为什么C ++ Map的[]运算符调用映射值的默认构造函数? [英] Why is the [] operator of C++ Map calling default constructor of mapped-value?

问题描述

我用g ++编译了以下代码，执行该行时将调用构造函数A():

I compiled the following code with g++, the construct function A() will be called when executing the line:

m["1"]

为什么会这样?我看不到在这里调用构造函数的任何必要.

Why is this happening? I don't see any necessarily of calling the constructor here.

struct A
{
   int mem;
   A(int arg){}
   A(){}
};
int main()
{
   unordered_map<string, A> m;
   m["1"]; // will call A(), but why?
   m.find("1")->second; // will not call A()
}

推荐答案

这是 operator[] .如果他没有找到您要查看的值，则使用默认构造函数创建该条目.

That's the design of operator[]. If he doesn't find the value you're looking at, the entry is created with default constructor.

如果要查看元素是否存在而无需创建它，可以使用 find() .

If you want to look if an element exists without necessary creating it, you could use find() instead.

如果要像使用operator[]一样处理元素，但是如果找不到该元素而不是创建缺少的条目，则抛出异常，则您最好使用

If you want to address an element like you do with operator[] but throwing an exception if the element is not found instead of creating the missing entry, you would prefer at()

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