如何为集合提供带迭代器的const接口？ [英] How to provide const interface with iterators to a collection?

问题描述

我想用这样的签名创建一个函数：

I would like to create a function with a signature like this:

// Set found to be an iterator to the location of key in map or end()
// if not found.
bool lookup(const Key &key,
            const std::map<Key, Value> &map,
            std::map<Key, Value>::const_iterator &found);

但是我想在地图和迭代器不是const的情况下调用它以便我可以修改找到的值：

But I would like to also call it in cases where the map and iterator are not const so that I can modify the found value:

const Key key;
std::map<Key, Value> map;
std::map<Key, Value>::iterator found;

if (lookup(key, map, found)) {
  found->second.modifingNonConstFunction()
}

但我不相信我可以传递 std :: map< Key，Value> :: iterator 对象期望引用 std :: map< Key，Value> :: const_iterator ，因为它们是不同的类型，而我通常可以在 const 是类似C ++声明的一部分，我可以将非const类型提升为const类型：

But I do not believe I can pass a std::map<Key, Value>::iterator object to a function expecting a reference to a std::map<Key, Value>::const_iterator since they are different types, whereas I normally could if the const was part of C++ declaration of the type like this and I could promote the non-const type to a const type:

void someFunction(const int &arg);

int notConstArg = 0;
someFunction(nonConstArg);

除了使用模板为 lookup（）提供两个定义，一个如 const 参数2和3以及另一个非const参数2和3所示，在C ++中有更好的方法来实现这一点类似于 const int& 如何在上面的例子中传递一个非const int 。换句话说，我可以只有一个功能而不是两个吗？

Other than by using templates to provide two definitions for lookup(), one as shown with const arguments 2 and 3 and another with non-const arguments 2 and 3, is there a better way in C++ to accomplish this more akin to how const int & can be passed a non-const int in the example above. In other words, can I just have a single function and not two?

推荐答案

不，我认为你不能做这没有重载/模板魔术。

No, I don't think you can do this without overloads/template magic.

编译器正在保护您免受以下情况的影响：

The compiler is protecting you from the following scenario:

typedef vector<int> T;

const T v;  // Can't touch me

void foo(T::const_iterator &it) {
    it = v.begin();  // v.begin() really is a const_iterator
}

int main() {
    T::iterator it;
    foo(it);
    *it = 5;   // Uh-oh, you touched me!
}

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