c++ - 模板转换运算符问题

Question

我进行了一些浏览，这是最相关的 link我可以找到，但它没有回答我的问题

问题:为什么模板替换失败，以下内容无法编译？

template <typename T>
struct A
{
   A() {};
   A(T value) : val(value){} 
   operator T() { return this->val;}
   T val;
};


A<std::string> test;
std::cout << "xxx" + std::string(test); //works fine
std::cout << "xxx" + test; //compiler error

错误消息:

error: no match for 'operator+' (operand types are 'const char [4]' and 'A<std::__cxx11::basic_string<char> >')
   19 |    std::cout << "xxx" + test;
      |                 ~~~~~ ^ ~~~~
      |                 |       |
      |                 |       A<std::__cxx11::basic_string<char> >
      |                 const char [4]

Answer 1

std::operator+(std::basic_string) 是一组运算符模板，需要对第二个操作数进行模板实参推导test 。但 template argument deduction 中不会考虑隐式转换(从 A<std::string> 到 std::string ) .

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

正如您所展示的，显式转换如 "xxx" + std::string(test);工作正常。您还可以显式指定模板参数(以丑陋的方式)来绕过模板参数推导。

operator+ <char, std::char_traits<char>, std::allocator<char>>("xxx", test);