Why does enable_if_t in template arguments complains about redefinitions?

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.