高分或RMB求改写一个C++的函数

[cbxmir]

Algorithm A3A8 -
/* An implementation of the GSM A3A8 algorithm. (Specifically, COMP128.)
*/
/* Copyright 1998, Marc Briceno, Ian Goldberg, and David Wagner.
* All rights reserved.
*/
/*
* For expository purposes only. Coded in C merely because C is a much
* more precise, concise form of expression for these purposes. See Judge
* Patel if you have any problems with this...
* Of course, it's only authentication, so it should be exportable for the
* usual boring reasons.
*/
typedef unsigned char Byte;
#include <stdio.h>
/* #define TEST */

/*
* rand[0..15]: the challenge from the base station
* key[0..15]: the SIM's A3/A8 long-term key Ki
* simoutput[0..11]: what you'd get back if you fed rand and key to a real
* SIM.
*
* The GSM spec states that simoutput[0..3] is SRES,
* and simoutput[4..11] is Kc (the A5 session key).
* (See GSM 11.11, Section 8.16. See also the leaked document
* referenced below.)
* Note that Kc is bits 74..127 of the COMP128 output, followed by 10
* zeros.
* In other words, A5 is keyed with only 54 bits of entropy. This
* represents a deliberate weakening of the key used for voice privacy
* by a factor of over 1000.
*
* Verified with a Pacific Bell Schlumberger SIM. Your mileage may vary.
*
* Marc Briceno <[e-mail]>, Ian Goldberg <[e-mail]>,
* and David Wagner <[e-mail]>
*/
void A3A8(/* in */ Byte rand[16], /* in */ Byte key[16],
/* out */ Byte simoutput[12]);
/* The compression tables. */
static const Byte table_0[512] = {
102,177,186,162, 2,156,112, 75, 55, 25, 8, 12,251,193,246,188,
109,213,151, 53, 42, 79,191,115,233,242,164,223,209,148,108,161,
252, 37,244, 47, 64,211, 6,237,185,160,139,113, 76,138, 59, 70,
67, 26, 13,157, 63,179,221, 30,214, 36,166, 69,152,124,207,116,
247,194, 41, 84, 71, 1, 49, 14, 95, 35,169, 21, 96, 78,215,225,
182,243, 28, 92,201,118, 4, 74,248,128, 17, 11,146,132,245, 48,
149, 90,120, 39, 87,230,106,232,175, 19,126,190,202,141,137,176,
250, 27,101, 40,219,227, 58, 20, 51,178, 98,216,140, 22, 32,121,
61,103,203, 72, 29,110, 85,212,180,204,150,183, 15, 66,172,196,
56,197,158, 0,100, 45,153, 7,144,222,163,167, 60,135,210,231,
174,165, 38,249,224, 34,220,229,217,208,241, 68,206,189,125,255,
239, 54,168, 89,123,122, 73,145,117,234,143, 99,129,200,192, 82,
104,170,136,235, 93, 81,205,173,236, 94,105, 52, 46,228,198, 5,
57,254, 97,155,142,133,199,171,187, 50, 65,181,127,107,147,226,
184,218,131, 33, 77, 86, 31, 44, 88, 62,238, 18, 24, 43,154, 23,
80,159,134,111, 9,114, 3, 91, 16,130, 83, 10,195,240,253,119,
177,102,162,186,156, 2, 75,112, 25, 55, 12, 8,193,251,188,246,
213,109, 53,151, 79, 42,115,191,242,233,223,164,148,209,161,108,
37,252, 47,244,211, 64,237, 6,160,185,113,139,138, 76, 70, 59,
26, 67,157, 13,179, 63, 30,221, 36,214, 69,166,124,152,116,207,
