C ++使用继承来调整算法 [英] C++ Using inheritance to tweak an algorithm

问题描述

由于Dijkstra的算法和Prim的算法是如此相似，我想创建一个基本的算法类(我称其为"Greedy"之类)，然后我想从Greedy继承并基于课.

Since Dijkstra's algorithm and Prim's algorithm are so similar, I'd like to make a base algorithm class (I'll call it "Greedy" or something) and then I want to inherit from Greedy and tweak the algorithm based on the class.

我认为可以归结为这一点.我想重用算法的很大一部分，但要调整一两个操作.

I think it boils down to this. I'd like to reuse a large portion of an algorithm but tweak an operation or two.

class BaseAlg
{
public:
    BaseAlg(std::vector<int> data)  //constructor sums a vector and stores result
    {
        int accum = 0;
        for (unsigned int i = 0; i < data.size(); ++i)
            accum += data[i];
        result = accum;
    }

protected:
    int result;
};

class Alg1  //A second, similar algorithm
{
public:
    Alg1(std::vector<int> data)
    {
        //I want to reuse BaseAlg except use *= instead of +=; 
    }
};

因此，执行此操作的一种方法是仅拥有一个类(BaseAlg)，并向该类的构造函数添加选择器"值.我将打开该选择器值，并在不同情况下执行+ =或* =.我觉得应该有一种通过继承来实现这种重用的方法，因为贪婪与Prim和贪婪与Dijkstra之间存在一种是"关系.但我不太清楚.有什么想法吗?

So one way to do this is to only have one class (BaseAlg) and to add a "Selector" value to the class's constructor. I would switch on that selector value and execute += or *= in different cases. I feel like there should be a way to implement this reuse with inheritance because there is an "is a" relationship between Greedy and Prim and Greedy and Dijkstra. But I can't quite figure it out. Any thoughts?

推荐答案

对于这种重用但也不是真正的继承"，您应该使用模板.

You should use templates for this kind of "reuse but nor really inheritance".

例如，在您的情况下，基本上可以归结为:

For instance, in your case, basically it boils down to this:

template<class Op, class Iter, class T> T reduce(const Op & op, Iter begin, Iter end, T init = T()) {
    T accum = init;
    for(Iter i = begin; i != end; ++i)
        accum = Op(accum, *i);
    return accum;
}

然后您可以像这样使用它:

You could then use it like this:

std::vector<int> data; // fill data
int sum = reduce(add<int>, data.begin(), data.end());
int prod = reduce(mul<int>, data.begin(), data.end(), 1);

您需要在其中定义 add 和 mult 的位置，

where you'd need to define add and mult like this:

template<class T> T add(T a, T b) { return a + b; }
template<class T> T mult(T a, T b) { return a * b; }

现在，这一切都是出于说明目的，正如Jerry Coffin所指出的那样，使用STL，您可以轻松做到:

Now, this was all for illustrative purpose, as Jerry Coffin pointed out, with the STL you can simply do:

#include <functional>
#include <numeric>
int sum = std::accumulate(data.begin(), data.end(), 0, std::plus<int>);

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