Does a templated constructor override the implicit copy constructor in C++?

问题

Does a templated constructor (such as the following) override the implicit copy constructor?

template <class T>
struct Foo
{
    T data;

    // ...

    template <class U>
    Foo(const Foo<U> &other) : data((T)doSomethingWith(other.data)) {}

    // ...
};

If so, does it still override it if other is passed by value rather than constant reference?

If so, is there any way around this without explicitly defining a copy constructor?

回答1:

No, that is not a copy constructor. Section 12.8 ([class.copy]) of the Standard requires that:

A non-template constructor for class X is a copy constructor if its first parameter is of type X&, const X&, volatile X& or const volatile X&, and either there are no other parameters or else all other parameters have default arguments.

The compiler will still implicitly generate a defaulted one.

You can make that explicit (requires C++11) by

Foo(const Foo<T>&) = default;

回答2:

Does a templated constructor (such as the following) override the implicit copy constructor?

No. The copy constructor is still implicitly declared, and is chosen in preference to the template.

Is there any way around this without explicitly defining a copy constructor?

No. If you don't want the implicit copy constructor, then you'll have to define one yourself.

回答3:

A templated constructor or assignment operator which looks like a templated [default constructor/copy constructor/move constructor/copy assignment operator/move assignment operator] is not really a [default constructor/copy constructor/move constructor/copy assignment operator/move assignment operator] and will not replace it or prevent it from being implicitly generated.

来源：https://stackoverflow.com/questions/11016262/does-a-templated-constructor-override-the-implicit-copy-constructor-in-c

标签

c++

templates

copy-constructor