c – 模板隐式转换

template <typename T>
class MyClass
{
    public:

        MyClass(const MyClass& other)
        {
            //Explicit,types must both match;
        }    

        template <typename U>
        MyClass(const MyClass<U>& other)
        {
            //Implicit,types can copy across if assignable.
            //<?> Is there a way to make this automatically happen for memberwise
            //copying without having to define the contents of this function?

            this->value = other.getValue();
        }

    privte:

        T value;
};

以下情况属实：

MyClass<float> a;

MyClass<float> b = a; //Calls explicit constructor which would have been auto-generated.

MyClass<int> c;

MyClass<float> d = c; //Calls implicit constructor which would not have been auto gen.

我的问题是：有没有办法获得这样的隐式复制构造函数,但不必定义它是如何工作的？默认情况下创建类时,都会提供复制,赋值和标准构造函数…如果可能的话,在使用模板时,能够获得隐式转换构造函数也是很好的.

我猜它不可能,但有人可以解释为什么那是我的话,如果是这样的话？

谢谢！