关于void*函数返回

一、

sample

#include<iostream>
using namespace std;

void* test(void* pass)
{
    return pass;
}

int _tmain(int argc, _TCHAR* argv[])
{
    char* a=(char*)malloc(sizeof(char));
    a="a";
    char* b = (char*)malloc(sizeof(char));
    b=(char*)test(a);
    cout<<"b:"<<b<<endl;
    cout<<"a:"<<a<<endl;

    getchar();
    return ;

}

注意  test的返回值为void*  要强制转换成char*指针

二、

#include <iostream>

using namespace std;

class CBase
{
public:
    CBase(int a,long b,char c)
    {
        a=a;
        b=b;
        c=c;
    }
private:
    int a;
    long b;
    char c;
};

int main()
{
    unsigned long num1 = ;
    unsigned long num2 = ;
    unsigned long num3 = ;
    unsigned long num4 = ;

    unsigned long num5 = ;

    CBase *base1 = new CBase(,,'d');
    CBase *base2 = new CBase(,,'c');

    num1 = *((unsigned long*)base1);
    num2 = *((unsigned long*)base2);

    num3 = unsigned long((unsigned long*)base1);
    num4 = unsigned long((unsigned long*)base2);

    cout << "num1:"<< num1<< endl;
    cout << "num2:"<< num2<< endl;

    cout<<endl;

    cout << "num3:"<< num3<< endl;
    cout << "num4:"<< num4<< endl;

    getchar();
    return ;

}

1）由于类的private里的数据排列时 先int，long然后是char.

　　指针Base1和Base2的指针自身的地址是不一样的；指向的内容不一样；

　　但是当指针强制转换成long（32位下是4个字节）型的时候；在再解引用时候；他们指向是分别是Base1类的前四个字节和Base2类的前四个字节；而由于他们的前四个字节

？ 解引用为什么不是10？

用c++的强转