194,247, 84, 41, 1, 71, 14, 49, 35, 95, 21,169, 78, 96,225,215,
243,182, 92, 28,118,201, 74, 4,128,248, 11, 17,132,146, 48,245,
90,149, 39,120,230, 87,232,106, 19,175,190,126,141,202,176,137,
27,250, 40,101,227,219, 20, 58,178, 51,216, 98, 22,140,121, 32,
103, 61, 72,203,110, 29,212, 85,204,180,183,150, 66, 15,196,172,
197, 56, 0,158, 45,100, 7,153,222,144,167,163,135, 60,231,210,
165,174,249, 38, 34,224,229,220,208,217, 68,241,189,206,255,125,
54,239, 89,168,122,123,145, 73,234,117, 99,143,200,129, 82,192,
170,104,235,136, 81, 93,173,205, 94,236, 52,105,228, 46, 5,198,
254, 57,155, 97,133,142,171,199, 50,187,181, 65,107,127,226,147,
218,184, 33,131, 86, 77, 44, 31, 62, 88, 18,238, 43, 24, 23,154,
159, 80,111,134,114, 9, 91, 3,130, 16, 10, 83,240,195,119,253
}, table_1[256] = {
19, 11, 80,114, 43, 1, 69, 94, 39, 18,127,117, 97, 3, 85, 43,
27,124, 70, 83, 47, 71, 63, 10, 47, 89, 79, 4, 14, 59, 11, 5,
35,107,103, 68, 21, 86, 36, 91, 85,126, 32, 50,109, 94,120, 6,
53, 79, 28, 45, 99, 95, 41, 34, 88, 68, 93, 55,110,125,105, 20,
90, 80, 76, 96, 23, 60, 89, 64,121, 56, 14, 74,101, 8, 19, 78,
76, 66,104, 46,111, 50, 32, 3, 39, 0, 58, 25, 92, 22, 18, 51,
57, 65,119,116, 22,109, 7, 86, 59, 93, 62,110, 78, 99, 77, 67,
12,113, 87, 98,102, 5, 88, 33, 38, 56, 23, 8, 75, 45, 13, 75,
95, 63, 28, 49,123,120, 20,112, 44, 30, 15, 98,106, 2,103, 29,
82,107, 42,124, 24, 30, 41, 16,108,100,117, 40, 73, 40, 7,114,
82,115, 36,112, 12,102,100, 84, 92, 48, 72, 97, 9, 54, 55, 74,
113,123, 17, 26, 53, 58, 4, 9, 69,122, 21,118, 42, 60, 27, 73,
118,125, 34, 15, 65,115, 84, 64, 62, 81, 70, 1, 24,111,121, 83,
104, 81, 49,127, 48,105, 31, 10, 6, 91, 87, 37, 16, 54,116,126,
31, 38, 13, 0, 72,106, 77, 61, 26, 67, 46, 29, 96, 37, 61, 52,
101, 17, 44,108, 71, 52, 66, 57, 33, 51, 25, 90, 2,119,122, 35
}, table_2[128] = {
52, 50, 44, 6, 21, 49, 41, 59, 39, 51, 25, 32, 51, 47, 52, 43,
37, 4, 40, 34, 61, 12, 28, 4, 58, 23, 8, 15, 12, 22, 9, 18,
55, 10, 33, 35, 50, 1, 43, 3, 57, 13, 62, 14, 7, 42, 44, 59,
62, 57, 27, 6, 8, 31, 26, 54, 41, 22, 45, 20, 39, 3, 16, 56,
48, 2, 21, 28, 36, 42, 60, 33, 34, 18, 0, 11, 24, 10, 17, 61,
29, 14, 45, 26, 55, 46, 11, 17, 54, 46, 9, 24, 30, 60, 32, 0,
20, 38, 2, 30, 58, 35, 1, 16, 56, 40, 23, 48, 13, 19, 19, 27,
31, 53, 47, 38, 63, 15, 49, 5, 37, 53, 25, 36, 63, 29, 5, 7
}, table_3[64] = {
1, 5, 29, 6, 25, 1, 18, 23, 17, 19, 0, 9, 24, 25, 6, 31,
28, 20, 24, 30, 4, 27, 3, 13, 15, 16, 14, 18, 4, 3, 8, 9,
20, 0, 12, 26, 21, 8, 28, 2, 29, 2, 15, 7, 11, 22, 14, 10,
17, 21, 12, 30, 26, 27, 16, 31, 11, 7, 13, 23, 10, 5, 22, 19
}, table_4[32] = {
15, 12, 10, 4, 1, 14, 11, 7, 5, 0, 14, 7, 1, 2, 13, 8,
10, 3, 4, 9, 6, 0, 3, 2, 5, 6, 8, 9, 11, 13, 15, 12
}, *table[5] = { table_0, table_1, table_2, table_3, table_4 };
/*

代码未完.接下楼
==========================
愿意支付RMB20.00或300分,限第一个回答并编译通过没问题的.

