在看一些C++的文章的时候，经常会看到关于delete与free的区别，delete不需要检查指针是否为NULL。如下面一篇文章 http://www.cnblogs.com/zhuyp1015/archive/2012/07/20/2601698.html 。

但是有时候，又会看到由于delete引起的double free or corruption，我们可以写一个很简单的程序，来测试delete会不会引起double free。

#include <iostream>
using namespace std;
#ifndef NULL
#define NULL 0
#endif
class A
public:
        cout << "\tA was constructed" << endl;    
    ~A(){
        cout << "\tA was deconstructed" << endl;    
int main()
    int* a = NULL;
    cout << "befor delete: " << a << endl;    
    delete a;
    cout << "delete once: " << a << endl;    
    delete a;
    cout << "delete twice: " << a << endl;    
    a = new int(3);
    cout << "after new an integer: " << a << endl;    
    delete a;
    cout << "delete once: " << a << endl;    
    delete a;
    cout << "delete twice: " << a << endl;            
    A* b = new A();
    cout << "after new A: " << b << endl;    
    delete b;
    cout << "delete once: " << b << endl;    
    delete b;
    cout << "delete twice: " << b << endl;        
    int* c = new int[3];
    cout << "after new an array: " << c << endl;    
    delete  [] c;
    cout << "delete once: " << c << endl;    
    delete  [] c;
    cout << "delete once: " << c << endl;    
    return 0;
下面是结果，
测试环境，ubuntu10.04，g++4.4.3.
出现double free了。从输出可以看到，如果指针ptr的值不为NULL，那么在第一次delete之后，ptr也不会被置为0，然后进行第二次delete的时候，就出现double free。
上面代码里有一段是创建一个类的实例，在构造函数和析构函数中有输出，测试的时候，在第二次delete的时候，析构函数又执行了一次，然后就出现double free了。
如何避免double free呢？
一种方法就是在delete之后将指针置为NULL。有其它方法吗？检查指针是否为0？这个。。。