Constexpr VS. Static Const: Which One to Prefer

constexpr vs. static const: Which one to prefer?

As long as we are talking about declaring compile-time constants of ~~scalar~~ integer or enum types, there's absolutely no difference between using const (static const in class scope) or constexpr.

Note that compilers are required to support static const int objects (declared with constant initializers) in constant expressions, meaning that they have no choice but to treat such objects as compile-time constants. Additionally, as long as such objects remain odr-unused, they require no definition, which further demonstrates that they won't be used as run-time values.

Also, rules of constant initialization prevent local static const int objects from being initialized dynamically, meaning that there's no performance penalty for declaring such objects locally. Moreover, immunity of integral static objects to ordering problems of static initialization is a very important feature of the language.

constexpr is an extension and generalization of the concept that was originally implemented in C++ through const with a constant initializer. For integer types constexpr does not offer anything extra over what const already did. constexpr simply performs an early check of the "constness" of initializer. However, one might say that constexpr is a feature designed specifically for that purpose so it fits better stylistically.

What's the difference between constexpr and const?

Basic meaning and syntax

Both keywords can be used in the declaration of objects as well as functions. The basic difference when applied to objects is this:

const declares an object as constant. This implies a guarantee that once initialized, the value of that object won't change, and the compiler can make use of this fact for optimizations. It also helps prevent the programmer from writing code that modifies objects that were not meant to be modified after initialization.
constexpr declares an object as fit for use in what the Standard calls constant expressions. But note that constexpr is not the only way to do this.

When applied to functions the basic difference is this:

const can only be used for non-static member functions, not functions in general. It gives a guarantee that the member function does not modify any of the non-static data members (except for mutable data members, which can be modified anyway).
constexpr can be used with both member and non-member functions, as well as constructors. It declares the function fit for use in constant expressions. The compiler will only accept it if the function meets certain criteria (7.1.5/3,4), most importantly ^(†):
- The function body must be non-virtual and extremely simple: Apart from typedefs and static asserts, only a single return statement is allowed. In the case of a constructor, only an initialization list, typedefs, and static assert are allowed. (= default and = delete are allowed, too, though.)
- As of C++14, the rules are more relaxed, what is allowed since then inside a constexpr function: asm declaration, a goto statement, a statement with a label other than case and default, try-block, the definition of a variable of non-literal type, definition of a variable of static or thread storage duration, the definition of a variable for which no initialization is performed.
- The arguments and the return type must be literal types (i.e., generally speaking, very simple types, typically scalars or aggregates)

Constant expressions

As said above, constexpr declares both objects as well as functions as fit for use in constant expressions. A constant expression is more than merely constant:

It can be used in places that require compile-time evaluation, for example, template parameters and array-size specifiers:

  template<int N>
  class fixed_size_list
  { /*...*/ };

  fixed_size_list<X> mylist;  // X must be an integer constant expression

  int numbers[X];  // X must be an integer constant expression

But note:
Declaring something as constexpr does not necessarily guarantee that it will be evaluated at compile time. It can be used for such, but it can be used in other places that are evaluated at run-time, as well.
An object may be fit for use in constant expressions without being declared constexpr. Example:
```
     int main()
     {
       const int N = 3;
       int numbers[N] = {1, 2, 3};  // N is constant expression
     }
```
This is possible because N, being constant and initialized at declaration time with a literal, satisfies the criteria for a constant expression, even if it isn't declared constexpr.

So when do I actually have to use constexpr?

An object like N above can be used as constant expression without being declared constexpr. This is true for all objects that are:
const
of integral or enumeration type and
initialized at declaration time with an expression that is itself a constant expression

_{[This is due to §5.19/2: A constant expression must not include a subexpression that involves "an lvalue-to-rvalue modification unless […] a glvalue of integral or enumeration type […]" Thanks to Richard Smith for correcting my earlier claim that this was true for all literal types.]}

For a function to be fit for use in constant expressions, it must be explicitly declared constexpr; it is not sufficient for it merely to satisfy the criteria for constant-expression functions. Example:

 template<int N>
 class list
 { };

 constexpr int sqr1(int arg)
 { return arg * arg; }

 int sqr2(int arg)
 { return arg * arg; }

 int main()
 {
   const int X = 2;
   list<sqr1(X)> mylist1;  // OK: sqr1 is constexpr
   list<sqr2(X)> mylist2;  // wrong: sqr2 is not constexpr
 }

When can I / should I use both, const and constexpr together?