[cbxmir]

* This code derived from a leaked document from the GSM standards.
* Some missing pieces were filled in by reverse-engineering a working SIM.
* We have verified that this is the correct COMP128 algorithm.
*
* The first page of the document identifies it as
* _Technical Information: GSM System Security Study_.
* 10-1617-01, 10th June 1988.
* The bottom of the title page is marked
* Racal Research Ltd.
* Worton Drive, Worton Grange Industrial Estate,
* Reading, Berks. RG2 0SB, England.
* Telephone: Reading [phone] Telex: 847152
* The relevant bits are in Part I, Section 20 (pages 66--67). Enjoy!
*
* Note: There are three typos in the spec (discovered by
* reverse-engineering).
* First, "z = (2 * x[n] + x[n]) mod 2^(9-j)" should clearly read
* "z = (2 * x[m] + x[n]) mod 2^(9-j)".
* Second, the "k" loop in the "Form bits from bytes" section is severely
* botched: the k index should run only from 0 to 3, and clearly the range
* on "the (8-k)th bit of byte j" is also off (should be 0..7, not 1..8,
* to be consistent with the subsequent section).
* Third, SRES is taken from the first 8 nibbles of x[], not the last 8 as
* claimed in the document. (And the document doesn't specify how Kc is
* derived, but that was also easily discovered with reverse engineering.)
* All of these typos have been corrected in the following code.
*/
void A3A8(/* in */ Byte rand[16], /* in */ Byte key[16],
/* out */ Byte simoutput[12])
{
Byte x[32], bit[128];
int i, j, k, l, m, n, y, z, next_bit;
/* ( Load RAND into last 16 bytes of input ) */
for (i=16; i<32; i++)
x[i] = rand[i-16];
/* ( Loop eight times ) */
for (i=1; i<9; i++) {
/* ( Load key into first 16 bytes of input ) */
for (j=0; j<16; j++)
x[j] = key[j];
/* ( Perform substitutions ) */
for (j=0; j<5; j++)
for (k=0; k<(1<<j); k++)
for (l=0; l<(1<<(4-j)); l++) {
m = l + k*(1<<(5-j));
n = m + (1<<(4-j));
y = (x[m]+2*x[n]) % (1<<(9-j));
z = (2*x[m]+x[n]) % (1<<(9-j));
x[m] = table[j][y];
x[n] = table[j][z];
}
/* ( Form bits from bytes ) */
for (j=0; j<32; j++)
for (k=0; k<4; k++)
bit[4*j+k] = (x[j]>>(3-k)) & 1;
/* ( Permutation but not on the last loop ) */
if (i < 8)
for (j=0; j<16; j++) {
x[j+16] = 0;
for (k=0; k<8; k++) {
next_bit = ((8*j + k)*17) % 128;
x[j+16] |= bit[next_bit] << (7-k);
}
}
}
/*
* ( At this stage the vector x[] consists of 32 nibbles.
* The first 8 of these are taken as the output SRES. )
*/
/* The remainder of the code is not given explicitly in the
* standard, but was derived by reverse-engineering.
*/
for (i=0; i<4; i++)
simoutput[i] = (x[2*i]<<4) | x[2*i+1];
for (i=0; i<6; i++)
simoutput[4+i] = (x[2*i+18]<<6) | (x[2*i+18+1]<<2)
| (x[2*i+18+2]>>2);
simoutput[4+6] = (x[2*6+18]<<6) | (x[2*6+18+1]<<2);
simoutput[4+7] = 0;
}

#ifdef TEST
int hextoint(char x)
{
x = toupper(x);
if (x >= 'A' && x <= 'F')
return x-'A'+10;
else if (x >= '0' && x <= '9')
return x-'0';
fprintf(stderr, "bad input.\n");
exit(1);
}
int main(int argc, char **argv)
{
Byte key[16], rand[16], simoutput[12];
int i;
if (argc != 3 || strlen(argv[1]) != 34 || strlen(argv[2]) != 34
|| strncmp(argv[1], "0x", 2) != 0
|| strncmp(argv[2], "0x", 2) != 0) {
fprintf(stderr, "Usage: %s 0x<key> 0x<rand>\n", argv[0]);
exit(1);
}
for (i=0; i<16; i++)
key[i] = (hextoint(argv[1][2*i+2])<<4)
| hextoint(argv[1][2*i+3]);
for (i=0; i<16; i++)
rand[i] = (hextoint(argv[2][2*i+2])<<4)
| hextoint(argv[2][2*i+3]);
A3A8(key, rand, simoutput);
printf("simoutput: ");
for (i=0; i<12; i++)
printf("%02X", simoutput[i]);
printf("\n");
return 0;
}
#endif

[蒋晟]

自己不会写DLL么？

[rainstormmaster]

是啊,直接输出dll,供VB调用即可,如果一定要改成vb的话,回帖

[laviewpbt]

这种代码该起来有什么难度啊，只要你有点C的基础就可以看懂。

[WM_JAWIN]

很好玩,难度不高.俺试试

[cbxmir]

看是看的懂.但是一堆FOR循环那把我搞晕了.我自己改了2次计算出来都不对.还是要靠大家了.
这个是手机的SIM卡的A3A8加密算法,破解手机卡的KI用的.我也是爱好玩玩的.

