如何对多种类型进行显式专业化? [英] How to do one explicit specialization for multiple types?

问题描述

考虑如下所示的模板函数，如何针对多种类型显式专门化一个版本的函数:

Considering a template function like below how is it possible to do explicitly specialize one version of function for multiple types:

template <typename T>
void doSomething(){
 //whatever
}

目的是拥有一个专业而不是多个专业，因为//相同的东西:

The intention is to have one specialization instead of multiple following ones because //something is the same:

void doSomething<int>(){
 //something
}
void doSomething<float>(){
 //something
}
void doSomething<double>(){
 //something
}

有什么方法可以实现一种专业化?

any method to achieve one specialization?

推荐答案

您无法进行模板功能的特殊化.但是您可以在帮助程序类中委派实现，可以从您的函数中使用它.一些基本代码:

You can't make template function specialization. But you could delegate the implementation in a helper class, that can be used from your function. Some skeleton code:

实施模板类并对其进行专门化:

Implement a template class and specialize it:

template< typename T, bool isArithmetic>
struct Something { void operator()() { ... } };

template< typename T, true>
struct Something { void operator()() { ... do something specialized for arithmetic types; } }

然后在模板函数中使用它:

Then use it in the template function:

template< typename T>
void myFunction()
{
   Something<T, IsArithmetic<T>::value>()();
}

IsArithmetic是提供有关类型T(选择器)的信息的类.例如，您可以在boost库中找到此类类型信息.

Where IsArithmetic is a class that provides the information about type T (selector). You can find such type info in boost libraries, for example.

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