本文共 2971 字，大约阅读时间需要 9 分钟。

如下是一份测试的代码，测试重载 new, new[], delete, delete[] 操作，申请内存的时候，其内部的操作。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace System;using namespace std;class people{private:	int arg;	int x, y;	char *name;public:	people()	{		cout << "in deault constructor! " << endl;	}		people(int arg1, char *name1)	{		arg = arg1;		name = new char[strlen(name1) + 1];		memcpy(name, name1, strlen(name1) + 1);	};	void show()	{		cout << "arg = " << arg << ", name = " << name << endl;	}	void* operator new(size_t size);	void* operator new[](size_t size);	void operator delete(void *p);	void operator delete[](void *p);};void *people::operator new(size_t size){	char *p = new char[size];	printf("p = %X, size = %d \n", p, size);	return p;}void *people::operator new[](size_t size){	char *p = new char[size];	printf("p[] = %X, size = %d \n", p, size);	return p;}void people::operator delete(void *p){	printf("delete p = %X \n", p);	delete p;}void people::operator delete[](void *p){	printf("delete []p = %X \n", p);	delete p;}int main(array
       
         ^args){	people p(12, "wkf");	p.show();	cout << "sizeof(people) = " << sizeof(people) << endl;	people *p1 = new people();	people *p2 = new people[3];	cout << "p1 = " << p1 << endl;	cout << "p2 = " << p2 << endl;		cout << "p2[0] = " << &p2[0] << endl;	cout << "p2[1] = " << &p2[1] << endl;	cout << "p2[2] = " << &p2[2] << endl;	memcpy(&p2[0], &p, sizeof(people));	delete p1;	delete []p2;	printf("hehe......\n");	getchar();	return 0;}
       
      
     
    
   

程序运行结果如下：

arg = 12, name = wkf

sizeof(people) = 16 p = 42CC40, size = 16 in deault constructor! p[] = 42CEB8, size = 48 in deault constructor! in deault constructor! in deault constructor! p1 = 0042CC40 p2 = 0042CEB8 p2[0] = 0042CEB8 p2[1] = 0042CEC8 p2[2] = 0042CED8 delete p = 42CC40 delete []p = 42CEB8 hehe......

其中，执行：

people *p2 = new people[3];

的时候，调用了重载 operator new[] 函数，将申请 3 个对象的内存空间，一个对象是 16 个字节，所以，

总共需要申请 3 * 16 = 48 个字节。

所以，输出如下：

p[] = 42CEB8, size = 48

可以看到，我们返回的地址是 43CEB8，同时，在 main(); 中打印返回的地址是：

p2 = 0042CEB8

p2[0] = 0042CEB8 p2[1] = 0042CEC8 p2[2] = 0042CED8

完全对应上。

但是，如果我们在该类中增加 析构函数，如下：

 ~people()  {   cout << "in un constructor! " << endl;  }

其他的不用修改，然后，编译程序，输出如下：

arg = 12, name = wkf

sizeof(people) = 16 p = 42CC40, size = 16 in deault constructor! p[] = 42CEB8, size = 52 in deault constructor! in deault constructor! in deault constructor! p1 = 0042CC40 p2 = 0042CEBC p2[0] = 0042CEBC p2[1] = 0042CECC p2[2] = 0042CEDC in un constructor! delete p = 42CC40 in un constructor! in un constructor! in un constructor! delete []p = 42CEB8 hehe......

可以看到，执行如下代码：

people *p2 = new people[3];

申请了 52 个字节，52 = 3 * 16 + 4；

就是多了 申请 4 个字节的空间。

输出：

p[] = 42CEB8, size = 52

可以知道，返回的是 42CEB8 的地址，但是，p2 得到的地址是 42CEBC 的地址。

其中，42CEB8、42CEB9、42CEBA、42CEBB 然后到 42CEBC，中间正好是 4 个字节。

最后，执行

delete []p2;

的时候，传递给 operator delete[](); 函数的地址是 42CEB8，并不是 p2 的地址，而是 operator new[](); 中

申请内存的时候，返回的地址，那么，在中间产生的 4 个字节空间是做什么？

难道只是引入了 一个析构函数，就多产生了 4 个字节的空间。

转载地址：https://mylinux.blog.csdn.net/article/details/9058553 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！