静态数据成员的定义 [英] Definition of the static data member

问题描述

我正在阅读 Scott Meyers 的 C++ 并遇到了这个例子:

class GamePlayer{私人的:静态常量 int NumTurns = 5;int score[NumTurns];//...};

<块引用>

您在上面看到的是 NumTurns 的声明，而不是定义.

为什么不是定义?看起来我们用 5 来初始化静态数据成员.

我只是不明白声明但未定义一个值为5的变量是什么意思.我们可以很好地取变量的地址.

A 类{民众:void foo(){ const int * p = &a;}私人的:静态常量 int a = 1;};int主(){一个;a.foo();}

演示

解决方案

因为它不是定义.静态数据成员必须在类定义之外定义.

[class.static.data]/2

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static 数据成员在其类定义中的声明不是一个定义并且可能是 cv 限定的不完整类型无效.static 数据成员的定义应出现在包含成员类定义的命名空间范围.

至于在没有实际定义的情况下获取静态成员的地址，它会编译，但不应该链接.

I'm reading Scott Meyers' C++ and come across this example:

class GamePlayer{
private:
    static const int NumTurns = 5;
    int scores[NumTurns];
    // ...
};

What you see above is a declaration for NumTurns, not a definition.

Why not a definition? It looks like we initialize the static data member with 5.

I just don't understand what it means to declare but not define a variable with the value 5. We can take the address of the variable fine.

class A
{
public:
    void foo(){ const int * p = &a; }
private:
    static const int a = 1;
};

int main ()
{
    A a;
    a.foo();
}

DEMO

解决方案

Because it isn't a definition. Static data members must be defined outside the class definition.

[class.static.data] / 2

The declaration of a static data member in its class definition is not a definition and may be of an incomplete type other than cv-qualified void. The definition for a static data member shall appear in a namespace scope enclosing the member’s class definition.

As for taking the address of your static member without actually defining it, it will compile, but it shouldn't link.

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