初始化模板化类的静态成员 [英] Initializing static members of a templated class

问题描述

我试图弄清楚为什么这个例子不能编译.我的理解是，如果未显式设置静态变量，则其默认值为0.在下面的五个示例中，其中四个示例的行为与我期望的一样，但是被注释掉的示例将无法编译.

I'm trying to figure out why this example doesn't compile. My understanding is that if a static variable is not explicitly set then it defaults to 0. In the five examples below four of them behave as I would expect, but the one that's commented out won't compile.

#include <iostream>
class Foo
{
public:
    static int i;
    static int j;
};
template <int n>
class Bar
{
public:
    Bar(int) { }
    static int i;
}; 

static int i;
int Foo::i;
int Foo::j = 1;
template <> int Bar<2>::i;
template <> int Bar<3>::i = 3;

int main(int argc, char** argv)
{
    std::cout << "i         " << i << std::endl;
    std::cout << "Foo::i    " << Foo::i << std::endl;
    std::cout << "Foo::j    " << Foo::j << std::endl;
    //std::cout << "Bar<2>::i " << Bar<2>::i << std::endl; // Doesn't compile?
    std::cout << "Bar<3>::i " << Bar<3>::i << std::endl;
    return 0;
}

int Bar<2>::i为什么不做与int Foo::i或static int i相同的事情?

Why doesn't int Bar<2>::i do the same thing as int Foo::i or static int i?

我忘记将模板<>添加到Bar< 2>和Bar< 3>声明中. (虽然不能解决问题，但仍然会出现链接器错误)

I had forgotten to add template<> to the Bar<2> and Bar<3> declarations. (doesn't solve the problem though, still getting linker errors)

推荐答案

在当前C ++标准的规则下，特化template <> int Bar<2>::i;只是声明，而不是定义. 要成为定义，必须指定一个初始化程序. (请参阅第14.7.3/15条)

Under the rules of the current C++ standard, the specialisation template <> int Bar<2>::i; is only a declaration and never a definition. To become a definition, you must specify an initialiser. (See clause 14.7.3/15)

除此之外，您还缺少一种非常常见的情况:模板的非专业静态成员的定义:

Apart from that, you were missing one very common case: the definition of a non-specialised static member of a template:

template <int n> int Bar<n>::i;

这为N不等于2或3的Bar<N>::i提供了定义.

This provides a definition for Bar<N>::i for N not equal to 2 or 3.

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