从返回迭代器的const和非const方法中删除代码重复 [英] Removing code duplication from const and non-const methods that return iterators

问题描述

我正在考虑这个问题上const和非const类方法。首选的答案取自Scott Meyers的Effective C ++，其中非const方法是用const方法实现的。

I'm considering this question on const and non-const class methods. The preferred answer is taken from Effective C++ by Scott Meyers, where the non-const method is implemented in terms of the const method.

进一步扩展，如果方法返回迭代器而不是引用，如何减少代码重复？修改关联问题中的示例：

Extending this further, how can code duplication be reduced if the methods return iterators instead of references? Modifying the example in the linked question:

class X
{
    std::vector<Z> vecZ;    
public:
    std::vector<Z>::iterator Z(size_t index)
    {
        // ...
    }
    std::vector<Z>::const_iterator Z(size_t index) const
    {
        // ...
    }
};

我不能用const方法实现非const方法，因为它不可能

I can't implement the non-const method in terms of the const method because it's not possible to convert directly from a const_iterator to an iterator without using the distance()/advance() technique.

在示例中，因为我们使用std :: vector作为一个const容器，它实际上可能从一个const_iterator转换到一个迭代器，因为它们可能被实现为指针。我不想依靠这个。是否有更通用的解决方案？

In the example, because we're using a std::vector as the container, it might actually be possible to cast from a const_iterator to an iterator because they may well be implemented as pointers. I don't want to rely on this. Is there a more general solution?

推荐答案

我相信，只有帮助者才能使用

I believe, it is only possible with the helper

typedef int Z;

class X
{
    std::vector<Z> vecZ;    
public:
    std::vector<Z>::iterator foo(size_t index)
    {
        return helper(*this);
    }
    std::vector<Z>::const_iterator foo(size_t index) const
    {
        return helper(*this);
    }

    template <typename T>
    static auto helper(T& t) -> decltype(t.vecZ.begin())
    {
        return t.vecZ.begin();
    }
};

EDIT
同样可以不使用c ++ 11

EDIT Same can be implemented without c++11

template <typename T>
struct select
{
    typedef std::vector<Z>::iterator type;
};
template <typename T>
struct select<const T&>
{
    typedef std::vector<Z>::const_iterator type;
};

template <typename T>
static
typename select<T>::type
helper(T t) 
{
    return t.vecZ.begin();
}

但是，我想你应该三思而后行approcach

But, well, I think you should think twice before using this approcach

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