Question

查看以下简化示例代码：

class Foo {
private:
    struct bar {
        int private_value;  // only used for internal stuff of class Foo

        int public_value;
    };
public:
    class iterator {
    public:
        iterator(Foo::bar* c)
            :curr{ c }
        {
        }

        int& operator*() { return curr->public_value; }
        int* operator->() { return &curr->public_value; }
    private:
        Foo::bar* curr;
    };


    void some_function();
};

void Foo::some_function()
{
    Foo::iterator it = get_iterator();  // get an iterator from function

    *it //  this gives only curr->public_value
        //  what to add to iterator to have acces  of private value from iterator class?
}

我想在函数中使用迭代器，并从中获取private_value。有什么方法可以在迭代器中创建函数，以便仅Foo类可以从迭代器访问private_value吗？我不允许在迭代器中返回整个* bar结构，因为只有public_value的用户才可以使用foo

Answer 1

只需在迭代器中使周围的类成为朋友即可

class Foo {
private:
    struct bar {
        int private_value;  // only used for internal stuff of class Foo

        int public_value;
    };
public:
    class iterator {
    public:
        iterator(Foo::bar* c)
            :curr{ c }
        {
        }

        int& operator*() { return curr->public_value; }
        int* operator->() { return &curr->public_value; }
    private:
        Foo::bar* curr;

        friend class Foo;           // friend
    };

    void some_function();
};

有时候您看不到树林中的森林...

在周围的类中访问迭代器的私有值

1 个答案: