copy constructor, assignment operator in template

[Martin Vorbrodt]

In "C++ Templates, The Complete Guide" i read that template copy-con is
never default copy constructor, and template assignment-op is never a copy
assignment operator. Could someone please explain how I could
declate/override the two.

Thanx

 

[Victor Bazarov]

In "C++ Templates, The Complete Guide" i read that template copy-con is
never default copy constructor, and template assignment-op is never a copy
assignment operator. Could someone please explain how I could
declate/override the two.

I am not sure how much you did understand, so, the text is talking about
this situation

struct A {
template<class T> A(T const&); // "copy"-constructor
};

This "copy" constructor will not be used if 'T' is 'A'. The compiler will
still generate another copy constructor,

A(A const&);

according to its usual rules of generating one, and will use it when copy
construction is performed from another 'A' object.

Same with the assignment op.

To make sure you _do_ have control over the copy construction process from
another 'A' object, define both the template and the non-template c-tors:

struct A {
template<class T> A(T const&);
A(A const&);
};

and do your "true" copy-construction in the latter and "pseudo" copy
construction in the former. Remember that there is no way to invoke one
c-tor from another, so if you need to do some common processing, pull it
out into a separate function.

V

 

[Martin Vorbrodt]

what about this:

template <typename T>
class C {
public:
C(const T& t);
C(const C& c); // DO I STILL NEED THIS IF T = C ?
};

 

[Victor Bazarov]

what about this:

template <typename T>
class C {
public:
C(const T& t);
C(const C& c); // DO I STILL NEED THIS IF T = C ?

Huh? T is C?

};

(a) Don't top-post.

(b) How can you have a template whose argument is itself?

[...]

V