vc socket的接收端解析一串字符串解决方案

vc socket的接收端解析一串字符串
字符串例子如下：原始数据例子:
02 18 00 4C 4A 49 41 4A 01 27 0C 22 38 4E 0A 5E A1 13 1F 3D 4C 48 40 FF 01 00 00 03

包头: 02

数据长度: 18 00 = 00 18 = 10进制= 24s

命令类型:4C

数据内容:4A 49 41 4A 01 27 0C 22 38 4E 0A 5E A1 13 1F 3D 4C 48 40 4E

终端编号: 4A 49 41 4A = 74736574

SIM卡号: 01 27 0C 22 38 4E = 013912345678

终端 IP : 0A 5E A1 13 =10.94.161.19

信号质量: 1F = 31 

司机工号: 3D 4C 48 40 4E = 61767264
LED版本号:FF 01 00 00 = 00 00 FF 01 = 10 进制化=511
包尾:03
大家有什么好的意见，想法，给俺指点指点，谢谢大家啦！

[解决办法]
慢慢按长度一段段截取
字符串转16进制
比较繁杂罢了
===============
16进制转换：

C/C++ code

#include <iostream.h>#include <afxwin.h>int HexstringToInt(CString result);void main(){        CString str = "01 0E 7B 22 18";        CString temp1 = str.Mid(6,2);    CString temp2 = str.Mid(9,2);    CString temp3 = str.Right(2);    CString result = temp3 + temp2 + temp1;        cout<< str <<endl;    cout<< result <<endl;         result = "14aFDEC8";    cout<< HexstringToInt(result) <<endl;    }int HexstringToInt(CString result){    int val = 0;    int sum = 0;    for (int i=0;i<result.GetLength();i++)    {        char ch = result.GetAt(i);        if (ch>=65&&ch<=70)        {            val = ch-55;        }                else if (ch>=97&&ch<=102)        {            val = ch-87;        }                else if (ch>=48&&ch<=57)        {            val = ch-48;        }        else        {            cout<< "ERROR Prameter" <<endl;            sum = 0 ;            break;                    }                val = val<<((result.GetLength()-1-i)*4);        sum += val;            }    return sum;}
[解决办法]
不知道有多少前人掉在TCP Socket
send(人多)send(病少)send(财富)
recv(人多病)recv(少财富)
陷阱里面啊！
http://topic.csdn.net/u/20120210/09/51109ed0-07b9-41f2-b487-a51597f2ca01.html