How do I make a "static assertion" which is only checked on one branch of an if constexpr?

Question

Consider the following code:

enum { a = 1, b = 2 - a};

void foo() {
    if constexpr(a == 1) { }
    else { static_assert(b != 1, "fail :-("); }
}

naively, one might expect the static_assert() to never be triggered, since the if constexpr condition holds. But - it is triggered (GodBolt).

What C++ mechanism should I use instead, to ensure failure only when the condition of an if statement evaluates to false?

static_assert(a == 1 || b != 1, "fail :-("); Since a static_assert isn't scoped in this case (... because it is static and not in a template), so doesn't matter that it is in the else block. — Eljay
– Eljay, Commented Mar 18 at 21:57
@Eljay: I "don't know" about a as I write the static asseration, or whatever alternative I use. Also, DRY. — einpoklum
– einpoklum, Commented Mar 18 at 22:07
How can I create a type-dependent expression that is always false? — 許恩嘉
– 許恩嘉, Commented Mar 18 at 23:44
AFAIK if constexpr has no meaning outside a template. It's purpose is to eliminate the branching when the template is instantiated (hence the need to have a constexpr condition). In your case, it's equivalent to a regular if but restricted to constexpr conditions only (which is then useless). — Fareanor
– Fareanor, Commented Mar 19 at 9:41

max66 · Accepted Answer · 2025-03-18 22:54:34Z

1

What C++ mechanism should I use instead, to ensure failure only when the condition of an if statement evaluates to false?

According to cppreference

Outside a template, a discarded statement is fully checked. if constexpr is not a substitute for the #if preprocessing directive:

so I suppose you can make your foo() function a template one

template <int = 0>
void foo() {
    if constexpr(a == 1) { }
    else { static_assert(b != 1, "Fail"); }
}

--- EDIT ---

The OP asks

it's quite problematic to change the function definition just for an assertion somewhere... can't I use some kind of templated gadget instead?

I don't see a way through a type trait.

But if you can use at least C++20, so template lambdas, you can wrap your code inside a template lambda defined inside the foo() function. And call the lambda.

I mean something as follows

enum { a = 1, b = 2 - a};

void foo() {
  auto l = []<int = 0>{
    if constexpr (a == 1) {
    }
    else {
      static_assert(b != 1, "fail :-(");
    }
  };

  l();
}

int main ()
{
  foo();
}

edited Mar 18 at 22:54

answered Mar 18 at 19:37

max66

67k11 gold badges76 silver badges121 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

einpoklum Mar 18 at 22:20

I guess that's an option, but it's quite problematic to change the function definition just for an assertion somewhere... can't I use some kind of templated gadget instead?

max66 Mar 18 at 22:56

@einpoklum - I've edited my answer. Yes... if C++20 is OK, you can use a sort of templated gadget. you can wrap your code in a templated lambda.

einpoklum Mar 18 at 23:19

It's still messing with the other branch of the if()... I want something which will replace static_assert(...), and nothing else. But - maybe I have no recourse other than macros. PS - If it's doable without choosing the error string, that might still be useful.

max66 Mar 18 at 23:34

@einpoklum - something inside the else branch that gives a compilation error when the else branch is compiled? The problem is that, without templates, the else branch is ever compiled (if I understand correctly what cppreference says). Maybe a static_assert, outside the if/then/else, that fail when the if condition fails?

Bob__ Mar 19 at 7:51

It can be shrunk a bit, if it helps: godbolt.org/z/cGevY6jE5

|

Collectives™ on Stack Overflow

How do I make a "static assertion" which is only checked on one branch of an if constexpr?

1 Answer 1

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related