when calling a function? In the example below there is an f(int) and a f(char) function. It seemed more logical that the compiler would coerce an unsigned char argument to a char and call f(char) since they have the same number of bits. It is calling f(int) instead, even when that means promoting the argument to a type with more bits. Any pointers to where the rule is defined? Standard or compiler/platform specific?

#include <iostream>

void f( int key )
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

void f( char key )
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

int main( int argc, char* argv[] )
{
    int a = 'a';
    char b = 'b';
    unsigned char c = 'c';

    f(a);
    f(b);
    f(c);

    return 0;
}

产生此输出：

void f(int)
void f(char)
void f(int)

推荐答案

因为 unsigned char 不能用 char 。例如，如果它们都是8位，并且你的unsigned char包含值 255 ，溢出 signed char - 和有符号整数溢出调用未定义的行为。无论如何，可打印字符通常预期存储在 char 而不是 unsigned char （和C字符串类型 char [] ，而不是 unsigned char [] 等）

Because unsigned char cannot be represented by char. For example, if they're both 8 bits, and your unsigned char contains the value 255, that overflows the signed char - and signed integer overflow invokes undefined behavior. Anyway, printable characters are generally expected to be stored in char and not unsigned char (and C strings are of type char[] and not unsigned char[], etc.)

因此，需要提升到 int 来表示大于 1 < （CHAR_BIT - 1）。

So the promotion to int is necessary to represent values greater than 1 << (CHAR_BIT - 1).

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