Can a nonstatic member of a template specialize to data or function?

Question

GCC, Clang, ICC, and MSVC all reject this code, but I don't find any violated rule in the latest working draft of the C++ standard.

Is the rule already in the standard, or is it in a defect report?

#include <type_traits>

template< typename t >
struct s {
    std::conditional_t< std::is_integral< t >::value, t, void() > mem;
};

s< int > a;
s< void * > b;

Answer 1

The code is invalid due to 14.3.1/3:

If a declaration acquires a function type through a type dependent on a template-parameter and this causes a declaration that does not use the syntactic form of a function declarator to have function type, the program is ill-formed.

The type of the declaration here is dependent on the template parameter t, and therefore cannot be a function type.