#include "CXKUsername.h"
#include "MD5.h"
#include<cstring> CXKUsername::CXKUsername(CString username, INT ver, long lasttimec)
:m_username(username),RADIUS("chongqingradius1"),LR("\r\n")
{
m_ver = ver;
m_lasttimec = lasttimec;
} long CXKUsername::GetLastTimeC()
{
return m_lasttimec;
}
CString CXKUsername::Realusername()
{
time_t m_time; //得到系统时间，从1970.01.01.00:00:00 开始的秒数
long m_time1c; //时间初处理m_time1c为结果,经过时间计算出的第一次加密
long m_time1convert; //对时间操作后的结果，此为格式字串的原始数据
unsigned char ss[4] =
{
0,0,0,0
}; //源数据1,对m_time1convert进行计算得到格式符源数据
unsigned char ss2[4] =
{
0,0,0,0
}; //md5加密参数的一部分,m_time1c的字符形式
CString strS1; //md5加密参数的一部分,ss2的整体形式
CString m_formatsring; //由m_timece算出的字符串,一般为可视字符
CString m_md5; //对初加密(m_timec字符串表示+m_username+radius)的MD5加密
CString m_md5use; //md5 Lower模式的前两位 //取得系统时间m_time
time(&m_time);
//时间初处理m_time1c为结果,经过时间计算出的第一次加密
//子函数////////////////////////////
{
LONG64 t;
t = m_time;
t *= 0x66666667;
t >>= 0x20;
t >>= 0x01;
m_time1c = (long) t;
}
//5秒内动态用户名一致处理
if (m_time1c <= m_lasttimec)
{
m_time1c = m_lasttimec + 1;
}
m_lasttimec = m_time1c;
{
long t;
t = m_time1c;
ss2[3] = (t & 0xFF);
ss2[2] = (t & 0xFF00) / 0x100 ;
ss2[1] = (t & 0xFF0000) / 0x10000;
ss2[0] = (t & 0xFF000000) / 0x1000000;
{
//strS1必须用自加得到，真接加出问题
for (int i = 0; i < 4; i++)
{
strS1 += ss2[i];
}
}
} /////////////////////////////////////
//倒置过程m_time1convert为结果
//子函数////////////////////////////
{
int t, t1, t2, t3;
t = m_time1c;
t1 = t;
t2 = t;
t3 = t;
t3 = t3 << 0x10;
t1 = t1 & 0x0FF00;
t1 = t1 | t3;
t3 = t;
t3 = t3 & 0x0FF0000;
t2 = t2 >> 0x10;
t3 = t3 | t2;
t1 = t1 << 0x08;
t3 = t3 >> 0x08;
t1 = t1 | t3;
m_time1convert = t1;
}
///////////////////////////////////// //源数据1,对m_time1convert进行计算得到格式符源数据
//子函数////////////////////////////
{
long t;
t = m_time1convert;
ss[3] = (t & 0xFF);
ss[2] = (t & 0xFF00) / 0x100 ;
ss[1] = (t & 0xFF0000) / 0x10000;
ss[0] = (t & 0xFF000000) / 0x1000000;
}
///////////////////////////////////// //格式符初加密
unsigned char pp[4] =
{
0,0,0,0
};
//子函数////////////////////////////
{
int i = 0, j = 0, k = 0;
for (i = 0; i < 0x20; i++)
{
j = i / 0x8;
k = 3 - (i % 0x4);
pp[k] *= 0x2;
if (ss[j] % 2 == 1)
{
pp[k]++;
}
ss[j] /= 2;
}