为什么编译器不执行类型转换？

Question

考虑下面的代码。

#include <iostream>
#include <string>

struct SimpleStruct
{
    operator std::string () { return value; }
    std::string value;
};

int main ()
{
    std::string s;    // An empty string.
    SimpleStruct x;   // x.value constructed as an empty string.

    bool less = s < x; // Error here.
    return 0;
}

此代码不能在 g++ 或 Microsoft Visual C++ 上编译。编译器给出的错误报告是no match for operator '<' in 's < x'。问题是为什么编译器不简单地将SimpleStruct x 转换成string 根据给定的operator string () 再使用operator < ( string, string )？

Answer 1

operator< for std::string 是一个函数模板。重载是：

  template<class charT, class traits, class Allocator>
    bool operator< (const basic_string<charT,traits,Allocator>& lhs,
            const basic_string<charT,traits,Allocator>& rhs);
  template<class charT, class traits, class Allocator>
    bool operator< (const basic_string<charT,traits,Allocator>& lhs,
            const charT* rhs);
  template<class charT, class traits, class Allocator>
    bool operator< (const charT* lhs,
            const basic_string<charT,traits,Allocator>& rhs);

您的调用与任何可用的重载都不匹配，因此它们都从候选列表中删除。由于没有选择函数模板作为解决调用的候选者，因此没有任何东西可以转换为 SimpleStruct。

template <class T>
class String
{
};

template <class T>
bool operator< (const String<T>&, const String<T>&) { return true; }


//if a suitable non-template function is available, it can be picked
//bool operator< (const String<char>&, const String<char>&) { return true; }

struct SimpleStruct
{
   operator String<char> () { return value; }
   String<char> value;
};

int main()
{
    String<char> s;
    SimpleStruct ss;
    s < ss; //the call doesn't match the function template, leaving only the commented-out candidate
}