函数组合在C ++ / C ++ 11 [英] function composition in C++ / C++11

问题描述

我目前正在编写一些需要大量函数组合的C ++ 11中的加密算法。有两种类型的组成我必须处理：

I am currently coding some cryptographic algorithms in C++11 that require a lot of function compositions. There are 2 types of composition I have to deal with :

1. 在自己上构造一个函数可变次数。在数学上，对于某个函数F，F ^ n（x）=（F ^ {n-1} o F）（x）= F ^ {n-1}（F（x））。

1. Compose a function on itself a variable number of times. Mathematically, for a certain function F, F^n(x) = (F^{n-1} o F)(x) = F^{n-1}(F(x)).

将不同的功能组合在一起。例如，对于一些相同类型的函数f，g，h，i，j和k，我将有f（g（h（i（j（k（x））））））。

Compose different functions together. For example, for some functions f,g,h,i,j and k of the same type, I'll have f(g(h(i(j(k(x)))))).

在我的例子中，我使用以下F的定义：

In my case, I'm using the following definition of F :

const std::vector<uint8_t> F(const std::vector<uint8_t> &x);

我想自己组合这个函数n次。我已经实现了一个简单的递归方式，这是工作正常的组合：

I would like to compose this function on itself n times. I have implemented the composition in a simple recursive way which is working fine :

const std::vector<uint8_t> compose(const uint8_t n, const std::vector<uint8_t> &x)
{
    if(n > 1)
       return compose(n-1, F(x));

    return F(x);
}

对于这种情况，是否有更有效的方法组合使用c ++ 11但不使用BOOST ？
当然可以使用这个表单：

For this case, is there a more efficient way an proper way to implement this composition using c++11 but without using BOOST ? It would be great to use this form if it is possible of course :

answer = compose<4>(F)(x); // Same as 'answer = F^4(x) = F(F(F(F(x))))'

对于第二种情况，我想实现可变数量的函数的组合。对于具有与F相同定义的给定函数集合F0，F1，...，Fn，是否存在有效和适当的方法来组合它们，其中n是变量？
我认为variadic模板将在这里有用，但我不知道如何使用它们。

For the second case, I would like to implement the composition of a variable number of functions. For a given set of functions F0, F1, ..., Fn having the same definition as F, is there an efficient and proper way to compose them where n is variable ? I think variadic template would be useful here, but I don't know how to use them in that case.

感谢您的帮助。

推荐答案

（未测试）：

template <typename F>
class Composer {
  int n_;
  F f_;
public:
  Composer(int n, F f) : n_(n), f_(f) {}

  template <typename T>
  T operator()(T x) const {
    int n = n_;
    while (n--) {
      x = f_(x);
    }
    return x;
  }
};

template <int N, typename F>
Composer<F> compose(F f) {
  return Composer<F>(N, f);
}

EDIT：这次测试）：

And for the second case (tested this time):

#include <iostream>

template <typename F0, typename... F>
class Composer2 {
    F0 f0_;
    Composer2<F...> tail_;
public:
    Composer2(F0 f0, F... f) : f0_(f0), tail_(f...) {}

    template <typename T>
    T operator() (const T& x) const {
        return f0_(tail_(x));
    }
};

template <typename F>
class Composer2<F> {
    F f_;
public:
    Composer2(F f) : f_(f) {}

    template <typename T>
    T operator() (const T& x) const {
        return f_(x);
    }
};

template <typename... F>
Composer2<F...> compose2(F... f) {
    return Composer2<F...>(f...);
}

int f(int x) { return x + 1; }
int g(int x) { return x * 2; }
int h(int x) { return x - 1; }

int main() {
  std::cout << compose2(f, g, h)(42);
  return 0;
}

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