C++中模板类可以有静态成员吗

Question

C++中的模板类可以有静态成员吗？既然它不存在并且在使用之前是不完整的，那么这可能吗？

Answer 1

是的。静态成员在

template< … > class { … }

块内声明或定义。如果声明但未定义，则必须有另一个声明提供该成员的定义。

template< typename T >
class has_static {
    // inline method definition: provides the body of the function.
    static void meh() {}

    // method declaration: definition with the body must appear later
    static void fuh();

    // definition of a data member (i.e., declaration with initializer)
    // only allowed for const integral members
    static int const guh = 3;

    // declaration of data member, definition must appear later,
    // even if there is no initializer.
    static float pud;
};

// provide definitions for items not defined in class{}
// these still go in the header file

// this function is also inline, because it is a template
template< typename T >
void has_static<T>::fuh() {}

/* The only way to templatize a (non-function) object is to make it a static
   data member of a class. This declaration takes the form of a template yet
   defines a global variable, which is a bit special. */
template< typename T >
float has_static<T>::pud = 1.5f; // initializer is optional

为模板的每个参数化创建一个单独的静态成员。不可能在模板生成的“所有”类之间共享单个成员。为此，您必须在模板之外定义另一个对象。部分专业化的特质类可能会有所帮助。