C ++模板，静态函数专门化 [英] C++ template, static function specialization

问题描述

我的模板出现语法错误

我想对模板类的静态函数进行部分专业化处理

I would like to partial specialize a static function of my template class

template <typename Foo, size_t bar = 26>
class MyClass
{
    MyClass();
    static void function();
};

#include "class.tpp"

class.tpp

template <typename Foo, bar>
MyClass<Foo, bar>::MyClass()
{ }

template <typename Foo>
inline
void
MyClass<Foo, 6>::function()
{
    // ...
}

template <typename Foo>
inline
void
MyClass<Foo, 26>::function()
{
    // ...
}

error: template definition of non-template

我只想为bar == 26和bar == 6实施MyClass<Foo, bar>::function

I just want to implement MyClass<Foo, bar>::function for bar == 26 and bar == 6

如何正确执行此操作? 谢谢

How to do that properly ? Thanks

推荐答案

该函数本身不是模板，它仅位于类模板中.您可以为这些情况专门设置类，但不能专门提供函数本身.

The function is not a template itself, it is only inside a class template. You can specialize the class for those cases, but not the function itself.

template <class Foo>
class MyClass<Foo, 26>
{
    static void function() { ... }
};

假设您已经像这样专门化了类，则只能在类内部声明函数，然后在外部进行定义，如下所示:

Provided you have specialized the class like so, you can only declare the function inside the class, and define it outside like so:

template <class Foo>
void MyClass<Foo, 26>::function() { ... }

如果您事先不专门研究它，则会因使用不完整的类型而出现编译错误.

If you don't specialize it beforehand, you'll get a compilation error for using an incomplete type.

您可能还会发现此问题和有关专门化单个功能的问题的答案.在相关的类模板中.

You might also find this question and answer on specializing a single function inside a class template relevant.

这篇关于C ++模板，静态函数专门化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！