c++ - 显式转换运算符 bool

Question

我有以下代码

  class C
    {
    public:
        C(bool v_):v(v_){};
        explicit operator bool() const {return v;}
        bool v = false;    
    };

 int main(){
  C a{true};
  C b{false};
  // C d = true; // doesn't compile, since requires explicit conversion
  C d = static_cast<bool>(true); // fine

  if(a || b) std::cout << "conversion happened!" << std::endl; // why it doesn't require explicit conversion here?

 }

它输出“转换发生了!”。我的问题是为什么 || 被认为是显式转换？

编辑

  // C d = true; // actually this line compiles because of the constructor

Answer 1

来自 [expr.log.or]:

The operands are both contextually converted to bool

允许上下文转换为 bool 以使用显式转换。这就是“上下文”概念的要点:在有意义的时候允许显式转换，否则就不允许，所以你不会不小心从 bool 转换中形成任意整数或指针值，但是当你实际要求一个 bool ，你明白了。

详情见 [conv]/4:

Certain language constructs require that an expression be converted to a Boolean value. An expression e appearing in such a context is said to be contextually converted to bool and is well-formed if and only if the declaration bool t(e); is well-formed, for some invented temporary variable t