为什么删除[]应该崩溃？ [英] Why delete[] should crash?

问题描述

int _tmain(int argc, _TCHAR* argv[])
{
    int *c= new int[5];
    c[0] = 100;
    c[1] = 200;
    c[2] = 300;
    c[3] = 400;
    c[4] = 500;
    int i=0;

    while(i<5)
    {
        printf("%d\n",*c++);
        i++;
    }

    delete[] c;
        return 0;
}

为什么当我删除指针时它断言使用堆块？ ：confused：

why it asserts Heap-Block-in-use when I delete the pointer? :confused:

推荐答案

当你执行while循环时，你增加c指针，所以最后不再指向最初分配的分配。



你必须在循环中使用临时变量;

When you do your while loop, you increment c pointer so at the end is not pointing to the originally allocated allocation anymore.

You have to use a temporary variable Inside the loop;

...
int i=0;
int *t = c;

while(i<5)
{
    printf("%d\n",*t++);
    i++;
}
...

非常感谢。

如何跟踪这个指针将运行到内存中的这些元素？

例如假设我有一个类似下面的函数

Thanks a lot.
How does it keep track that this pointer will run to these many elements in memory?
For example assume I have a function like below

int* supplyPtr()
{
   int* pSet  = new int[10];
   return ++pSet;
}

void main()
{

 int* p = supplyPtr();
 delete[] p;

}

在上面的示例中，当我从pSet返回pSet时，它如何跟踪p？

supplyPtr，它干净地删除了内存。

但是如果我传递++ pSet，它会断言。它如何关联指针及其在内存中的元素数量？你能解释一下吗？或者指向正确的资源？

In the above sample how does it keep track of p?
when I return pSet from the supplyPtr, it cleanly deletes the memory.
But if I pass ++pSet, it asserts. How does it associate a pointer and it''s number of elements in memory? Could you please explain? or point me to the right resource?

这篇关于为什么删除[]应该崩溃？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！