Are static_asserts to be evaluated if a member template isn't instantiated?

Question

I\'m thought I understood how a static_assert worked. But when I tried this on a g++ compiler, I started to wonder:

#include 
#incl         


        

Answer 1

Writing static_assert(false) is simply ill-formed, because it runs afoul of [temp.res]:

The program is ill-formed, no diagnostic required, if:
— no valid specialization can be generated for a template or a substatement of a constexpr if statement (6.4.1) within a template and the template is not instantiated

You can think of it as - since the static_assert's constant expression isn't dependent, the compiler can just immediately see that it's ill-formed and fire.

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You can either just not provide a definition of fn() for that specialization, or you could do something like:

template <class T> struct always_false : std::false_type { };
static_assert(always_false<some_dependent_type>::value, "...");

This would make it hypothetically possible for a valid specialization to be generated, even if nobody should ever specialize always_false.