二级指针分析 -- (int *)*(&b)

Posted 2022-10-31 狱典司

看代码说话！

# include <bits/stdc++.h>
using namespace std;

//typedef void(*Fun)(void) //(C风格)
using Fun = void(*)(void); //(C++11风格)

class Base
public:
	virtual void func1() cout << "virtual func1 is called " << endl; 
	virtual void func2() cout << "virtual func2 is called " << endl; 
	virtual void func3() cout << "virtual func3 is called " << endl; 
;


int main()
	Base b;
	cout << "vfptr指针（内容指向vftable地址）：" <<  (int *)(&b) << endl;
	
	Fun pFun = (Fun)*((int*)*(int*)(&b)+0);  pFun();
	pFun = (Fun)*((int*)*(int*)(&b)+1);  pFun();
	// 可调用对象包装器 
	function<void()> f = (Fun)*(((int*)*(int*)(&b)+2));  f();
	
	
	
	cout << endl;
	
	int a = 10;
	int *ptr = &a;
	int **ptr2 = &ptr;
	
	cout << "&a = " << &a << endl;
	cout << "ptr = "<< ptr << endl;
	cout << "&ptr = "<< &ptr << endl;
	cout << "*ptr = "<< *ptr << endl;
	
	cout << endl;
	cout << "ptr2 = " << ptr2 << endl;
	cout << "&ptr2 = "<< &ptr2 << endl;
	cout << "*ptr2 = "<< *ptr2 << endl;
	
	cout << endl;
	cout << "(int *)(ptr2) = " << (int *)(ptr2) << endl;
	cout << "*(int *)(ptr2) = " << *(int *)(ptr2) << endl;
	cout << "(int *)*(int *)(ptr2) = " << (int *)*(int *)(ptr2) << endl;
	cout << "*(int *)*(int *)(ptr2) = " << *(int *)*(int *)(ptr2) << endl;
	
	return 0; 	
 

输出如下：

vfptr指针（内容指向vftable地址）：0x6afee8
virtual func1 is called
virtual func2 is called
virtual func3 is called

&a = 0x6afee4
ptr = 0x6afee4
&ptr = 0x6afee0
*ptr = 10

ptr2 = 0x6afee0
&ptr2 = 0x6afedc
*ptr2 = 0x6afee4

(int *)(ptr2) = 0x6afee0
*(int *)(ptr2) = 7012068
(int *)*(int *)(ptr2) = 0x6afee4
*(int *)*(int *)(ptr2) = 10

可以看出来，类似(int *)*(&b)这样的操作，主要是通过二级指针访问一级指针存储内容 ，上示代码中关于ptr和ptr2的相关操作解释：

1. (int *)(ptr2)是指二级指针中存放的一级指针的地址；
2. * 是取出 一级指针的地址 上的值，也就是一级指针本身存储的值；
3. （int *）是将从一级指针取到内容指针化，因为是一级指针存储的还是一块地址，否则直接按cout输出会输出整型 。