constexpr 与静态 const:更喜欢哪一个? [英] constexpr vs. static const: Which one to prefer?

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问题描述

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?


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


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 << '
';
    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.

解决方案

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.

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