指针 - 重复对象实例

问题描述

如果我有

  Class * a1 = new Class（）; 
 Class * b1 = a1; 
 
 delete b1; 
 delete a1; //这将给出一个双重免费或腐败的消息; 
  

如果我删除指针b，它和删除指针一样吗？因为两者都指向同一个Class的实例。因此，我如何将a1的实例复制到b1，这样当我删除b1时，a1不会被删除。

  a1 = new Class（）; 
 Class * b1 = a1; 
 
 //如何复制实例a1指向
 //，这样当我删除b1时，a1仍然存在。 
 
 delete b1; 
 delete a1; //这仍然有效
  

谢谢。

解决方案

有没有理由使用指针和分配？其它简单的

 类a1; 
 ... 
类b1 = a1; 
  

这里不需要删除。

如果你需要保持结构，因为你需要做

 类* a1 = new Class（）; 
 Class * b1 = new Class（* a1）; 
  

或

  Class * a1 = new Class（）; 
 Class * b1 = new Class（）; 
 * b1 = * a1; 
  

这假设你有一个有效的拷贝构造函数（例如＃1）或赋值运算符）

ps：尝试使用 std :: unique_ptr ，而不是原始指针，

If I have

Class *a1 = new Class();
Class *b1 = a1;

delete b1;
delete a1; //this will give a double free or corruption message;

if I delete pointer b, it's the same as deleting pointer a right? Since the two are pointing at the same instance of Class. So, how can I copy the instance of a1 to b1 so that when I delete b1, a1 is NOT deleted.

Class *a1 = new Class();
Class *b1 = a1;

//how do I duplicate the instance a1 is pointing
//so that when I delete b1, a1 still exists.

delete b1;
delete a1; //this is still valid

Thanks.

解决方案

Is there a reason you are using pointers and allocation? Else its as simple as

Class a1;
...
Class b1 = a1;

There is no need here for a delete.

If you need to keep the structure as it is you need to do

Class *a1 = new Class();
Class *b1 = new Class(*a1);

or

Class *a1 = new Class();
Class *b1 = new Class();
*b1 = *a1;

This assumes you have a valid copy-constructor ( e.g #1) or assignment operator (e.g #2)

p.s: try to use std::unique_ptr instead of raw pointers to be safer.

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