如果两个过载功能相似，如何做出更好的设计 [英] How to make a better design if two overload functions are similar

问题描述

我正在开发一个c ++项目，我有一个例子如下：

I'm working on a c++ project and I have a case as below:

void func(int a, int b)
{
    if (some_bool)
    {
        generateString(generateFunc()); // different
    }
    for (auto it : myVector)
    {
        // do something
        for (int i = 0; i < a + b; ++i)
        {
            if (some_bool2)
            {
                myData = generateInt();
            }
            else
            {
                myData2 = generateInt2(); // different
            }
        }
    }
}


void func(int a, int b, string str)
{
    if (some_bool)
    {
        generateString(generateFunc(str));  // different
    }
    for (auto it : myVector)
    {
        // do something
        for (int i = 0; i < a + b; ++i)
        {
            if (some_bool2)
            {
                myData = generateInt();
            }
            else
            {
                myData2 = convertStrToInt(str);  //different
            }
        }
    }
}

如你所见，我有两个重载功能。

他们的逻辑结构是一样的，但是一些细节不是。

As you see, I have two overloading functions.
Their logical structures are the same but some details are not.

我考虑到是否有一些可以合并它们的技术，以便我可以有更好的设计。因为现在我必须改变两次，如果我需要做一些改变。

I'm considering if there is some technique which can merge them so that I can have a better design. Because for now I have to change two times if I need to do some change.

推荐答案

你可以使用模板，做一些像这个：

You can use templates and do something like this:

template<typename F, typename... A>
void func(int a, int b, F &&f, A&&... a)
{
    if (some_bool)
    {
        generateString(generateFunc(std::forward<A>(a)...));  // different
    }
    for (auto it : myVector)
    {
        // do something
        for (int i = 0; i < a + b; ++i)
        {
            if (some_bool2)
            {
                myData = generateInt();
            }
            else
            {
                myData2 = std::forward<F>(f)(std::forward<A>(a)...);  //different
            }
        }
    }
}

然后调用您的函数，如下所示：

Then call your function as it follows:

func(0, 0, generateInt2);

// or

func(0, 0, convertStrToInt, std::string{0});

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