Question

是否可以声明变量extern constexpr并在另一个文件中定义它？

我试了但是编译器给出了错误：

声明constexpr变量＆＃39; i＆＃39;不是定义

在.h：

extern constexpr int i;

in .cpp：

constexpr int i = 10;

Answer 1

不，你不能这样做，这是标准所说的内容（第7.1.5节）：

1 constexpr说明符仅适用于a的定义变量或变量模板，函数的声明或函数模板，或者静态数据成员的声明字面类型（3.9）。如果有任何函数声明，函数模板或变量模板有一个constexpr说明符，然后是它的全部声明应包含constexpr说明符。 [注：明确专业化可能与模板声明有所不同到constexpr说明符。无法声明函数参数 constexpr。 - 结束说明]

标准给出的一些例子：

  constexpr void square(int &x);  // OK: declaration
  constexpr int bufsz = 1024;  // OK: definition
  constexpr struct pixel {  // error: pixel is a type
    int x;
    int y;
    constexpr pixel(int);  // OK: declaration
  };

  extern constexpr int memsz; // error: not a definition

Answer 2

没有。外部constexpr没有任何意义。请阅读http://en.cppreference.com/w/cpp/language/constexpr

即。 “它必须立即构建或赋值。”

Answer 3

C ++ 17 inline变量

这项令人敬畏的C ++ 17功能使我们能够：

为每个常量方便地只使用一个内存地址
将其存储为constexpr
从一个标题中一行完成

main.cpp

#include <cassert>

#include "notmain.hpp"

int main() {
    // Both files see the same memory address.
    assert(&notmain_i == notmain_func());
    assert(notmain_i == 42);
}

notmain.hpp

#ifndef NOTMAIN_HPP
#define NOTMAIN_HPP

inline constexpr int notmain_i = 42;

const int* notmain_func();

#endif

notmain.cpp

#include "notmain.hpp"

const int* notmain_func() {
    return &notmain_i;
}

编译并运行：

g++ -c -o notmain.o -std=c++17 -Wall -Wextra -pedantic notmain.cpp
g++ -c -o main.o -std=c++17 -Wall -Wextra -pedantic main.cpp
g++ -o main -std=c++17 -Wall -Wextra -pedantic main.o notmain.o
./main

GitHub upstream。

C ++标准保证地址相同。 C++17 N4659 standard draft 10.1.6“内联说明符”：

6具有外部链接的内联函数或变量在所有翻译单元中应具有相同的地址。

cppreference https://en.cppreference.com/w/cpp/language/inline解释说，如果未提供static，则它具有外部链接。

另请参阅：How do inline variables work?

Answer 4

I agree with 'swang' above, but there is a consequence. Consider:

ExternHeader.hpp

extern int e; // Must be extern and defined in .cpp otherwise it is a duplicate symbol.

ExternHeader.cpp

#include "ExternHeader.hpp"
int e = 0;

ConstexprHeader.hpp

int constexpr c = 0; // Must be defined in header since constexpr must be initialized.

Include1.hpp

void print1();

Include1.cpp

#include "Include1.hpp"
#include "ExternHeader.hpp"
#include "ConstexprHeader.hpp"
#include <iostream>

void print1() {
    std::cout << "1: extern = " << &e << ", constexpr = " << &c << "\n";
}

Include2.hpp

void print2();

Include2.cpp

#include "Include2.hpp"
#include "ExternHeader.hpp"
#include "ConstexprHeader.hpp"
#include <iostream>

void print2() {
    std::cout << "2: extern = " << &e << ", constexpr = " << &c << "\n";
}

main.cpp

#include <iostream>
#include "Include1.hpp"
#include "Include2.hpp"

int main(int argc, const char * argv[]) {
    print1();
    print2();
    return 0;
}

Which prints:

1: extern = 0x1000020a8, constexpr = 0x100001ed0
2: extern = 0x1000020a8, constexpr = 0x100001ed4

IE the constexpr is allocated twice whereas the extern is allocated once. This is counterintuitive to me, since I 'expect' constexpr to be more optimized than extern.

Edit: const and constexpr have the same behaviour, with regard to allocation, therefore from that point of view the behaviour is as expected. Though, as I said, I was surprised when I came across the behaviour of constexpr.

Answer 5

是的有点是......

//===================================================================
// afile.h

#ifndef AFILE
#define AFILE

#include <cstddef>
#include <iostream>

enum class IDs {

  id1,
  id2,
  id3,
  END

};

// This is the extern declaration of a **constexpr**, use simply **const**
extern const int ids[std::size_t(IDs::END)];

// These functions will demonstrate its usage

template<int id> void Foo() { std::cout << "I am " << id << std::endl; }

extern void Bar();

#endif // AFILE

//===================================================================
// afile.cpp

#include "afile.h"

// Here we define the consexpr. 
// It is **constexpr** in this unit and **const** in all other units
constexpr int ids[std::size_t(IDs::END)] = {

  int(IDs::id1),
  int(IDs::id2),
  int(IDs::id3)

};

// The Bar function demonstrates that ids is really constexpr
void Bar() {

  Foo<ids[0]      >();
  Foo<ids[1] + 123>();
  Foo<ids[2] / 2  >();

}

//===================================================================
// bfile.h

#ifndef BFILE
#define BFILE

// These functions will demonstrate usage of constexpr ids in an extern unit

extern void Baz();
extern void Qux();


#endif // BFILE

//===================================================================
// bfile.cpp

#include "afile.h"

// Baz demonstrates that ids is (or works as) an extern field
void Baz() {

  for (int i: ids) std::cout << i << ", ";
  std::cout << std::endl;

}

// Qux demonstrates that extern ids cannot work as constexpr, though
void Qux() {

#if 0 // changing me to non-0 gives you a compile-time error...

  Foo<ids[0]>();

#endif

  std::cout << "Qux: 'I don't see ids as consexpr, indeed.'" 
            << std::endl;

}

//===================================================================
// main.cpp

#include "afile.h"
#include "bfile.h"

int main(int , char **)
{

  Bar();
  Baz();
  Qux();

  return 0;
}

Answer 6

您可能想要的是extern和constexpr初始化，例如：

// in header
extern const int g_n;

// in cpp
constexpr int g_n = 2;

这是支持，尽管在Visual Studio 2017中仅通过一致性模式：

如何申报constexpr extern？

6 个答案: