静态元组类成员的constexpr具有链接器错误 [英] constexpr of static tuple class member has linker error

问题描述

我有以下代码：

#include <iostream>
#include <tuple>

class T
{
    public:
        using Names = std::tuple<char const*, char const*>;
        static constexpr Names names {"First", "Second"};
};

int main()
{
    std::cout << std::get<0>(T::names);
}

由于名称为我希望 constexpr 可以工作。但是我得到一个链接器错误：

As names is a constexpr I expected this to work. But I get a linker error:

编译器：

> g++ --version
Configured with: --prefix=/Applications/Xcode.app/Contents/Developer/usr --with-gxx-include-dir=/usr/include/c++/4.2.1
Apple LLVM version 6.0 (clang-600.0.56) (based on LLVM 3.5svn)
Target: x86_64-apple-darwin14.0.0
Thread model: posix

错误：

> g++ -std=c++1y pl.cpp
Undefined symbols for architecture x86_64:
  "T::names", referenced from:
      _main in pl-377031.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

[实时演示]

推荐答案

在类中声明 static 数据成员绝不是定义 ¹ 。

每当变量被odr-used ² 使用时，都需要一个定义。
std :: get<> 接受每个引用的参数，并将变量绑定到引用odr-立即使用它 ³ 。

A declaration of a static data member in class is never a definition¹.
A definition is necessary whenever a variable is odr-used². std::get<> takes arguments per reference, and binding a variable to a reference odr-uses it immediately³.

只需在外部定义名称：

constexpr T::Names T::names; // Edit: This goes *outside* the class "as is"!

演示 。

^1） [ basic.def] / 2：

¹⁾ [basic.def]/2:

声明是一个定义，除非[..]声明了 static 数据
类定义中的成员（9.2，9.4）

A declaration is a definition unless [..] it declares a static data member in a class definition (9.2, 9.4)

^2） [basic.def.odr] / 4：

²⁾ [basic.def.odr]/4:

每个程序都应仅包含每个非内联$ b $的一个定义b在该程序中使用的函数或变量；不需要诊断
。

Every program shall contain exactly one definition of every non-inline function or variable that is odr-used in that program; no diagnostic required.

^3）根据[basic.def.odr] / 3：

³⁾ According to [basic.def.odr]/3:

变量 x 的名称显示为可能值的表达式
ex 由 ex 使用，除非应用左值到右值转换
（ 4.1）到 x 会产生一个常量表达式（5.19），该表达式不会调用
任何非平凡函数，如果 x 是对象， ex 是
的元素，是表达式 e ，其中将
左值到右值转换（4.1）应用于 e 或 e 是
的舍弃值表达式（第5条）。

A variable x whose name appears as a potentially-evaluated expression ex is odr-used by ex unless applying the lvalue-to-rvalue conversion (4.1) to x yields a constant expression (5.19) that does not invoke any non-trivial functions and, if x is an object, ex is an element of the set of potential results of an expression e, where either the lvalue-to-rvalue conversion (4.1) is applied to e, or e is a discarded-value expression (Clause 5).

此处的id表达式 T :: names 指的是有问题的变量。包含 T :: names 的所有可能结果的唯一超表达式 e 是 T： ：names 本身，因为函数调用的可能结果集，即 std :: get< 0>（T :: names），是空的。但是，显然不应用从左值到右值的转换，并且显然也不会丢弃 T :: names 的值（因为将其传递给函数）。

因此它很常用，需要定义。

Here the id-expression T::names refers to the variable in question. The only superexpression e that contains all the potential results of T::names is T::names itself, because the set of potential results of a function call, i.e. std::get<0>(T::names), is empty. However, the lvalue-to-rvalue conversion is clearly not applied, and the value of T::names is also clearly not discarded (as it is passed to a function).
Thus it is odr-used and requires a definition.

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