CODEWITHSUNDEEP
c++copy-constructorc++03
Tigran Saluev
Tigran Saluev

Reputation: 3582

C++03: keep default copy constructor when using template constructor

Imagine I create a class with a template constructor to define implementations later:

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

How can I avoid overriding compiler's implicitly generated copy constructor A(const A&)? In C++11 I can do something like

#include <type_traits>
struct A {
    template<typename T, class = 
        typename std::enable_if<std::is_same<A, T>::value>::type>
    A(const T& arg);
};

and it works. But C++03 doesn't support default template arguments. Any workaround here?

Upvotes: 1

Views: 236

Answers (2)

juanchopanza
juanchopanza

Reputation: 227468

You don't need to do anything. The compiler generated coppy constructor will kick in as needed. A copy constructor cannot be a template. For example,

#include <iostream>

struct A {
    template<typename T>
    A(const T& arg) { std::cout << "template\n"; }
};

int main()
{
    A a(42);  // template ctor
    A b(a);   // copy ctor
    A c = b;  // copy ctor
}

Output:

template

Upvotes: 3

Sebastian Redl
Sebastian Redl

Reputation: 72044

You shouldn't need one. The constructor in question will instantiate to A(const A&) when given an A, which is the same as the actual copy constructor, so the non-template constructor will be preferred.

Upvotes: 1

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