Why does enable_if_t in template arguments complains about redefinitions?

Question

I have the following case that works using std::enable_if :

template<typename T,
         typename std::enable_if<std::is_same<int, T>::value>::type* = nullptr>
void f() { }

template<typename T,
         typename std::enable_if<std::is_same<double, T>::value>::type* = nullptr>
void f() { }

Now, I saw in cppreference the new syntax, much cleaner in my opinion : typename = std::enable_if_t<std::is_same<int, T>::value>>

I wanted to port my code :

template<typename T,
         typename = std::enable_if_t<std::is_same<int, T>::value>>
void g() { }

template<typename T,
         typename = std::enable_if_t<std::is_same<double, T>::value>>
void g() { }

But now GCC (5.2) complains :

error: redefinition of 'template<class T, class> void g()'
       void g() { }

Why is that so ? What can I do to have the new, more concise syntax in this case if this is possible ?

Your second code uses default template arguments. They are not part of the function signature, hence you're declaring two function templates with the same signature = redefinition. The corresponding use of `enable_if_t` is literally `std::enable_if_t::value>* = nullptr`. — dyp, Jul 19 '15 at 11:07
Well, you didn't rewrite your original code exactly. You forgot the `* = nullptr`. — Kerrek SB, Jul 19 '15 at 11:08
You can add a dummy template parameter to one of the declarations, like `, typename = void` after `enable_if` — Piotr Skotnicki, Jul 19 '15 at 11:33
@dyp: I think you can put your comment as an answer ! @KerrekSB: no, that's the point of `enable_if_t` @PiotrS. : ok — Jean-Michaël Celerier, Jul 19 '15 at 11:59
Probably a duplicate of: http://stackoverflow.com/q/15427667/ or of http://stackoverflow.com/q/28674543/ or of http://stackoverflow.com/q/8743159/ — dyp, Jul 19 '15 at 12:03
I went the `typename std::enable_if_t::value>* = nullptr` route. 4 characters gained, oh well :p — Jean-Michaël Celerier, Jul 19 '15 at 14:20
@Jean-MichaëlCelerier You don't need the `typename` in front of `std::enable_if_t`. More characters gained. — bogdan, Jul 19 '15 at 15:08
Even more characters gained with `std::enable_if_t::value, int> = 0` or `std::enable_if_t, int> = 0`. — Oktalist, Jul 19 '15 at 18:21
@Oktalist Just tried but my compiler (gcc 4.9) complains with `*=0` and not with `*=nullptr`. — Jean-Michaël Celerier, Jul 21 '15 at 10:22
That's why I didn't say `*=0`. I supplied `int` as the second argument to `enable_if_t` instead of the default `void`, so `0` is an integer literal, not a null pointer literal. — Oktalist, Jul 21 '15 at 12:07

Yakk - Adam Nevraumont · Accepted Answer · 2018-09-28T17:29:38.170

Let's remove some code.

template<
  class T,
  class U/* = std::enable_if_t<std::is_same<int, T>::value>*/
 >
void g() { }

template<
  class T,
  class U/* = std::enable_if_t<std::is_same<double, T>::value>*/
 >
void g() { }

would you be surprised if the compiler rejected the two above templates?

They are both template functions of "type" template<class,class>void(). The fact that the 2nd type argument has a different default value matters not. That would be like expecting two different print(string, int) functions with different default int values to overload. ;)

In the first case we have:

template<
  typename T,
  typename std::enable_if<std::is_same<int, T>::value>::type* = nullptr
>
void f() { }

template<
  typename T,
  typename std::enable_if<std::is_same<double, T>::value>::type* = nullptr
>
void f() { }

here we cannot remove the enable_if clause. Updating to enable_if_t:

template<
  class T,
  std::enable_if_t<std::is_same<int, T>::value, int>* = nullptr
>
void f() { }

template<
  class T,
  std::enable_if_t<std::is_same<double, T>::value, int>* = nullptr
>
void f() { }

I also replaced a use of typename with class. I suspect your confusion was because typename has two meanings -- one as a marker for a kind of template argument, and another as a disambiguator for a dependent type.

Here the 2nd argument is a pointer, whose type is dependent on the first. The compiler cannot determine if these two conflict without first substituting in the type T -- and you'll note that they will never actually conflict.

score 12 · Answer 2 · answered Jul 19 '15 at 18:26

enable_if_t<B> is just an alias for typename enable_if<B>::type. Let's substitute that in g so we can see the real difference between f and g:

template<typename T,
         typename std::enable_if<std::is_same<int, T>::value>::type* = nullptr>
void f() { }

template<typename T,
         typename std::enable_if<std::is_same<double, T>::value>::type* = nullptr>
void f() { }

template<typename T,
         typename = typename std::enable_if<std::is_same<int, T>::value>::type>
void g() { }

template<typename T,
         typename = typename std::enable_if<std::is_same<double, T>::value>::type>
void g() { }

In the case of f, we have two function templates both with template parameters <typename, X*>, where the type X is dependent on the type of the first template argument. In the case of g we have two function templates with template parameters <typename, typename> and it is only the default template argument which is dependent, so C++ considers that they are both declaring the same entity.

Either style can be used with the enable_if_t alias:

template<typename T,
         std::enable_if_t<std::is_same<int, T>::value>* = nullptr>
void f() { }

template<typename T,
         std::enable_if_t<std::is_same<double, T>::value>* = nullptr>
void f() { }

template<typename T,
         typename = std::enable_if_t<std::is_same<int, T>::value>>
void g() { }

template<typename T,
         typename = std::enable_if_t<std::is_same<double, T>::value>>
void g() { }

If I could accept multiples answers I would have accepted yours as well because it is equally valid! thanks ! — Jean-Michaël Celerier, Jul 21 '15 at 08:00

score 4 · Answer 3 · answered Jul 19 '15 at 16:51

For a function return type, you're looking for the following:

template<typename T> std::enable_if_t< conditional, instantiation result > foo();

Example:

#include <iostream>

// when T is "int", replace with 'void foo()'   
template<typename T>
std::enable_if_t<std::is_same<int, T>::value, void> foo() {
    std::cout << "foo int\n";
}

template<typename T>
std::enable_if_t<std::is_same<float, T>::value, void> foo() {
    std::cout << "foo float\n";
}

int main() {
    foo<int>();
    foo<float>();
}

http://ideone.com/TB36gH

Why does enable_if_t in template arguments complains about redefinitions?

4 Answers4

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