C++与运算符重载转换模板

Question

我有一个模板类，我尝试通过运算符重载为模板verision转换为另一种

enum MyTypes {A,B,C} 

template<MyTypes T> 
MyClass { 
    const static MyType type_ = T; 
    template<MyTypes U>  
    MyClass<U> convert(MyTypes t) { 
     MyType<U> ret = MyType<U>(); 
     .... 
     return r; 
    } 
    template<MyTypes U>  
    MyClass<U> operator()() { 
     return convert(U); 
    } 
} 

然而，这种收益率（海合会，C11）

conversion from MyClass<0u> to non-scalar type MyClass<1u> requested 

除去模板功能，并试图

MyClass<A> operator()() { 
    MyClass<A> a = MyClass<A>(); 
    ... 
    return a; 
} 

抛出

the error operator cannot be overloaded 

基本上，我想实现的是，如果我有

MyClass<A> a = MyClass<A>; 
MyClass<B> b = a; 

它创建基于一个和转换一个新的MyClass的。任何想法这里我的错误是什么？

编辑： 我抛出了一个模板功能，只需留下操作

template<MyTypes U>  
MyClass<U> operator()() { 
    MyClass<U> ret = MyClass<U>(); 
    ... 
    return ret; 
} 

而是试图做

MyClass<B> = a 

Answer 1

下转换时，这仍然产生

conversion from MyClass<0u> to non-scalar type MyClass<1u> requested 

该值并允许分配：

#include <iostream> 
#include <string> 

enum MyTypes { A, B, C }; 

template<MyTypes T> 
struct MyClass{ 
    const static MyTypes type_ = T; 
    std::string history{"started as " + std::to_string(T)}; 

    template<MyTypes U> 
    operator MyClass<U>() { 
     return {history+" then became " + std::to_string(U)}; 
    } 
}; 

int main() 
{ 
    MyClass<A> a; 
    MyClass<B> b = a; 
    MyClass<C> c = b; 

    std::cout << a.history << '\n'; 
    std::cout << b.history << '\n'; 
    std::cout << c.history << '\n'; 
} 

输出：

started as 0 
started as 0 then became 1 
started as 0 then became 1 then became 2