Is this program ill-formed despite SFINAE?

Question

template  void f() {
    return 0;  // returning value from function returning `void`
}

int main()
{
    // Not instantiating or calling any f

        

Answer 1

Reading more closely, that standard passage says:

If a substitution results in an invalid type or expression, type deduction fails. An invalid type or expression is one that would be ill-formed if written using the substituted arguments. [..]

return 0 is not an expression, so SFINAE does not apply.

The passage goes on:

Only invalid types and expressions in the immediate context of the function type and its template parameter types can result in a deduction failure.

return 0 has nothing to do with the function type or its template parameter types, so SFINAE still does not apply.

Answer 2

The program is ill-formed as per 14.6/8:

If no valid specialization can be generated for a template definition, and that template is not instantiated, the template definition is ill-formed, no diagnostic required.

That is whether you instantiate the template or not, the template definition is ill-formed as there is no possible instantiation that will succeed.

Note that this is completely unrelated to SFINAE: Substitution Failure is not an Error is part of the substitution process, and never takes into account the contents of the template.