我必須發佈我的代碼,因爲我無法知道爲什麼會出現此錯誤!
我知道我可以使用的模板..C++:重載的'basic_string()'調用不明確
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 */
}
錯誤是:
../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>]
我能不明白的地方是含糊!
你說的「作品」,但實際上這是行不通的。它可以編譯(儘管它不應該這樣做 - 使用'-pedantic -Wall -Werror')。但是它會在運行時失敗,因爲您正在返回對本地對象的引用。 – 2014-09-19 09:31:12
實際上,您的代碼中存在與您的錯誤無關的其他問題,即Handler類中的轉換運算符用於轉換爲* reference *類型。所以這意味着你返回一個局部變量的引用,這無疑會導致[* undefined behavior *](http://en.wikipedia.org/wiki/Undefined_behavior)。 – 2014-09-19 09:32:39
沒有Joachim,返回的參考是我想要的:這樣我可以訪問一個元素並修改它。 – 2014-09-19 09:42:56