Question

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?

Answer 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;