Is It Possible to Use a `constexpr` Template Variable as the Default for a Formal Template Argument

Question

Using clang 3.6.0, I am unable to compile the following code example.

#include <type_traits>

template <typename T> constexpr bool IS_SCALAR = ::std::is_scalar<T>::value;
template <typename T, bool = IS_SCALAR<T>>
struct Class_Breaks
{
};

template <typename T, bool = ::std::is_scalar<T>::value>
struct Class_Works
{
};

void function()
{
    Class_Breaks<int> break_error;
    Class_Breaks<int, IS_SCALAR<int>> breaks_ok;
    Class_Works<int> ok;
}

But, the following error messages are returned:

1>  [ 66%] Building CXX object CMakeFiles/Core.dir/tests.cpp.obj
1>D:\Projects\Core\Core\tests.cpp(4,30): error : non-type template argument is not a constant expression
1>  template <typename T, bool = IS_SCALAR<T>>
1>                               ^
1>  D:\Projects\Core\Core\tests.cpp(16,18) :  note: while checking a default template argument used here
1>          Class_Breaks<int> break_error;
1>          ~~~~~~~~~~~~~~~~^
1>  1 error generated.

Answer 1

As mentioned by @StenSoft, it is a known bug. If you need to make it work because you have a constexpr template variable you'd like to use as a default, you can wrap the default value into an std::intergral_constant:

template<
    typename T,
    bool = std::integral_constant< bool, IS_SCALAR<T> >::value
>

Live example