模板类复制构造函数参数是否带有模板参数？

Question

Hi is the following two are equivalent:

template<class T>
class name {
public:
    name() {/*...*/}
    name(name const &o) {/*...*/} // WITHOUT TEMPLATE ARGUMENT
    /*...*/
};

---

template<class T>
class name {
public:
    name() {/*...*/}
    name(name<T> const &o) {/*...*/} // WITH TEMPLATE ARGUMENT SPECIFIED
    /*...*/
};

So my question is this: Do I have to write classname with or without the template argument list in the copy constructor or not? If I don't write the template arguments does it mean that then other versions (with different template argument) can be passed to the copy constructor?

So if I want to achive that from the class A with template argument eg: int , the copy constructor of it only accept A with the same template argument (in our example: int ), do I have to put the template arguments (< T, K, ...>) there?

Answer 1

Do I have to write classname with or without the template argument list in the copy constructor or not?

They are equivalent. Within the template's scope, name is the injected class name , denoting the class name<T> . It can also be used as the template name, if you specify template arguments.

If I don't write the template arguments does it mean that then other versions (with different template argument) can be passed to the copy constructor?

No, without arguments it specifically denotes name<T> . To allow conversion from other specialisations, you'd need a constructor template:

template <typename T2> name(name<T2> const & other);

Note that this won't act as a copy constructor: you'll need to declare that separately if you don't want the implicitly generated one.

So if I want to achive that [...] it only accept A with the same template argument [...], do I have to put the template arguments (< T, K, ...>) there?

No, what you have is fine in that case.

Answer 2

在模板定义中， name是name<T>的简写（请参见第14.6.1节）