std :: map :: operator []

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我当时正在做一个简单的地图程序，但最终遇到了这个问题. C ++文档说:

I was doing a simple map program but ended up with this question. The c++ doc says this:

访问元素 如果k与容器中元素的键匹配，则该函数返回对其映射值的引用. 如果k与容器中任何元素的键都不匹配，则该函数将使用该键插入一个新元素，并返回对其映射值的引用.请注意，即使没有为该元素分配任何映射值(该元素是使用其默认构造函数构造的)，这也会始终使容器大小增加一.

Access element If k matches the key of an element in the container, the function returns a reference to its mapped value. If k does not match the key of any element in the container, the function inserts a new element with that key and returns a reference to its mapped value. Notice that this always increases the container size by one, even if no mapped value is assigned to the element (the element is constructed using its default constructor).

我真正不了解的部分是它说元素是使用其默认构造函数构造的".

The part I don't really get is where it says "the element is constructerd using its default constructor".

我尝试了一下，并做了

std::map<string, int> m; m["toast"];

我只是想看看映射的"toast"元素将是什么值.它最终为零，但是，为什么呢?基本类型是否具有默认构造函数?还是正在发生什么?

I just wanted to see what value would the mapped element of "toast" be. And it ended up being zero, but, why? does the primitive types have a default constructor? or what is happening?

推荐答案

使用其默认构造函数"的声明令人困惑.更准确地说，对于 std :: map :: operator [] ，如果密钥不存在，则插入的值将为值已初始化.

The statement of "using its default constructor" is confusing. More precisely, for std::map::operator[], if the key does not exist, the inserted value will be value-initialized.

使用默认分配器时，这将导致从密钥复制密钥，并且映射值值已初始化.

对于int，它表示零初始化.

4)否则，该对象将被初始化为零.

4) otherwise, the object is zero-initialized.