由多个对象指向的内存的多个C ++删除(Multiple C++ deletion of a memory pointed by multiple objects)

由多个对象指向的内存的多个C ++删除(Multiple C++ deletion of a memory pointed by multiple objects)

另一个C ++指针删除问题在以下示例中：

class Foo { public: int *p; ~Foo() { delete p; p = NULL; } }; Foo *f1 = new Foo(); Foo *f2 = new Foo(); f1->p = new int(1); f2->p = f1->p; delete f2; // ok delete f1; // no error?

为什么在调用“delete f1”时没有出错？ 我没有删除同一个地址（* p）两次吗？

如果我直接删除最后2行代码中的指针，我会得到错误。

delete f2->p; // ok delete f1->p; // error!! *** glibc detected *** double free or corruption (fasttop) ***

Another C++ pointer deletion question is in the following example:

class Foo { public: int *p; ~Foo() { delete p; p = NULL; } }; Foo *f1 = new Foo(); Foo *f2 = new Foo(); f1->p = new int(1); f2->p = f1->p; delete f2; // ok delete f1; // no error?

Why I did not get error when calling "delete f1"? didn't I delete the same address (*p) twice?

If I directly delete the pointers in the last 2 lines of code, I will get error.

delete f2->p; // ok delete f1->p; // error!! *** glibc detected *** double free or corruption (fasttop) ***

最满意答案

这是一件非常糟糕的事情。 但是C ++不一定会在这里做任何事情。 这是“未定义”的行为。 这并不意味着它会崩溃，但它很可能会导致糟糕的事情发生。

编辑：此外，在你的第二个例子中，它崩溃的事实只是“未定义”行为的一部分。 关于反应将会是什么不确定。

It is a VERY bad thing to do. However C++ won't necessarily do anything here. It is "un-defined" behaviour. That doesn't mean it will crash but it will, most probably, cause bad shit (tm) to happen.

Edit: Furthermore in your 2nd example the fact it crashes is just a part of "undefined" behaviour. It is undefined as to what the reaction will be.