c++ - C++:重载的 ‘basic_string()’的调用不明确

我必须向您发布我的代码，因为我无法了解为什么会收到此错误!
我知道我可以使用模板。

class Mappable {
    typedef boost::variant<int, bool, unsigned short, float, char, timeval,
            double, std::string, size_t> MultiType;
private:
    class Handler {
        friend class Mappable;
    public:
        template<typename T>
        operator T&() {
            T t = 0;
            try {
                t = boost::get<T>(it);
            } catch (...) {
            }
            return t;
        }
        template<class T>
        Handler& operator=(const T& rhs) {
            it = rhs;
            return *this;
        }
    private:
        Handler(MultiType& it, const std::string& key);
        MultiType& it;
    };
public:
    Mappable(const std::string& tableName);
    virtual ~Mappable();
    Handler operator[](const std::string& key) {
        return Handler(map_[key], key);
    }
    std::string keyTypeToString(const std::string& key) {
        std::stringstream ss;
        ss << boost::get<T>(map_[key]);
        return ss.str();
    }
private:
    typedef std::map<std::string, MultiType> MultiTypeMap;
    std::string valueFromKey(const MultiTypeMap::iterator& it);
    template<class T>
    MultiTypeMap map_;
};

/* Main.cpp */

int main() {
   Mappable m;
   m["x"] = 2;
   m["data"] = "my data string value"; /* Correctly works */
   cout << (int)(m["x"]); /* Correctly works */
   cout << (string)(m["data"]); /* Error */
   cout << m.keyTypeToString<string>("data"); /* Correctly works */
}

错误是:

../src/data/Mappable.h:98:41: error: call of overloaded ‘basic_string(Mappable::Handler)’ is ambiguous
../src/data/Mappable.h:98:41: note: candidates are:
/usr/include/c++/4.6/bits/basic_string.tcc:214:5: note: std::basic_string<_CharT, _Traits, _Alloc>::basic_string(const _CharT*, const _Alloc&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]
/usr/include/c++/4.6/bits/basic_string.tcc:171:5: note: std::basic_string<_CharT, _Traits, _Alloc>::basic_string(const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>, std::basic_string<_CharT, _Traits, _Alloc> = std::basic_string<char>]
/usr/include/c++/4.6/bits/basic_string.tcc:179:5: note: std::basic_string<_CharT, _Traits, _Alloc>::basic_string(const _Alloc&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]

我不明白歧义在哪里!

最佳答案

您可以隐式转换为任何类型。

template<typename T> operator T&();

这意味着，basic_string的任何单参数隐式构造函数都是可行的(char const *，std::string等)。

口头禅是:隐式转换是邪恶的。

在C++ 11中，可以将explicit关键字添加到转换中。这意味着您将显式调用static_cast<>

    template<typename T>
    explicit operator T&() { // throws on type mismatch
        return boost::get<T>(it);
    }