Question

我正在玩容器类型，深入细节并尝试构建一个不可修改的（或不可变的）地图。

为此，我构建了一些像copy-constructor来将另一个unordered_map中的所有元素添加到我的UnmodifiableMap（它基本上是std :: unordered_map的包装器）。为了使其不可修改，我只提供const迭代器和只读方法。但我对构造函数感到困惑，我确信我会错过一些东西，也许有人可以在这里指出我的问题。也许这是一种完全错误的方式，但到目前为止他的尝试是：

template <typename Key, typename T,
    typename HashFcn = std::hash<Key>,
    typename EqualKey = std::equal_to<Key>,
    typename Alloc = std::allocator<std::pair<const Key, T> > > 
class UnmodifiableMap {
public:
    // The actual data
    typedef std::unordered_map<T, Key, HashFcn, EqualKey> base;

    typedef Key key_type;
    typedef T data_type;
    typedef T mapped_type;
    typedef std::pair<const key_type, data_type> value_type;
    typedef HashFcn hasher;
    typedef EqualKey key_equal;
    typedef Alloc allocator_type;

    typedef typename base::size_type size_type;
    typedef typename base::const_iterator const_iterator;
    typedef typename base::iterator iterator;

private:
    base _map;

    /**
     * Constructs an empty unordered_map 
     */
    UnmodifiableMap(
        size_type n = 0, const hasher& hf = hasher(),
        const key_equal& eql = key_equal(),
        const allocator_type &alloc = allocator_type())
            : _map(n, hf, eql, alloc) {}

public:
    /** Constructs a copy of unordered_map */
    UnmodifiableMap(const base& other)
    : _map(static_cast<const base&>(other)) {}

    ~UnmodifiableMap() {}   

    iterator begin() { return _map.begin(); }
    iterator end() { return _map.end(); }
    const_iterator begin() const { return _map.begin(); }
    const_iterator end() const { return _map.end(); }

    bool empty() const { return _map.empty(); }

    bool contains(const key_type& key) const { 
        return _map.find(key) != _map.end(); }
};

这里是主体：

int main(int argc, char **argv) {
    typedef std::unordered_map<int, std::string> Items;
    Items map;

    map[1] = "first string";
    map[4] = "string 4";
    map[5] = "string 5";
    map[22] = "string 22";
    map[12] = "string 12";
    map[18] = "string 18";

    typedef UnmodifiableMap<int, std::string> ReadOnlyItems;
    ReadOnlyItems readonlymap(map);

    return 0;
}

我得到的错误是

Unmodifiable_map.cpp: In function ‘int main(int, char**)’:
Unmodifiable_map.cpp:56:25: error: no matching function for call to ‘UnmodifiableMap<int, std::basic_string<char> >::UnmodifiableMap(Items&)’
Unmodifiable_map.cpp:56:25: note: candidates are:
Unmodifiable_map.h:45:2: note: UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::UnmodifiableMap(const base&) [with Key = int, T = std::basic_string<char>, HashFcn = std::hash<int>, EqualKey = std::equal_to<int>, Alloc = std::allocator<std::pair<const int, std::basic_string<char> > >, UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::base = std::unordered_map<std::basic_string<char>, int, std::hash<int>, std::equal_to<int>, std::allocator<std::pair<const std::basic_string<char>, int> > >]
Unmodifiable_map.h:45:2: note:   no known conversion for argument 1 from ‘Items {aka std::unordered_map<int, std::basic_string<char> >}’ to ‘const base& {aka const std::unordered_map<std::basic_string<char>, int, std::hash<int>, std::equal_to<int>, std::allocator<std::pair<const std::basic_string<char>, int> > >&}’
Unmodifiable_map.h:37:2: note: UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::UnmodifiableMap(UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::size_type, const hasher&, const key_equal&, const allocator_type&) [with Key = int, T = std::basic_string<char>, HashFcn = std::hash<int>, EqualKey = std::equal_to<int>, Alloc = std::allocator<std::pair<const int, std::basic_string<char> > >, UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::size_type = long unsigned int, UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::hasher = std::hash<int>, UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::key_equal = std::equal_to<int>, UnmodifiableMap<Key, T, HashFcn, EqualKey, Alloc>::allocator_type = std::allocator<std::pair<const int, std::basic_string<char> > >]
Unmodifiable_map.h:37:2: note:   no known conversion for argument 1 from ‘Items {aka std::unordered_map<int, std::basic_string<char> >}’ to ‘long unsigned int’
Unmodifiable_map.h:14:7: note: UnmodifiableMap<int, std::basic_string<char> >::UnmodifiableMap(const UnmodifiableMap<int, std::basic_string<char> >&)
Unmodifiable_map.h:14:7: note:   no known conversion for argument 1 from ‘Items {aka std::unordered_map<int, std::basic_string<char> >}’ to ‘const UnmodifiableMap<int, std::basic_string<char> >&’
Unmodifiable_map.h:14:7: note: UnmodifiableMap<int, std::basic_string<char> >::UnmodifiableMap(UnmodifiableMap<int, std::basic_string<char> >&&)
Unmodifiable_map.h:14:7: note:   no known conversion for argument 1 from ‘Items {aka std::unordered_map<int, std::basic_string<char> >}’ to ‘UnmodifiableMap<int, std::basic_string<char> >&&’

希望有人可以对此有所了解。另外我想我需要在复制构造函数中做更多的工作，可以复制元素（比如swap函数）和operator=的实现？！

Answer 1

您正在包装类中的T类型定义中交换Key和base的角色：

template <typename Key, typename T,
    typename HashFcn = std::hash<Key>,
    typename EqualKey = std::equal_to<Key>,
    typename Alloc = std::allocator<std::pair<const Key, T> > > 
class UnmodifiableMap {
public:
    // typedef std::unordered_map<T, Key, HashFcn, EqualKey> base; // ERROR
    typedef std::unordered_map<Key, T, HashFcn, EqualKey> base; // OK
    ...

因此，您的基础地图最终成为unordered_map<string, int>而不是unordered_map<int, string>，并且编译器会抱怨不匹配。

另请注意，构造函数中有一个不必要的static_cast：

UnmodifiableMap(const base& other)
// : _map(static_cast<const base&>(other)) {} // NOT NEEDED. Just do:
   : _map(other) {}

包装unordered_map以构建不可修改（不可变）的映射

1 个答案: