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成员模板变量专门化

[英]member template variable specializing

A class can contain a member template variable which must be static: 一个类可以包含一个必须是静态的成员模板变量:

class B
{   
    public:
        template <typename X>
            static X var;

        B() { std::cout << "Create B " << __PRETTY_FUNCTION__ << std::endl; }

        template <typename T>
        void Print() { std::cout << "Value is " << var<T> << std::endl; }
};

It must as all static members be declared outside the class scope: 它必须在类范围之外声明所有静态成员:

The following compiles and works as expected: 以下编译并按预期工作:

 template<typename T> T B::var=9; // makes only sense for int,float,double...

But how to specialize such a var like the following non working code ( error messages with gcc 6.1): 但是如何将这样的var专门化为以下非工作代码(使用gcc 6.1的错误消息):

template <> double B::var<double>=1.123; 

Fails with: 失败:

main.cpp:49:23: error: parse error in template argument list
 template <> double B::var<double>= 1.123;
                       ^~~~~~~~~~~~~~~~~~
main.cpp:49:23: error: template argument 1 is invalid
main.cpp:49:23: error: template-id 'var<<expression error> >' for 'B::var' does not match any template declaration
main.cpp:38:22: note: candidate is: template<class X> T B::var<T>
             static X var;

template <> double B::var=1.123;

Fails with 失败了

   template <> double B::var=1.123;
                       ^~~
main.cpp:38:22: note: does not match member template declaration here
             static X var;

What is the correct syntax here? 这里的语法是什么?

I suppose you should add a space 我想你应该添加一个空格

template <> double B::var<double> = 1.123;
                                 ^ here

Otherwise (if I'm not wrong) >=1.123 is confused with "equal or greather than 1.123" 否则(如果我没错) >=1.123与“等于或大于1.123”相混淆

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