https://gcc.gnu.org/bugzilla/show_bug.cgi?id=64455
Bug ID: 64455
Summary: A constexpr variable template can't be used with
enable_if
Product: gcc
Version: 5.0
Status: UNCONFIRMED
Keywords: rejects-valid
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: ville.voutilainen at gmail dot com
CC: jason at redhat dot com
#include <type_traits>
template<typename Type>
constexpr bool IsType = true;
template<class T>
struct X {
typedef typename std::enable_if<IsType<T>,T>::type type;
};
int main()
{
X<int>::type t;
}
dionne.cpp:8:37: error: the value of ‘IsType<T>’ is not usable in a constant
expression
typedef typename std::enable_if<IsType<T>,T>::type type;
^
dionne.cpp:4:16: note: ‘IsType<T>’ used in its own initializer
constexpr bool IsType = true;
^
dionne.cpp:8:48: error: the value of ‘IsType<T>’ is not usable in a constant
expression
typedef typename std::enable_if<IsType<T>,T>::type type;
^
dionne.cpp:4:16: note: ‘IsType<T>’ used in its own initializer
constexpr bool IsType = true;
^
dionne.cpp:8:48: note: in template argument for type ‘bool’
typedef typename std::enable_if<IsType<T>,T>::type type;
^
Clang accepts the code.
