在表达式 bool << integer 中，bool 是提升为 int，还是提升为与整数相同的类型？

Question

考虑以下代码：

// E1 << E2 with E1 bool, and E2 an integer
true << static_cast<char>(1);
true << static_cast<short int>(1);
true << static_cast<int>(1);
true << static_cast<unsigned int>(1);
true << static_cast<long long int>(1); 
true << static_cast<long long unsigned int>(1);

运算时，E1是提升为与E2同类型，还是先提升为int再提升为std::common_type<E1, E2>::type？

换句话说就是：

true << static_cast<char>(9);

已定义或未定义的行为？

Answer 1

来自[expr.shift]：

The operands shall be of integral or unscoped enumeration type and integral promotions are performed. The type of the result is that of the promoted left operand.

来自[conv.prom]：

A prvalue of type bool can be converted to a prvalue of type int, with false becoming zero and true becoming one.

bool 的积分提升产生 int。所以true << (any integral type)的结果类型是int。积分提升后的true << static_cast<char>(9)等同于1 << 9，明确定义为512。

在表达式 bool << integer 中，bool 是提升为 int，还是提升为与整数相同的类型？

In the expression bool << integer, is bool promoted to int, or to the same type as the integer?

c++

integer-promotion

implicit-conversion