使用delete [] p时堆的损坏;

Question

我使用delete [] p,会发生堆损坏.

Rectangle *p[3];

for (int i = 0; i<3; i++)
  p[i] = new Rectangle();

p[0]->set(3, 4); 
p[1]->set(10, 8); 
p[2]->set(5, 6); 

for(int i=0; i < 3; i++) 
    cout << "Area is " << p[i]->area() << endl; 

delete []p; 

我将删除[]更改为

for (int i = 0; i<3; i++)
    delete p[i]; 

它有效,但为什么我不能使用delete [] p？删除[] p只删除p [0],删除[] p中的含义是什么？

Answer 1

delete []仅在分配之后使用new [],并且delete仅在之后使用new.

这就是C++的工作原理.如果你想简单地分配一个3的数组Rectangle,请使用:

Rectangle *p = new Rectangle[3];

...

delete [] p;

Answer 2

delete [] x 是动态分配的数组,而不是动态分配元素的指针数组,这是你拥有的:

Rectangle* pA = new Rectalgle[3]; // dynamically allocated array
....
delete [] pA;  // OK

Rectangle* p[3]; // array of pointers. 
// Each pointer must be dealt with individually.
delete p[0]; // Might be OK, if p[0] points to dynamically allocated Rectangle.