请考虑以下代码.
#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;
}
该代码不会在g或Microsoft Visual C上编译.编译器提供的错误报告与运算符’<'在' X'.问题是为什么编译器不会根据给定的操作符string()简单地将SimpleStruct x转换为字符串,然后使用operator< (string,string)？
解决方法
操作符LT;对于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,Allocator>& rhs);
template<class charT,const charT* rhs);
template<class charT,class Allocator>
bool operator< (const charT* lhs,Allocator>& rhs);
您的呼叫与任何可用的超载不匹配,因此它们都从候选人列表中删除.由于没有选择功能模板作为解决调用的候选项,所以没有任何可以将SimpleStruct转换为.
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
}
