理解c共享指针

嗨我正在制作我自己的引用计数智能指针,但在我开始之前有两个概念我不太明白.

>当我创建一个共享指针指向一个对象时,我必须为一个结构/类分配内存,该结构/类将包含诸如引用计数(最初为一个)之类的信息,也可能是一个互斥量以及递增和递减.当我使用时,假设=运算符使另一个共享指针也指向该对象,我也将指向此结构/类的指针传递给该新指针,这样我就可以增加计数.我的问题是,如果我使第三个共享指针指向此对象(不使用复制构造函数或=运算符),那么此指针将不知道结构,因此将引用计数为1,如果我然后删除指针,计数将达到0并且对象将被销毁,实际上,此对象还有另外两个指针？
>如果共享指针的引用计数为1,然后创建多个线程,如果一个线程结束/销毁它,那么可能仍在运行的其他线程会发生什么？

解决方法:

I get that when a shared pointer is created to point to an object I must allocate memory for a struct/class that will contain information such as the reference count (initially one) and maybe a mutex as well for incrementing and decrementing.

是的,你需要一个柜台.
只有在计划多线程时才需要互斥锁.我会集中精力让计数工作首先担心锁定后记.

When I use, say the =operator to make another shared pointer also point to this object, I will also pass the pointer to this struct/class to that new pointer so I can increment the count.

关于共享指针的观点是它们取得了指针的所有权.一旦创建了共享指针,就不应该有指向同一对象的RAW指针的实例.因此,当您复制或分配时,您正在通过共享指针执行所有操作.

My question is, if I make a third shared pointer point to this object(not using the copy constructor or =operator), then this pointer won’t know about the struct and therefore will have a reference count of 1.

你的假设是正确的.这就是为什么当你创建共享指针时,你不应该保留指针的副本.这是引入std :: make_shared()的另一个原因,它分配内存并立即将其包装在智能指针中,因此没有RAW指针返回给用户代码.

if I then delete the pointer, the count will reach 0 and the object will be destroyed when in fact, there are two other pointers for this object?

这正是问题所在.这就是为什么你不应该创建或传递RAW指针到已经被管理的对象.

If a shared pointer has a reference count of 1, and then multiple threads are created, if one thread ends/destroys it, what happens with the other threads that may still be running?

如果您的引用计数有效并且您只有管理RAW指针的共享指针,它应该按预期工作.但如果你在一个线程中删除它,它将在所有线程中被销毁.

标签：c,shared-ptr,pointers,smart-pointers
来源： https://codeday.me/bug/20190824/1704797.html