如何使用具有特定方法的类来实例化C ++模板

Question

Let's say we have a class

class X {
public:
  void my_method() { std::cout << "I'm X"; }
} 

and we have a template class:

template<typename T> 
class Y
{
public:
    void put(typename T item) { item.my_method(); }

};

I want to execute item.my_method(); if the class Y is instantiated with a X class (if not there would be a compile error). How can I work this out?

Answer 1

Not sure I fully understand the question because what you have works.

class X
{
public:
    void my_method() { std::cout << "I'm X"; }
};

class Z
{
};

template <typename T>
class Y
{
public:
    void put(T item) { item.my_method(); }
};

int main(int argc, char* argv[])
{
    // This compiles fine
    X theX;
    Y<X> theXY;
    theXY.put( theX );

    // ERROR: This fails to compile
    Z theZ;
    Y<Z> theYZ;
    theYZ.put( theZ );
}

When Y is used with a class that does not have the my_method() member, it fails to compile.

Answer 2

What you want is template specialization :

template<>
class Y <X>
{
public:
    void put(X item) { item.my_method(); }
};

Answer 3

I believe you won't get a compiler error if you instantiate Y with a class other than X as long as you don't call put (something unique to templates).

You could use template specialization if put needs to do different things for different types.

Answer 4

Simply un-templatize the put method and create it for type X alone. And put a check inside the method if T is X .

template<typename T> struct IsX;  // unimplemented for rest of types
template<> struct IsX<X> { typedef X yes; };  // implemented for 'X' alone

template<typename T> 
class Y
{
public:
  void put (X item) { typedef typename IsX<T>::yes check; item.my_method(); }
                    //^^^^^^^^^^^^^^^ compile time ensure that it's for 'X only
};