“未定义的引用"到从静态方法访问的模板类的静态成员

quot;undefined referencequot; to static member of template class accessed from static method(“未定义的引用到从静态方法访问的模板类的静态成员)

问题描述

我有一个静态类方法，需要访问指针 MyTypePointer，因此必须将其声明为静态.由于它是一个模板类，我必须将方法放在头文件中，但我无法在头文件中定义 MyTypePointer.

I have a static class method that needs access to a pointer MyTypePointer that therefore has to be declared static. Since it is a template class I have to put the methods inside the header file but I can't define MyTypePointer in the header.

因此我收到 "undefined Reference" 错误，因为 MyTypePointer 未声明.我怎样才能使这项工作/声明 MyTypePointer.

So I get the "undefined Reference" error since MyTypePointer is not declared. How can I make this work / declare MyTypePointer.

myclass.h

template <typename A, typename B>
class PathfindingClass {
    typedef std::vector<GenericEdgeClass<A, B>*> MyType;
    static MyType *MyTypePointer;
};

template <typename A, B>
void MyClass<A, B>::MyMethod(int x, int y) { 
    //do something with MyTypePointer
}

非常感谢.

推荐答案

这是一个迟到的完整参考答案，因为这个问题被链接为另一个问题的参考.

It is a late answer for full reference, as this question is linked as a reference to another question.

静态字段声明但未定义的最小损坏示例可能是:

A minimal broken example of a static field declared but not defined could be :

template<typename T>
class A
{
public:
    static T val;
    static void init()
    {
        val=0;
    }
};

int main()
{
    // A::init();
    A<double>::init();
    return 0;
}

修复只是在类定义之后定义静态字段:

The fix is just to define the static field after the class definition :

template<typename T>
class A
{
public:
    static T val;
    static void init()
    {
        val=0;
    }
};

template<typename T> T A<T>::val; // only change here

int main()
{
    // A::init();
    A<double>::init();
    return 0;
}

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