C++ 类模板中的静态成员初始化

Static member initialization in a class template

提问by Alexandre C.

I'd like to do this:

我想这样做：

template <typename T>
struct S
{
    ...
    static double something_relevant = 1.5;
};

but I can't since something_relevantis not of integral type. It doesn't depend on T, but existing code depends on it being a static member of S.

但我不能，因为something_relevant不是整数类型。它不依赖于T，但现有代码依赖于它是S.

Since S is template, I cannot put the definition inside a compiled file. How do I solve this problem ?

由于 S 是模板，我不能将定义放在编译文件中。我该如何解决这个问题？

回答by sbi

Just define it in the header:

只需在标题中定义它：

template <typename T>
struct S
{
    static double something_relevant;
};

template <typename T>
double S<T>::something_relevant = 1.5;

Since it is part of a template, as with all templates the compiler will make sure it's only defined once.

由于它是模板的一部分，与所有模板一样，编译器将确保它只定义一次。

回答by xaxxon

Since C++17, you can now declare the static member to be inline, which will define the variable in the class definition:

从 C++17 开始，您现在可以将静态成员声明为inline，这将在类定义中定义变量：

template <typename T>
struct S
{
    ...
    static inline double something_relevant = 1.5;
};

live: https://godbolt.org/g/bgSw1u

直播：https: //godbolt.org/g/bgSw1u

回答by Prasoon Saurav

This will work

这将工作

template <typename T>
 struct S
 {

     static double something_relevant;
 };

 template<typename T>
 double S<T>::something_relevant=1.5;