[cbxmir]

继续?没有朋友改写么?

[VirtualDesktop]

晕死，你的代码那么长，谁那么有闲情改。。。
还有，代码里面有移位运算，你确保你的移位算法没错吗？

[cbxmir]

to:VirtualDesktop(^_^)

我确保没错。

[guyehanxinlei]

路过，顶一下-_-

[kmlxk0]

VirtualDesktop 是这方面的高手啊....

[cbxmir]

如果需要什么报酬可以提出来。这个我也是爱好在搞的。破解手机的SIM卡。
通过A3A8算法随机与Key计算。如果运算的结果产生碰撞。则可以求出一组KEY。在通过其他的工具就可以求出SIM全部的KI。就可以把这张SIM卡复制了。

01IP.COM
有几个人搞出来。但是都不愿意公布方法。思路都不给。但都是VB写的。现有工具已经能解开80%的移动手机卡了。
我想自己实验下搞出来。但是一定要有这个算法。所以请大家帮忙了。

[cbxmir]

怎么CSDN现在问东西都好难啊。

[xie_jian_jun]

'由于没太多时间,只给你一个移位操作的函数,请指教
'Bitwise shift direction enumeration
Public Enum SHIFT_DIR
LEFT_SHIFT
RIGHT_SHIFT
End Enum


'The return value of this function should be as a unsigned long type value(ULONG)
'Bitwise Shift function
Public Function ShiftBit(ByVal value As Long, ByVal ShiftNum As Integer, ByVal ShiftDir As SHIFT_DIR) As Currency

Static BitArr(0 To 31) As Byte
Dim i As Integer
Dim lVal As Currency

'Initialize the bit array
For i = 0 To 31
BitArr(i) = 0
Next

'get each bit and put it into the array
For i = 0 To 30
If (value And (2 ^ i)) > 0 Then
BitArr(i) = 1
Else
BitArr(i) = 0
End If
Debug.Print BitArr(i);
Next

'to avoid overflow when execute the AND operation with number which is larger than 2147483648(2^31)
If value >= 2147483648# Or value < 0 Then
BitArr(31) = 1
Else
BitArr(31) = 0
End If
Debug.Print BitArr(31)

'execute bitwise shift operation
If (ShiftNum > 31 Or ShiftNum < 1) Then
ShiftBit = 0
Exit Function
End If

Select Case ShiftDir
Case LEFT_SHIFT
For i = 31 To ShiftNum Step -1
BitArr(i) = BitArr(i - ShiftNum)
Next
For i = 0 To ShiftNum - 1
BitArr(i) = 0
Next
Case RIGHT_SHIFT
For i = 0 To 31 - ShiftNum
BitArr(i) = BitArr(i + ShiftNum)
Next
For i = 32 - ShiftNum To 31
BitArr(i) = 0
Next
Case Else
ShiftBit = value
Exit Function
End Select

'make each bit in bitarr to be a long value and return it as a result
lVal = 0
For i = 0 To 31
Debug.Print BitArr(i);
lVal = lVal + (2 ^ i) * BitArr(i)
Next

ShiftBit = lVal
End Function

'Example
'Dim a As Long
'a = ShiftBit(12345678, 15, LEFT_SHIFT)


'Call ShiftBit(12345678, 15, LEFT_SHIFT) in visual basic
'equal to execute the statement "(unsigned long)(12345678 << 15)" in visual c++
'And call ShiftBit(12345678, 15, RIGHT_SHIFT) in visual basic
'equal to execute the statement "(unsigned long)(12345678 >> 15)" in visual c++

[xie_jian_jun]

更正一下上面的代码,将
If (ShiftNum > 31 Or ShiftNum < 1) Then
ShiftBit = 0
Exit Function
End If
更正为
If (ShiftNum > 31 Or ShiftNum < 0) Then
ShiftBit = 0
Exit Function
End If