c++ - C++中按位运算符的结果

Question

测试几个编译器(Comeau、g++)确认某些“整数类型”的按位运算符的结果是一个 int:

void foo( unsigned char );
void foo( unsigned short );

unsigned char a, b;

foo (a | b);

我希望“a | b”的类型是一个无符号字符，因为两个操作数都是无符号字符，但编译器说结果是一个整数，并且对 foo() 的调用是不明确的。为什么语言被设计成结果是一个 int，或者这个实现依赖？

谢谢，

Answer 1

这实际上是标准的 C++ 行为 (ISO/IEC 14882):

5.13/1 Bitwise inclusive OR operator

The usual arithmetic conversions are performed; the result is the bitwise inclusive OR function of its operands. The operator applies only to integral or enumeration operands.

5/9 Usual arithmetic conversions

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

• If either operand is of type long double, the other shall be converted to long double.
• Otherwise, if either operand is double, the other shall be converted to double.
• Otherwise, if either operand is float, the other shall be converted to float.
• Otherwise, the integral promotions shall be performed on both operands.
• ...

4.5/1 Integral Promotions

An rvalue of type char, signed char, unsigned char, short int, or unsigned short int can be converted to an rvalue of type int if int can represent all the values of the source type; otherwise, the source rvalue can be converted to an rvalue of type unsigned int.

我认为这与 int 有关，它被认为是执行环境的“自然”大小，以允许高效的算术(参见 Charles Bailey's answer)。