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c++constexpr
Lorah Attkins
Lorah Attkins

Reputation: 5856

constexpr variable in source file global scope

What is the proper way to declare a constexpr constant in a source file? I'm split between two ways:

constexpr int ORDER = 1;

vs

namespace {
constexpr int ORDER = 1;
} // unnamed namespace

The reason I question the usefulness of wrapping into an unnamed namespace is because at global scope, constexpr implies static. So similar to how in header files writing

static constexpr int ORDER = 1;

makes static just a repetition, I'm assuming that the same should apply in source files as well, hence internal linkage should be guaranteed for "constexpr variables declared in a source file's global scope".

Is this the case? Are there different suggestions?

Upvotes: 4

Views: 2143

Answers (1)

Nikos Athanasiou
Nikos Athanasiou

Reputation: 31577

It's not required to enclose constexpr variables, declared in a source file, in an unnamed namespace. Since the end goal is to achieve internal linkage you have to remember this:

The name of an entity that belongs to a namespace scope has internal linkage if it is the name of

  • a variable, variable template, function, or function template that is explicitly declared static; or
  • a non-template variable of non-volatile const-qualified type, unless
    • it is explicitly declared extern, or
    • it is inline or exported, or
    • it was previously declared and the prior declaration did not have internal linkage; or
  • a data member of an anonymous union.

i.e since "constexpr implies const and const on namespace scope implies internal linkage", it's superfluous to say

static constexpr int ORDER = 1;

or

namespace {
  static constexpr int ORDER = 1;
}

If you want a tangible proof of the internal linkage property, consider this compilation error

Upvotes: 4

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