Why does this substitution failure create an error?

user_4696802 0
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In a template specialization I have a template argument with an enable_if with parameter that results in the enable_if not having a 'type' member, and so that template specialization should fail, but not create an error:

#include <type_traits>


template <typename value_t_arg, typename T = void>
struct underlyingtype
{
    using underlyingtype_t = value_t_arg;
};

template <typename value_t_arg>
struct underlyingtype < value_t_arg, typename std::enable_if<false>::type>
// std::enable_if<false> has no 'type' member, and so substitution should fail, 
// but no create an error, right?
{
    //using underlyingtype_t = value_t_arg::integral_t;
};

Why is there an error created here?

enable-if template-meta-programming sfinae c++
2 Answers

Your code is ill-formed (no diagnostic required) because the condition is always false regardless of the template argument, meaning the specialization would be ill-formed for every possible template argument.

[temp.res.general]/6.1

The program is ill-formed, no diagnostic required, if:

— no valid specialization can be generated for a template ... and the template is not instantiated, ...

Partial specializations appear to count as "templates" for the purposes of this section.

user_27520750

It's SFINAE, not FINAE. For the compiler to be happy with std::enable_if<false>::type you'd need the latter, which would mean something nonsensical.

user_58252940
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