是否有可能在C ++中使用匿名类？ [英] Is it possible to use anonymous classes in C++?

问题描述

我在Quora的 C ++ 代码中见过匿名类。它已成功编译并运行。

代码位于：

  #include< iostream> 
 
 $ auto 
 
 
 
 
 $ public 
 int val; 
} a; 
 
 a.val = 5; 
 
返回a; 
} 
 
 int main（）
 {
 std :: cout<< func（）。val<<的std :: ENDL; 
返回0; 
 
  

所以，它在C ++中有效吗？

另外，我很想知道，是否可以在C ++中使用匿名类？

是这种形式的联合：

<$ p $

p> union {...};


它定义了一个未命名类型的未命名对象。它的成员被注入到周围的范围中，所以可以在不使用< object>。前缀的情况下引用它们，否则将是必要的。

从这个意义上说，不存在匿名类（不是C ++联合中的联合是类）。

另一方面，

，无名类（包括结构和联合）没有什么特别的。

  union {...} x; 
 class {...} y; 
 typedef struct {...} z; 
  

x 和 y 是未命名类型的命名对象。 z 是一个typedef名称，它是未命名结构的别名。他们不被称为匿名，因为这个词是保留给上述联合的形式。

  []（）{} 
  

Lambdas是未命名类类型的未命名对象，但它们也不被称为匿名对象。

I have seen anonymous classes in C++ code on Quora. It's successfully compiled and run.

Code here:

#include <iostream>

auto func()
{
    class // no name
    {
      public:
             int val;
    } a;

    a.val = 5;

    return a;
}

int main()
{
    std::cout << func().val << std::endl;
    return 0;
}

So, Is it valid in C++?

Also, I am curious to know, Is it possible to use anonymous classes in C++?

解决方案

In C++, an anonymous union is a union of this form:

 union { ... } ;

It defines an unnamed object of an unnamed type. Its members are injected in the surrounding scope, so one can refer to them without using an <object>. prefix that otherwise would be necessary.

In this sense, no anonymous classes (that are not unions --- in C++ unions are classes) exist.

On the other hand, unnamed classes (including structs and unions) are nothing unusual.

union { ... } x;
class { ... } y;
typedef struct { ... } z;

x and y are named object of unnamed types. z is a typedef-name that is an alias for an unnamed struct. They are not called anonymous because this term is reserved for the above form of a union.

[](){}

Lambdas are unnamed objects of unnamed class types, but they are not called anonymous either.

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