Variadic template constructor speciliazation in template class

Question

I want to be able to specialize the ctor of a class the following way:

template<typename T>
class Foo {
public:
  template<typename... Ts>
  Foo(Ts... & args) {
    // ...
  }

  template<>
  Foo(int i) {
    // ...
  }
};

I get the following error:

error: explicit specialization in non-namespace scope ‘class Foo’

If I try to move the specialization outside the class, like this:

template<typename T>
class Foo {
public:
  template<typename... Ts>
  Foo(Ts &... args) {
    // ...
  }
};

template<typename T>
template<int>
Foo<T>::Foo(int i) {
    // ...
}

I get the following errors:

error: prototype for ‘Foo::Foo(int)’ does not match any in class ‘Foo’

error: candidate is: template template Foo::Foo(Ts& ...)

How do I do this correctly?

Mark Garcia Mark Garcia · Accepted Answer · 2014-02-08T09:32:44

You can just overload the constructor instead:

template<typename T>
class Foo {
public:
  template<typename... Ts>
  Foo(Ts&... args) {
    // ...
  }

  // template<>  <- REMOVE THIS
  Foo(int i) {
    // ...
  }
};

Overload resolution will prefer the non-template overload so doing Foo<MyType> f(1234); would choose Foo<MyType>::Foo(int);.

LIVE EXAMPLE (I've modified the variadic to be const for it to accept temporaries for the sake of the example).

Note that the position of the type modifier in your variadic function is wrong. It should be with the type, on the left side of ...:

Foo(Ts&...  args)

Variadic template constructor speciliazation in template class

2 Answers