A. In object declarations. This is never necessary when both keywords refer to the same object to be declared. constexpr implies const.

constexpr const int N = 5;

is the same as

constexpr int N = 5;

However, note that there may be situations when the keywords each refer to different parts of the declaration:

static constexpr int N = 3;

int main()
{
  constexpr const int *NP = &N;
}

Here, NP is declared as an address constant-expression, i.e. a pointer that is itself a constant expression. (This is possible when the address is generated by applying the address operator to a static/global constant expression.) Here, both constexpr and const are required: constexpr always refers to the expression being declared (here NP), while const refers to int (it declares a pointer-to-const). Removing the const would render the expression illegal (because (a) a pointer to a non-const object cannot be a constant expression, and (b) &N is in-fact a pointer-to-constant).

B. In member function declarations. In C++11, constexpr implies const, while in C++14 and C++17 that is not the case. A member function declared under C++11 as

constexpr void f();

needs to be declared as

constexpr void f() const;

under C++14 in order to still be usable as a const function.

Which should I prefer for a constant within a function: constexpr const or enum?

You can accidentally or on purpose force ODR-existence of bar if it was a constexpr int bar = 456;, this is not possible with enum : int { bar = 456 };.

This may or may not be an advantage on either side.

For example

int baz(int const* ptr ) {
  if (ptr) return 7; return -1;
}
int foo(int x)
{
  // enum : int { bar = 456 };
  constexpr int bar = 456;
  return x + baz(&bar);
}

the enum version doesn't compile, the constexpr int one does. A constexpr int can be an lvalue, an enumerator (one of the listed enum constants) cannot.

The enum values aren't actually an int, while the constexpr int is actually an int. This may matter if you pass it to

template<class T>
void test(T) {
  static_assert(std::is_same<T,int>::value);
}

one will pass the test; the other will not.

Again, this could be an advantage, a disadvantage, or a meaningless quirk depending on how you are using the token.

When and why would you use static with constexpr?

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).

Difference between constexpr and static constexpr global variable

In your current example there is no difference: On variable declarations, constexpr implies const, and a const variable at namespace scope has internal linkage by default (so adding static does not change anything).

In C++14, you cannot declare a variable as constexpr and have it have external linkage unless you only ever do this in one single translation unit. The reason is that constexpr variables require an initializer, and a declaration with initializer is a definition, and you must only have a single definition.

However, what you can do is use a normal integral constant, which you can declare (not define) as extern, and in the translation unit where it is defined it can even be used as a constant expression:

lib.h:

extern const int a;

lib.cpp:

#include "lib.h"

const int a = 10;

int b[a] = {1, 2, 3};   // OK in this translation unit

In C++17, there is a new feature "inline variables" which lets you say:

inline constexpr int a = 10;

And this is an "inline definition" that can appear repeatedly, and each definition defines the same entity (just like all the other "inline" entities in the language).

What is the difference between a static const and constexpr variable?

The main difference that I know is, the value of constexpr must be known in compile-time while a const static can be assigned in run-time.

const static int x = rand();

Replacing constants: when to use static constexpr and inline constexpr?

In C++17, the proper way to replace those old idioms (e.g. #define) in headers in namespace scope is to use constexpr inline variables -- and not static (which is implied: they already have internal linkage).

While typically you won't encounter ODR issues (because integer compile-time constants such as those you describe are rarely ODR-used and there is a provision for their typical usage within inline functions), it is best to mark them as inline now that we have the feature in the language and avoid all problems.

See Should `const` and `constexpr` variables in headers be `inline` to prevent ODR violations? for the technical details about it.

Constexpr VS. Static Const: Which One to Prefer