Static_assert inside if constexpr not discarded [duplicate]

Question

P0292R1 constexpr if has been included, on track for C++17. It seems useful (and can replace use of SFINAE), but a comment regarding static_assert being ill-formed, no diagnostic required in the false branch scares me:

Disarming static_assert declarations in the non-taken branch of a
constexpr if is not proposed.

void f() {
  if constexpr (false)
    static_assert(false);   // ill-formed
}

template<class T>
void g() {
  if constexpr (false)
    static_assert(false);   // ill-formed; no 
               // diagnostic required for template definition
}

I take it that it's completely forbidden to use static_assert inside constexpr if (at least the false / non-taken branch, but that in practice means it's not a safe or useful thing to do).

How does this come about from the standard text? I find no mentioning of static_assert in the proposal wording, and C++14 constexpr functions do allow static_assert (details at cppreference: constexpr).

Is it hiding in this new sentence (after 6.4.1) ? :

When a constexpr if statement appears in a templated entity, during an instantiation of the enclosing template or generic lambda, a discarded statement is not instantiated.

From there on, I assume that it is also forbidden, no diagnostic required, to call other constexpr (template) functions which somewhere down the call graph may call static_assert.

Bottom line:

If my understanding is correct, doesn't that put a quite hard limit on the safety and usefulness of constexpr if as we would have to know (from documentation or code inspection) about any use of static_assert? Are my worries misplaced?

Update:

This code compiles without warning (clang head 3.9.0) but is to my understanding ill-formed, no diagnostic required. Valid or not?

template< typename T>
constexpr void other_library_foo(){
    static_assert(std::is_same<T,int>::value);
}

template<class T>
void g() {
  if constexpr (false)
    other_library_foo<T>(); 
}

int main(){
    g<float>();
    g<int>();
}

Answer 1

This is talking about a well-established rule for templates - the same rule that allows compilers to diagnose template<class> void f() { return 1; }. [temp.res]/8 with the new change bolded:

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 ([stmt.if]) within a template and the template is not instantiated, or
• [...]

No valid specialization can be generated for a template containing static_assert whose condition is nondependent and evaluates to false, so the program is ill-formed NDR.

static_asserts with a dependent condition that can evaluate to true for at least one type are not affected.

Answer 2

C++20 makes static_assert in the else branch of if constexpr much shorter now, because it allows template lambda parameters. So to avoid the ill-formed case, we can now define a lambda with a bool template non-type parameter that we use to trigger the static_assert. We immediately invoke the lambda with (), but since the lambda won't be instantiated if its else branch is not taken, the assertion will not trigger unless that else is actually taken:

template<typename T>
void g()
{
    if constexpr (case_1)
        // ...
    else if constexpr (case_2)
        // ...
    else
        []<bool flag = false>()
            {static_assert(flag, "no match");}();
}

Answer 3

Edit: I'm keeping this self-answer with examples and more detailed explanations of the misunderstandings that lead to this questions. The short answer by T.C. is strictly enough.

After rereading the proposal and on static_assert in the current draft, and I conclude that my worries were misguided. First of all, the emphasis here should be on template definition.

ill-formed; no diagnostic required for template definition

If a template is instantiated, any static_assert fire as expected. This presumably plays well with the statement I quoted:

... a discarded statement is not instantiated.

This is a bit vague to me, but I conclude that it means that templates occurring in the discarded statement will not be instantiated. Other code however must be syntactically valid. A static_assert(F), [where F is false, either literally or a constexpr value] inside a discarded if constexpr clause will thus still 'bite' when the template containing the static_assert is instantiated. Or (not required, at the mercy of the compiler) already at declaration if it's known to always be false.

Examples: (live demo)

#include <type_traits>

template< typename T>
constexpr void some_library_foo(){
    static_assert(std::is_same<T,int>::value);
}

template< typename T>
constexpr void other_library_bar(){
    static_assert(std::is_same<T,float>::value);
}

template< typename T>
constexpr void buzz(){
    // This template is ill-formed, (invalid) no diagnostic required,
    // since there are no T which could make it valid. (As also mentioned
    // in the answer by T.C.).
    // That also means that neither of these are required to fire, but
    // clang does (and very likely all compilers for similar cases), at
    // least when buzz is instantiated.
    static_assert(! std::is_same<T,T>::value);
    static_assert(false); // does fire already at declaration
                          // with latest version of clang
}

template<class T, bool IntCase>
void g() {
  if constexpr (IntCase){
    some_library_foo<T>();

    // Both two static asserts will fire even though within if constexpr:
    static_assert(!IntCase) ;  // ill-formed diagnostic required if 
                              // IntCase is true
    static_assert(IntCase) ; // ill-formed diagnostic required if 
                              // IntCase is false

    // However, don't do this:
    static_assert(false) ; // ill-formed, no diagnostic required, 
                           // for the same reasons as with buzz().

  } else {
    other_library_bar<T>();
  }      
}

int main(){
    g<int,true>();
    g<float,false>();

    //g<int,false>(); // ill-formed, diagnostic required
    //g<float,true>(); // ill-formed, diagnostic required
}

The standard text on static_assert is remarkably short. In standardese, it's a way to make the program ill-formed with diagnostic (as @immibis also pointed out):

7.6 ... If the value of the expression when so converted is true, the declaration has no effect. Otherwise, the program is ill-formed, and the resulting diagnostic message (1.4) shall include the text of the string-literal, if one is supplied ...

Answer 4

Your self-answer and possibly the one by T.C. are not quite correct.

First of all, the sentence "Both two static asserts will fire even though within if constexpr" is not correct. They won't because the if constexpr condition depends on a template parameter.
You can see that if you comment out the static_assert(false) statements and the definition of buzz() in your example code: static_assert(!IntCase) won't fire and it will compile.

Furthermore, things like AlwaysFalse<T>::value or ! std::is_same_v<T, T> are allowed (and have no effect) inside a discarded constexpr if, even if there's no T for which they evaluate to true.
I think that "no valid specialization can be generated" is bad wording in the standard (unless cppreference is wrong; then T.C. would be right). It should say "could be generated", with further clarification of what is meant by "could".

This is related to the question whether AlwaysFalse<T>::value and ! std::is_same_v<T, T> are equivalent in this context (which is what the comments to this answer are about).
I would argue that they are, since it's "can" and not "could" and both are false for all types at the point of their instantiation.
The crucial difference between std::is_same and the non-standard wrapper here is that the latter could theoretically be specialized (thanks, cigien, for pointing this out and providing the link).

The question whether ill-formed NDR or not also crucially depends on whether the template is instantiated or not, just to make that entirely clear.