20CN网络安全小组论坛 - 编程破解 - 谁能帮我解读这段C程序？！

论坛: 编程破解标题: 谁能帮我解读这段C程序？！

作者: cloret [cloret]

论坛用户

static unsigned short bytebit[8]        = {
        0200, 0100, 040, 020, 010, 04, 02, 01 };

/* Use the key schedule specified in the Standard (ANSI X3.92-1981).
*/

static unsigned char pc1[56] = {
        56, 48, 40, 32, 24, 16,  8,      0, 57, 49, 41, 33, 25, 17,

         9,  1, 58, 50, 42, 34, 26,     18, 10,  2, 59, 51, 43, 35,
        62, 54, 46, 38, 30, 22, 14,      6, 61, 53, 45, 37, 29, 21,
        13,  5, 60, 52, 44, 36, 28,     20, 12,  4, 27, 19, 11,  3 };

static unsigned char totrot[16] = {
        1,2,4,6,8,10,12,14,15,17,19,21,23,25,27,28 };
void deskey(key, edf)
unsigned char *key;
short edf;
{
        register int i, j, l, m, n;
        unsigned char pc1m[56], pcr[56];
        unsigned long kn[32];

        for ( j = 0; j < 56; j++ ) {
                l = pc1[j];
                m = l & 07;
                pc1m[j] = (key[l >> 3] & bytebit[m]) ? 1 : 0;
                }

        .........................
}

这是数据加密标准的一段源程序
这段功能是实现密匙置换
我想问一下
pclm[j] = (key[1 >> 3] & bytebit[m] ? 1: 0;
是什么意思
其中key []在main()为
char key[8] = {0x01,0x23,0x45,0x67,0x89,0xab,0xcd,0xef};

麻烦各位帮帮忙！！

地主发表时间: 06/18 20:10

回复: oyiboy [oyiboy]

论坛用户

pclm[j] = (key[1 >> 3] & bytebit[m] ? 1: 0;

?:是if ... else ...的快捷方式
即是当(key[1 >> 3] & bytebit[m]=true是返回1，反之返回0

B1层发表时间: 06/19 11:29

回复: 286 [unique]

版主

pclm[j] = (key[1 >> 3] & bytebit[m] ? 1: 0;
l>>3 等价于 l=l/8

pclm[j]=(key[1 >> 3] & bytebit[m] ? 1: 0等价于
if ((key[l>>3] & bytebit[m]) == true)
pclm[j]=1;
else
pclm[j]=0;

B2层发表时间: 06/19 18:02

论坛: 编程破解