1 Revision History

Since [P2647R0], updated wording, added an example and feature-test macro.

2 Introduction

Consider this example:

char xdigit(int n) {
  static constexpr char digits[] = "0123456789abcdef";
  return digits[n];
}

This function is totally fine. But when we try to extend it to work at compile time too, we run into a problem:

constexpr char xdigit(int n) {
  static constexpr char digits[] = "0123456789abcdef";
  return digits[n];
}

This is ill-formed. And it’s ill-formed for no good reason. We should make it valid.

2.1 History

Originally, no static variables could be declared in constexpr functions at all. That was relaxed in [P2242R3], which restructured the wording to be based on the actual control flow: rather than the function body not being allowed to contain a static variable declaration, now the rule is that control flow needs to not pass through an initialization of a static variable.

This makes sense for the perspective of a static (or, worse, thread_local) variable, whose initializer could run arbitrary code. But for a static constexpr variable, which must, by definition, be constant-initialization, there’s no question about whether and when to run what initialization. It’s a constant.

2.2 Workarounds

There are several workarounds for getting the above example to work. We could eschew the static variable entirely and directly index the literal, but this only works if we need to use it exactly one time:

constexpr char xdigit(int n) {
  return "0123456789abcdef"[n];
}

We could move the static variable into non-local scope, though we wanted to make it local for a reason - it’s only relevant to this particular function:

static constexpr char digits[] = "0123456789abcdef";
constexpr char xdigit(int n) {
  return digits[n];
}

We could make the variable non-static, but compilers have difficulty optimizing this, leading to much worse code-gen:

constexpr char xdigit(int n) {
  constexpr char digits[] = "0123456789abcdef";
  return digits[n];
}

Or we could wrap the variable into a consteval lambda, which is the most general workaround for this problem, whose only downside is that… we’re writing a consteval lambda because we can’t write a variable:

constexpr char xdigit(int n) {
  auto get_digits = []() consteval { return "0123456789abcdef"; };
  return get_digits()[n];
}

Having a local static constexpr variable is simply the obvious, most direct solution to the problem. We should permit it.

3 Proposal

Change 7.7 [expr.const]/5.2:

• 5 An expression E is a core constant expression unless the evaluation of E, following the rules of the abstract machine ([intro.execution]), would evaluate one of the following:

  • (5.1) …

  • (5.2) a control flow that passes through a declaration of a variable with static ([basic.stc.static]) or thread ([basic.stc.thread]) storage duration, unless that variable is usable in constant expressions;

[Example:

constexpr char test()
{
   static const int x = 5;
   static constexpr char c[] = "Hello World";
   return *(c+x);
}
static_assert(' ' == test());

-end example]

3.1 Feature-test macro

Change 15.11 [cpp.predefined]:

- __cpp_­constexpr 202207L
+ __cpp_­constexpr 20XXXXL

4 References