C++ – When and Why to Use Static with Constexpr

c++constexprstatic

As a disclaimer, I have done my research on this before asking. I found a similar SO question but the answer there feels a bit "strawman" and didn't really answer the question for me personally. I've also referred to my handy cppreference page but that doesn't offer a very "dumbed down" explanation of things most times.

Basically I'm still ramping up on constexpr, but at the moment my understanding is that it requires expressions to be evaluated at compile time. Since they may only exist at compile time, they won't really have a memory address at runtime. So when I see people using static constexpr (like in a class, for example) it confuses me… static would be superfluous here since that is only useful for runtime contexts.

I've seen contradiction in the "constexpr does not allow anything but compile-time expressions" statement (particularly here at SO). However, an article from Bjarne Stroustrup's page explains in various examples that in fact constexpr does require the evaluation of the expression at compile time. If not, a compiler error should be generated.

My previous paragraph seems a bit off-topic but it's a baseline necessary to understand why static can or should be used with constexpr. That baseline, unfortunately, has a lot of contradicting information floating around.

Can anyone help me pull all of this information together into pure facts with examples and concepts that make sense? Basically along with understanding how constexpr really behaves, why would you use static with it? And through what scopes/scenarios does static constexpr make sense, if they can be used together?

Best Answer

There is one significant difference for function-level static variables, and that has to do with lambda-capture:

void odr_use(int const&);

int main() {
    int non_static = 42;
    static int is_static = 42;
    []{
        odr_use(non_static); // error
        odr_use(is_static);  // OK
    }();
}

You are allowed to odr-use function-local static variables in lambdas without capturing them. That has nothing to do with constexpr - however, it usually makes little sense to enforce capture of constexpr variables. Therefore, static + constexpr gives a little more comfort in accessing constants from lambdas, consider:

#include <string_view>
int main()
{
    constexpr std::string_view x = "foo";
    []{ x.data(); }; // error: odr-use of non-captured variable
}

In this example, the string view and its contents are constants. The use of a member function however triggers odr-use, which requires us to capture the variable. Alternatively, use static + constexpr.

Odr-use means "use according to the One-Definition Rule", and it boils down to "is the address of the object required for that operation". For member functions, the address is required to form the this-pointer.

Below you'll find the conceptual differences which also explain the effect mentioned above.

constexpr variables are not compile-time values

A value is immutable and does not occupy storage (it has no address), however objects declared as constexpr can be mutable and do occupy storage (under the as-if rule).

Mutability

Most objects declared as constexpr are immutable, but it is possible to define a constexpr object that is (partially) mutable as follows:

struct S {
    mutable int m;
};

int main() {
    constexpr S s{42};
    int arr[s.m];       // error: s.m is not a constant expression
    s.m = 21;           // ok, assigning to a mutable member of a const object
}

Storage

The compiler can, under the as-if rule, choose to not allocate any storage to store the value of an object declared as constexpr. Similarly, it can do such optimizations for non-constexpr variables. However, consider the case where we need to pass the address of the object to a function that is not inlined; for example:

struct data {
    int i;
    double d;
    // some more members
};
int my_algorithm(data const*, int);

int main() {
    constexpr data precomputed = /*...*/;
    int const i = /*run-time value*/;
    my_algorithm(&precomputed, i);
}

The compiler here needs to allocate storage for precomputed, in order to pass its address to some non-inlined function. It is possible for the compiler to allocate the storage for precomputed and i contiguously; one could imagine situations where this might affect performance (see below).

Standardese

Variables are either objects or references [basic]/6. Let's focus on objects.

A declaration like constexpr int a = 42; is gramatically a simple-declaration; it consists of decl-specifier-seq init-declarator-list ;

From [dcl.dcl]/9, we can conclude (but not rigorously) that such a declaration declares an object. Specifically, we can (rigorously) conclude that it is an object declaration, but this includes declarations of references. See also the discussion of whether or not we can have variables of type void.

The constexpr in the declaration of an object implies that the object's type is const [dcl.constexpr]/9. An object is a region of storage[intro.object]/1. We can infer from [intro.object]/6 and [intro.memory]/1 that every object has an address. Note that we might not be able to directly take this address, e.g. if the object is referred to via a prvalue. (There are even prvalues which are not objects, such as the literal 42.) Two distinct complete objects must have different addresses[intro.object]/6.

From this point, we can conclude that an object declared as constexpr must have a unique address with respect to any other (complete) object.

Furthermore, we can conclude that the declaration constexpr int a = 42; declares an object with a unique address.

static and constexpr

The IMHO only interesting issue is the "per-function static", à la

void foo() {
    static constexpr int i = 42;
}

As far as I know -- but this seems still not entirely clear -- the compiler may compute the initializer of a constexpr variable at run-time. But this seems pathological; let's assume it does not do that, i.e. it precomputes the initializer at compile-time.

The initialization of a static constexpr local variable is done during static initializtion, which must be performed before any dynamic initialization[basic.start.init]/2. Although it is not guaranteed, we can probably assume that this does not impose a run-time/load-time cost. Also, since there are no concurrency problems for constant initialization, I think we can safely assume this does not require a thread-safe run-time check whether or not the static variable has already been initialized. (Looking into the sources of clang and gcc should shed some light on these issues.)

For the initialization of non-static local variables, there are cases where the compiler cannot initialize the variable during constant initialization:

void non_inlined_function(int const*);

void recurse(int const i) {
    constexpr int c = 42;
    // a different address is guaranteed for `c` for each recursion step
    non_inlined_function(&c);
    if(i > 0) recurse(i-1);
}

int main() {
    int i;
    std::cin >> i;
    recurse(i);
}

Conclusion

As it seems, we can benefit from static storage duration of a static constexpr variable in some corner cases. However, we might lose the locality of this local variable, as shown in the section "Storage" of this answer. Until I see a benchmark that shows that this is a real effect, I will assume that this is not relevant.

If there are only these two effects of static on constexpr objects, I would use static per default: We typically do not need the guarantee of unique addresses for our constexpr objects.

For mutable constexpr objects (class types with mutable members), there are obviously different semantics between local static and non-static constexpr objects. Similarly, if the value of the address itself is relevant (e.g. for a hash-map lookup).

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