为什么没有默认构造函数的对std :: map :: operator []的调用不能编译为值类型? [英] Why doesn't a call to std::map::operator[] compile for a value type without default constructor?

问题描述

请考虑以下类 MyStruct :

struct MyStruct
{
    int x;
    int y;

    MyStruct(int i, int j):
    x(i), y(j)
    {
    }
};

请注意， MyStruct 没有默认的析构函数.

Note that MyStruct doesn't have a default destructor.

以下代码中的赋值 m ["AAAA"] = MyStruct(1，1)无法编译:

The assignment m["AAAA"] = MyStruct(1, 1) in the code below doesn't compile:

int main(int, char**)
{
    map<string, MyStruct> m;
    m["AAAA"] = MyStruct(1, 1);

    return 0;
}

为什么我需要 MyStruct 的默认构造函数?为什么上面的代码无法编译?

Why I need default constructor for MyStruct? Why does the code above not compile?

推荐答案

为什么需要默认构造函数?

Why I need default constructor?

您可以使用 std :: map< std :: string，MyStruct> 对象 m 的下标运算符(即 [] )code>，方法如下:

You could use the subscript operator (i.e. []) of the std::map<std::string, MyStruct> object, m, in the following way:

auto value = m["AAAA"];

如果 std :: map< std :: string，MyStruct> 没有与键"AAAA"关联的 MyStruct 对象，则容器将创建一个默认构造的容器，并将其与键"AAAA" 关联.因此，如果 MyStruct 没有默认的构造函数，则对 operator [] 的调用将不会编译.

If the std::map<std::string, MyStruct> doesn't have a MyStruct object associated with the key "AAAA", then the container will create a default constructed one and associate it to the key "AAAA". For this reason, if MyStruct doesn't have a default constructor, then the call to the operator[] will not compile.

从技术上讲，下面的语句是做什么的:

Technically, what the statement below does:

m["AAAA"] = MyStruct(1, 1); 

是向容器 m 关联的 MyStruct 对象返回 lvalue ( MyStruct& )密钥"AAAA" .如果没有这样的关联，则容器会为此关联创建一个默认构造的 MyStruct 对象.最后，该返回的对象是赋值运算符的目标.

is to return an lvalue (MyStruct&) to the MyStruct object the container m has associated to the key "AAAA". If there is no such an association, the container creates a default-constructed MyStruct object for this association. Finally, this returned object is the target for the assignment operator.

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