C ++计数实例/直方图使用std :: map

问题描述

我已经看到类似于以下的示例代码：

I have seen sample code similar to the following:

std::string s = "Hello World!";
std::map<char, std::size_t> h;

for (std::string::const_iterator i=s.cbegin(); i!=s.cend(); ++i)
{
    ++h[*i];
}

assert(h['l'] == 3);

这似乎依赖于每个字母的第一次出现的值类型为零。即使使用类似 std :: size_t 的东西也是如此，它没有默认的构造函数将其重置为零？

This seems to rely on the value type being zeroed on the first occurence of each letter. Is this guaranteed even when using something like a std::size_t which has no default constructor resetting it to zero?

推荐答案

确实这就是 map 的工作原理： [] -operator正在变化并且如果它不存在，将创建映射类型的对象。由于 size_t value-初始化为零，所以您都可以。

Indeed that's how map works: The []-operator is mutating and will create the object of mapped type if it does not exist yet. Since size_t value-initializes to zero, you're all fine.

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