使用C ++和map/unordered_map创建直方图:不存在的键的默认值 [英] Create A histogram using C++ with map/unordered_map: the default value for a non-existant key

问题描述

我正在定义一个小的函数来创建整数向量的直方图.最初，我定义了以下函数，该函数首先在分配或递增值之前测试键是否在映射中.

I am defining a small function to create a histogram of a vector of integers, Initially I defined the following function that first test whether the key exists in the map before assigning or increment the value.

map<int, int> histogram(vector<int> &a){
     map<int, int> hist;
     for (auto &x : a){
         hist[x] = hist.count(x) == 0 ? 1 : hist[x] + 1; // check key existence 
     }
     return hist;
}

后来，我发现以下代码也可以运行，而无需检查密钥是否存在.因此，不存在的密钥的默认值应为零.我想知道在引用不存在的键时是否可以保证此行为具有默认的零值?

Later, I found the following code also works without checking the existence of the key. Therefore the default value for a non-existent key is supposed to be ZERO. I am wondering is this behavior guaranteed to have a default zero value when referencing a key that does not exist?

map<int, int> histogram(vector<int> &a){
     map<int, int> hist;
     for (auto &x : a){
         hist[x]++;        // without key existence checking. 
     }
     return hist;
}

推荐答案

是的，保证 [] 插入的值为零.从C ++ 11 23.4.4.3/1:

Yes, the value inserted by [] is guaranteed to be zero. From C++11 23.4.4.3/1:

效果:如果地图中没有等效于 x 的键，则将 value_type(x，T())插入地图

Effects: If there is no key equivalent to x in the map, inserts value_type(x, T()) into the map.

T()指定值初始化，对于数字类型，这意味着它以零值初始化.

T() specifies value-initialisation which, for numeric types, means it's initialised with the value zero.

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