constexpr vs. static const: Which one to prefer?

问题:

For defining compile-time constants of integral types like the following (at function and class scope), which syntax is best?

static const int kMagic = 64; // (1)
constexpr int kMagic = 64;    // (2)

(1) works also for C++98/03 compilers, instead (2) requires at least C++11. Are there any other differences between the two? Should one or the other be preferred in modern C++ code, and why?

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EDIT

I tried this sample code with Godbolt's CE:

int main()
{
#define USE_STATIC_CONST
#ifdef USE_STATIC_CONST
  static const int kOk = 0;
  static const int kError = 1;
#else
  constexpr int kOk = 0;
  constexpr int kError = 1;
#endif
  return kOk;
}

and for the static const case this is the generated assembly by GCC 6.2:

main::kOk:
        .zero   4
main::kError:
        .long   1
main:
        push    rbp
        mov     rbp, rsp
        mov     eax, 0
        pop     rbp
        ret

On the other hand, for constexpr it's:

main:
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], 0
        mov     DWORD PTR [rbp-8], 1
        mov     eax, 0
        pop     rbp
        ret

Although at -O3 in both cases I get the same (optimized) assembly:

main:
        xor     eax, eax
        ret

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EDIT #2

I tried this simple code (live on Ideone):

#include <iostream>
using namespace std;

int main() {
    const int k1 = 10;
    constexpr int k2 = 2*k1;
    cout << k2 << '\n';
    return 0;
}

which shows that const int k1 is evaluated at compile-time, as it's used to calculate constexpr int k2.

However, there seems to be a different behavior for doubles. I've created a separate question for that here.

回答1:

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.

回答2:

constexpr variable is guaranteed to have a value available at compile time. whereas static const members or const variable could either mean a compile time value or a runtime value. Typing constexpr express your intent of a compile time value in a much more explicit way than const.

One more thing, in C++17, constexpr static data member variables will be inline too. That means you can omit the out of line definition of static constexpr variables, but not static const.

---

As a demand in the comment section, here's a more detailed explanation about static const in function scope.

A static const variable at function scope is pretty much the same, but instead of having a automatic storage duration, it has static storage duration. That mean it's in some way the equivalent of declaring the variable as global, but only accessible in the function.

It is true that a static variable is initialize at the first call of the function, but since it's const too, the compiler will try to inline the value and optimize out the variable completely. So in a function, if the value is known at compile time for this particular variable, then the compiler will most likely optimize it out.

However, if the value isn't known at compile time for a static const at function scope, it might silently make your function (a very small bit) slower, since it has to initialize the value at runtime the first time the function is called. Plus, it has to check if the value is initialized each time the function is called.

That's the advantage of a constexpr variable. If the value isn't known at compile time, it's a compilation error, not a slower function. Then if you have no way of determine the value of your variable at compile time, then the compiler will tell you about it and you can do something about it.