問題描述
map<string, string> dada;
dada["dummy"] = "papy";
cout << dada["pootoo"];
我很困惑，因為我不知道它是否被認為是未定義的行為，如何知道我何時請求不存在的密鑰，我是否只使用 find ?
I'm puzzled because I don't know if it's considered undefined behaviour or not, how to know when I request a key which does not exist, do I just use find instead ?
推薦答案
map::operator[] 在數據結構中搜索與給定鍵對應的值，并返回對它的引用.
The map::operator[] searches the data structure for a value corresponding to the given key, and returns a reference to it.
如果它找不到一個，它會透明地為它創建一個默認的構造元素.(如果您不想要這種行為，您可以改用 map::at 函數.)
If it can't find one it transparently creates a default constructed element for it. (If you do not want this behaviour you can use the map::at function instead.)
您可以在此處獲取 std::map 方法的完整列表:
You can get a full list of methods of std::map here:
http://en.cppreference.com/w/cpp/container/map
這是當前 C++ 標準中 map::operator[] 的文檔...
Here is the documentation of map::operator[] from the current C++ standard...
效果:如果映射中沒有與 x 等效的鍵，則將 value_type(x, T()) 插入映射中.
Effects: If there is no key equivalent to x in the map, inserts value_type(x, T()) into the map.
要求:key_type 應為 CopyConstructible，mapped_type 應為 DefaultConstructible.
Requires: key_type shall be CopyConstructible and mapped_type shall be DefaultConstructible.
返回:對 *this 中 x 對應的映射類型的引用.
Returns: A reference to the mapped_type corresponding to x in *this.
復雜度:對數.
T&運算符[](key_type&& x);
效果:如果映射中沒有與 x 等效的鍵，則將 value_type(std::move(x), T()) 插入映射中.
Effects: If there is no key equivalent to x in the map, inserts value_type(std::move(x), T()) into the map.
要求:mapped_type 應為 DefaultConstructible.
Requires: mapped_type shall be DefaultConstructible.
返回:對 *this 中 x 對應的映射類型的引用.
Returns: A reference to the mapped_type corresponding to x in *this.
復雜度:對數.
這篇關于如果我在鍵不存在的情況下讀取地圖的值會發生什么?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！