Tomilov Anatoliy
Tomilov Anatoliy

Reputation: 16711

Explicitly defaulting a templated constructor

I tried to make default constructor = default; conditionally depending on class template argument's property using following technique:

#include <type_traits>
#include <utility>
#include <iostream>

#include <cstdlib>

template< typename T >
struct identity
{

};

template< typename T >
struct has_property
    : std::false_type
{

};

template< typename T >
struct S
{
    template< typename X = T,
              typename = std::enable_if_t< !has_property< X >::value > >
    S(identity< X > = {})
    { std::cout << __PRETTY_FUNCTION__ << std::endl; }

    template< typename X = T,
              typename = std::enable_if_t< has_property< X >::value > >
#if 0
    S() = default;
#else
    S()
    { std::cout << __PRETTY_FUNCTION__ << std::endl; }
#endif
};

struct A {};
struct B {};

template<>
struct has_property< B >
    : std::true_type
{

};

int main()
{
    S< A >{};
    S< B >{};
    return EXIT_SUCCESS;
}

But for #if 1 it gives an error:

main.cpp:32:11: error: only special member functions may be defaulted
    S() = default;
          ^

Is not template< ... > S() a declaration of a default constructor for S?

Can I implement such a dispatching using coming Concepts in future?

Upvotes: 3

Views: 219

Answers (1)

M.M
M.M

Reputation: 141554

I think this is covered by [dcl.fct.def.default]:

A function that is explicitly defaulted shall:

  • be a special member function,
  • have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T”, where T is the name of the member function’s class) as if it had been implicitly declared,

In your case, the implicitly-declared default constructor would not have been a function template, so you cannot explicitly default it.

GCC 5.x gives the error for your code, error: a template cannot be defaulted.

Upvotes: 2

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