从std容器复制frm任意源对象 [英] Copying from std container frm arbitrary source object

问题描述

我创建了一个只读迭代器，它允许我更方便地使用它在for循环，然后使用std迭代器，类似于使用FOREACH（不太好，但足够好）：）

I created a read only iterator which allows me to use it in a for loop more conveniently then with the std iterators, similar to what boost does with the FOREACH (not as good but good enough :))

现在看起来像这样：

for(ReadOnylIterator<MyClass *> class = parent.getIterator(); class.end(); ++class)
   class->function();

我现在的问题是，我必须在父实现一个函数返回迭代器。由于std容器具有相同的语法，我想知道是否可以在接受任何std ::容器并创建副本本身的Iterator上定义一个复制构造函数/赋值运算符，而不是要求返回

The problem that I have now is, that I must implement a function on the parent which returns the iterator. Since the std containers have all the same syntax, I was wondering if it is possible to define a copy constructor/assignment operator on the Iterator that accepts any of the std:: containers and creates the copy itself, instead of requiring the class to return it.

当然，我想避免自己定义所有的，因为它们有很多：

Of course I want to avoid having to define all of them myself like as there are lots of them:

ReadOnlyIterator<T> &operator=(std::list<T> const &v)
ReadOnlyIterator<T> &operator=(std::vector<T> const &v)
...

有办法吗？当我看向量的源，我没有看到一个共同的基类，所以我认为它可能是不可能的。

Is there a way to do this? When I look at the source of the vector I don't see a common base class, so I think it might not be posssible.

我不明白为什么赋值运算符不起作用。

I don't understnad why the assignment operator doesn't work.

在我的代码中，我测试如下：

In my code I test it like this:

std::vector<SimpleClass *>t;
ReadOnlyIterator<SimpleClass *> &it = t;

我得到

error C2440: 'initializing' : cannot convert from 'std::vector<_Ty>' to 'ReadOnlyIterator<T> &'

推荐答案

如果我正确理解，工作：

If I understand you correctly, this should work:

#include <type_traits>

template <typename Container>
typename std::enable_if<std::is_same<T, typename Container::value_type>::value, ReadOnlyIterator<T>&>::type operator= (const Container &v);

上述代码使用C ++ 11。如果您无法访问，可以使用Boost的等效功能。

The above code uses C++11. If you don't have access to that, you could use the equivalent functionality from Boost.

Live example

如果您既不能使用C ++ 11也不能使用Boost，你可以自己编写必要的类：

In case you can use neither C++11 nor Boost, you can code the necessary classes yourself:

template <typename T, typename U>
struct is_same
{
    enum { value = 0 };
};

template <typename T>
struct is_same<T, T>
{
    enum { value = 1 };
};

template <bool, typename>
struct enable_if
{};

template <typename T>
struct enable_if<true, T>
{
    typedef T type;
};

要在构造函数中使用它，请如下定义：

To use this in a constructor, define it like this:

template <typename Container>
ReadOnlyIterator(const Container &v, typename enable_if<is_same<T, typename Container::value_type>::value, void>::type * = 0) {}

实例

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