A5算法的C语言,C语言程序,流密码,序列密码

/*

* In writing this program, I've had to guess a few pices of information:

*

* 1. Which bits of the key are loaded into which bits of the shift register

* 2. Which order the frame sequence number is shifted into the SR (MSB

* first or LSB first)

* 3. The position of the feedback taps on R2 and R3 (R1 is known).

* 4. The position of the clock control taps. These are on the `middle' one,

* I've assumed to be 9 on R1, 11 on R2, 11 on R3.

*/

/*

* Look at the `middle' stage of each of the 3 shift registers.

* Either 0, 1, 2 or 3 of these 3 taps will be set high.

* If 0 or 1 or one of them are high, return true. This will cause each of

* the middle taps to be inverted before being used as a clock control. In

* all cases either 2 or 3 of the clock enable lines will be active. Thus,

* at least two shift registers change on every clock-tick and the system

* never becomes stuck.

*/

typedef struct

{

unsigned long r1, r2, r3;

}a5_ctx

static int threshold(r1, r2, r3)

unsigned int r1;

unsigned int r2;

unsigned int r3;

{

int total;

total = (((r1 >> 9) & 0x1) == 1) +

(((r2 >> 11) & 0x1) == 1) +

(((r3 >> 11) & 0x1) == 1);

if (total > 1)

return (0);

else

return (1);

}

unsigned long clock_r1(ctl, r1)

int ctl;

unsigned long r1;

{

unsigned long feedback;

/*

* Primitive polynomial x**19 + x**5 + x**2 + x + 1

*/

ctl ^= ((r1 >> 9) & 0x1);

if (ctl)

{

feedback = (r1 >> 18) ^ (r1 >> 17) ^ (r1 >> 16) ^ (r1 >> 13);

r1 = (r1 << 1) & 0x7ffff;

if (feedback & 0x01)

r1 ^= 0x01;

}

return (r1);

}

unsigned long clock_r2(ctl, r2)

int ctl;

unsigned long r2;

{

unsigned long feedback;

/*

* Primitive polynomial x**22 + x**9 + x**5 + x + 1

*/

ctl ^= ((r2 >> 11) & 0x1);

if (ctl)

{

feedback = (r2 >> 21) ^ (r2 >> 20) ^ (r2 >> 16) ^ (r2 >> 12);

r2 = (r2 << 1) & 0x3fffff;

if (feedback & 0x01)

r2 ^= 0x01;

}

return (r2);

}

unsigned long clock_r3(ctl, r3)

int ctl;

unsigned long r3;

{

unsigned long feedback;

/*

* Primitive polynomial x**23 + x**5 + x**4 + x + 1

*/

ctl ^= ((r3 >> 11) & 0x1);

if (ctl)

{

feedback = (r3 >> 22) ^ (r3 >> 21) ^ (r3 >> 18) ^ (r3 >> 17);

r3 = (r3 << 1) & 0x7fffff;

if (feedback & 0x01)

r3 ^= 0x01;

}

return (r3);

}

int keystream(key, frame, alice, bob)

unsigned char *key; /* 64 bit session key */

unsigned long frame; /* 22 bit frame sequence number */

unsigned char *alice; /* 114 bit Alice to Bob key stream */

unsigned char *bob; /* 114 bit Bob to Alice key stream */

{

unsigned long r1; /* 19 bit shift register */

unsigned long r2; /* 22 bit shift register */

unsigned long r3; /* 23 bit shift register */

int i;

/* counter for loops */

int clock_ctl; /* xored with clock enable on each shift register */

unsigned char *ptr; /* current posit