将整数或类型作为模板参数传递? [英] Passing an integer or a type as a template parameter?

问题描述

这是我正在尝试做的一个示例案例(这是一个测试"案例，只是为了说明问题):

Here is an example case of what I'm trying to do (it is a "test" case just to illustrate the problem) :

#include <iostream>
#include <type_traits>
#include <ratio>

template<int Int, typename Type> 
constexpr Type f(const Type x)
{
    return Int*x;
}

template<class Ratio, typename Type, 
         class = typename std::enable_if<Ratio::den != 0>::type>
constexpr Type f(const Type x)
{
    return (x*Ratio::num)/Ratio::den;
}

template</*An int OR a type*/ Something, typename Type>
constexpr Type g(const Type x)
{
    return f<Something, Type>(x);
}

int main()
{
    std::cout<<f<1>(42.)<<std::endl;
    std::cout<<f<std::kilo>(42.)<<std::endl;
}

如您所见，f() 函数有两个版本:第一个版本采用 int 作为模板参数，第二个采用一个std::ratio.问题如下:

As you can see, there are two versions of the f() function : the first one takes an int as a template parameter, and the second one takes a std::ratio. The problem is the following :

我想通过 g() 来包装"这个函数，它可以将 int 或 std::ratio 作为第一个模板参数并调用 f() 的好版本.

I would like to "wrap" this function through g() which can take an int OR a std::ratio as first template parameter and call the good version of f().

如何在不编写两个 g() 函数的情况下做到这一点?换句话说，我必须写什么来代替 /*An int OR a type*/ ?

How to do that without writing two g() functions ? In other words, what do I have to write instead of /*An int OR a type*/ ?

推荐答案

这是我的方法，但我对你的界面做了一些改动:

Here's how I would do it, but I've changed your interface slightly:

#include <iostream>
#include <type_traits>
#include <ratio>

template <typename Type>
constexpr
Type
f(int Int, Type x)
{
    return Int*x;
}

template <std::intmax_t N, std::intmax_t D, typename Type>
constexpr
Type
f(std::ratio<N, D> r, Type x)
{
    // Note use of r.num and r.den instead of N and D leads to
    //   less probability of overflow.  For example if N == 8 
    //   and D == 12, then r.num == 2 and r.den == 3 because
    //   ratio reduces the fraction to lowest terms.
    return x*r.num/r.den;
}

template <class T, class U>
constexpr
typename std::remove_reference<U>::type
g(T&& t, U&& u)
{
    return f(static_cast<T&&>(t), static_cast<U&&>(u));
}

int main()
{
    constexpr auto h = g(1, 42.);
    constexpr auto i = g(std::kilo(), 42.);
    std::cout<< h << std::endl;
    std::cout<< i << std::endl;
}

42
42000

注意事项:

1. 我已经利用 constexpr 不 通过模板参数传递编译时常量(这就是 constexpr).

1. I've taken advantage of constexpr to not pass compile-time constants via template parameters (that's what constexpr is for).

g 现在只是一个完美的转发器.但是我无法使用 std::forward 因为它没有用 constexpr 标记(可以说是 C++11 中的一个缺陷).所以我选择使用 static_cast 代替.完美转发在这里有点矫枉过正.但要彻底熟悉它是一个很好的习语.

g is now just a perfect forwarder. However I was unable to use std::forward because it isn't marked up with constexpr (arguably a defect in C++11). So I dropped down to use static_cast<T&&> instead. Perfect forwarding is a little bit overkill here. But it is a good idiom to be thoroughly familiar with.

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