constexpr versus anonymous namespace

Question

What is the difference between the following 2 patterns in creating const values?

constexpr int some_val = 0;

vs

namespace {
   const int some_val = 0;
}

I'm used to the 2nd method but is the 1st equivalent?

*Why* are you using the second method? In particular, what do you think is the effect of using an unnamed namespace here? I suspect that you’re not quite clear on its effect (= nonexistent). — Konrad Rudolph, Mar 25 '19 at 11:45
Unnamed namespaces would compare to `static`, not `constexpr`. — Jarod42, Mar 25 '19 at 12:32
Why you have put the second alternative in the namespace? Is there any reason behind that? The title of your Q is somehow like comparing apples and oranges. — TonySalimi, Mar 25 '19 at 12:41
@Jarod42 that would be my intention with the 2nd method. Is constexpr used primarily by the compiler to substitute the value? — regomodo, Mar 25 '19 at 13:16
Similar question: [constexpr vs static const](https://stackoverflow.com/questions/41125651/constexpr-vs-static-const-which-one-to-prefer). — 1201ProgramAlarm, Mar 25 '19 at 14:45

score 1 · Answer 1 · answered Mar 25 '19 at 15:00

unnamed namespace acts as static: linkage of the variable.

namespace {
   const int some_val = 0;
}

is equivalent to:

static const int some_val = 0;

constexpr doesn't change that: Demo

Now we can compare const vs constexpr:

constexpr variable are immutable values known at compile time (and so can be used in constant expression)
const variable are immutable values which might be initialized at runtime.

so you might have

int get_int() {
    int res = 0; 
    std::cin >> res;
    return res;
}

const int value = get_int();

but not

constexpr int value = get_int(); // Invalid, `get_int` is not and cannot be constexpr

Finally some const values are considered as constexpr as it would be for:

const int some_val = 0; // equivalent to constexpr int some_val = 0;

1 Answers1