c++ - 为什么 const char* 隐式转换为 bool 而不是 std::string？

Question

#include <iostream>
#include <string>

struct mystruct{
     mystruct(std::string s){
        
        std::cout<<__FUNCTION__ <<" String "<<s;
    }
    
     explicit mystruct(bool s) {
        
        std::cout<<__FUNCTION__<<" Bool "<<s;
    }
};


int main()
{
    
    const char* c ="hello";
    
    mystruct obj(c);

    return 0;
}

输出:

mystruct Bool 1

为什么const char*隐式转换为 bool而不是 std::string ，虽然构造函数需要 explicit类型 ？

隐式转换优先级在这里如何应用？

Answer 1

因为来自 const char* 的隐式转换至 bool被限定为标准转换，而 const char*至 std::string是用户定义的转换。前者排名更高，在overload resolution中获胜。 .

A standard conversion sequence is always better than a user-defined conversion sequence or an ellipsis conversion sequence.

顺便说一句: mystruct obj(c);执行 direct initialization , explicit转换构造函数，包括 mystruct::mystruct(bool)也被考虑。结果， c转换为 bool然后传递给 mystruct::mystruct(bool)作为构造 obj 的参数.

Direct-initialization is more permissive than copy-initialization: copy-initialization only considers non-explicit constructors and non-explicit user-defined conversion functions, while direct-initialization considers all constructors and all user-defined conversion functions.

关于 explicit specifier ,

1. Specifies that a constructor or conversion function (since C++11) or deduction guide (since C++17) is explicit, that is, it cannot be used for implicit conversions and copy-initialization.