[英]In map<string, int>, is it guaranteed that the int is initialized to zero?
For example, count the occurrence the words in a book, I saw somebody simply wrote: 例如,计算书中单词的出现次数,我看到有人简单地写道:
map<string, int> count;
string s;
while (cin >> s) count[s]++;
Is this the correct way of doing so? 这是正确的方法吗? I tested on my machine and seems so.
我在我的机器上测试过,看起来如此。 But is the initialization to zero guaranteed?
但保证初始化为零吗? If it is not, I would imagine a code like this:
如果不是,我会想象这样的代码:
map<string, int> count;
string s;
while (cin >> s)
if (count.find(s) != count.end()) count[s]++;
else count[s] = 1;
Yes, operator[]
on a std::map
will initialize the value with T()
, which in the case of int
, is zero. 是的,
std::map
上的operator[]
将使用T()
初始化该值,在int
的情况下,该值为零。
This is documented on section 23.4.4.3 of the C++ standard: 这在C ++标准的第23.4.4.3节中有记录:
T& operator[](const key_type& x);
Effects : If there is no key equivalent to
x
in the map, insertsvalue_type(x, T())
into the map.效果 :如果地图中没有等效于
x
键,则将value_type(x, T())
插入到地图中。
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