C++ 重载与有符号和无符号 int 不同

Question

使用重载函数定义编译 C++ 时，为什么有符号和无符号整数类型的提升行为不同？这是预期的行为吗？

在下面的示例中，main 中对“fail”的调用是模棱两可的，但对“pass”的调用不是。

unsigned int fail(unsigned int a) {
  return a;
}

unsigned int fail(unsigned short a) {
  return a;
}

int pass(int a) {
  return a;
}

int pass(short a) {
  return a;
}


int main(){

  unsigned char a;
  char b;
  fail(a);
  pass(b);

  return 0;
}

示例输出(来自 clang，VS 编译器给出了类似的内容):

fail.cpp:22:3: error: call to 'fail' is ambiguous
  fail(a);
  ^~~~
fail.cpp:1:14: note: candidate function
unsigned int fail(unsigned int a) {
             ^
fail.cpp:5:14: note: candidate function
unsigned int fail(unsigned short a) {
             ^
1 error generated.

Answer 1

根据integral promotion (强调我的):

The following implicit conversions are classified as integral promotions:

• signed char [...] can be converted to int;
• unsigned char [...] can be converted to int if it can hold its entire value range, and unsigned int otherwise;
• char can be converted to int or unsigned int depending on the underlying type: signed char or unsigned char (see above);

Note that all other conversions are not promotions; for example, overload resolution chooses char -> int (promotion) over char -> short (conversion).

在您的情况下，鉴于 int 能够同时保存 signed char 和 unsigned char 的整个值范围，只有 >int pass(int a)是一种提升，比剩下的三个转化更可取，转化之间没有偏好。