将数据拼接和分解的DWORD与char*.

想将 first转为char*但是又不想将它转为char* 的ASCII表示，想以内存原样转过去。
是不是有哪些内存复制的方法可以实现的呢？

class data{
private:
   DWORD first;
   char * second;
   char * combine;
public:
   void combineBoth();   // combineBoth()里面实现的是 combine = first + second 的数据。
   void separateEither();  // separateEither()里面实现的是　first = combine(0~3), second = combine(4~n);
};

实际上就是想将内存地址获取一下，最好是能combine的内存地址的下几个位置就到了second的内存地址中去了。

 

解决问题的源代码：

#include <iostream>
#include <Windows.h>
#include <string.h>
using namespace std;

class data
{
public:
   DWORD first;     // 四个字节
   char * second;
   char * combine;
   char * getSecond;
public:
   data();
   void combineBoth();   // combineBoth()里面实现的是 combine = first + second 的数据。
   void separateEither();  // separateEither()里面实现的是　first = combine(0~3), second = combine(4~n);
   void displayAllData();
};

data::data(){
    first = 987654321;
    second = new char[10];
    combine = new char[14];
    getSecond = combine + 4;
    const char * temp = "abcdefghi";    // 这里会有结束符9+1个字符
    strcpy(second, temp);
}

void data::combineBoth(){
    memcpy( (void*)combine, (void *) &first, 4);
    memcpy( (void*)(combine+4), (void*)second, 10);
}

void data::separateEither(){
    memcpy( (void *) &first, (void*)combine,  4);
    memcpy( (void*)second, (void*)(combine+4),  10);
}

void data::displayAllData()
{   cout << "first:" << first << endl;
    cout << "second:" << second << endl;
    cout << "combine:";
    for (int i = 0; i < 14; ++i){
        cout << *(char*)(combine+i);
    }
    cout << endl;
    cout << "getSecond:" << getSecond << endl;
}


int main(){
    cout << "初始化\n";
    data * myData = new data();
    myData -> displayAllData();

    cout << "\n合并\n";
    myData -> combineBoth();
    myData -> displayAllData();

    cout << "\n分解\n";
    myData->first = 0;
    myData -> separateEither();
    myData -> displayAllData();

    system("pause");
    return 0;
}

运行结果：

初始化

first:987654321
second:abcdefghi
combine:屯屯屯屯屯屯屯
getSecond:屯屯屯屯屯������������铪铪铪

合并

first:987654321
second:abcdefghi
combine:�h?abcdefghi
getSecond:abcdefghi

分解

first:987654321
second:abcdefghi
combine:�h?abcdefghi
getSecond:abcdefghi

我的提问网址：http://topic.csdn.net/u/20120908/09/bdbb6f55-58bf-4425-ae41-0f69b7963933.html?75984