C ++ 11中的默认函数有什么意义？ [英] What's the point in defaulting functions in C++11?

问题描述

C ++ 11增加了告诉编译器创建默认实施任何特殊成员职能。虽然我可以看到删除一个函数的值，其中的值明确默认一个函数？

C++11 adds the ability for telling the compiler to create a default implementation of any of the special member functions. While I can see the value of deleting a function, where's the value of explicitly defaulting a function? Just leave it blank and the compiler will do it anyway.

我可以看到的唯一一点是，默认构造函数只有在没有其他构造函数存在时才被创建：

The only point I can see is that a default constructor is only created when no other constructor exists:

class eg {
public:
    eg(int i);
    eg() = default; 
};

但是这真的比现在更好吗？

But is that really better than how you do it now?

class eg {
public:
    eg(int i);
    eg() {}
};

还是我缺少一个用例？

推荐答案

默认的构造函数将有一个声明，该声明将受正常访问规则的约束。例如。你可以使默认的拷贝构造函数保护。没有这些新的声明，默认生成的成员是public的。

A defaulted constructor will have a declaration, and that declaration will be subject to the normal access rules. E.g. you can make the default copy constructor protected. Without these new declarations, the default generated members are public.

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