-1
，我发现这个代码在网上和IM试图编译它，但GCC不断告诉我，有一个undefined reference to [email protected]。我不知道它来自哪里，所以即将发布整个代码。在网上搜索后，我只找到答案，在那里人们写错了int main()，并修复它使其工作。这在这个代码中起作用。的WinMain @ 16错误
/*
* The GOST 28147-89 cipher
*
* This is based on the 25 Movember 1993 draft translation
* by Aleksandr Malchik, with Whitfield Diffie, of the Government
* Standard of the U.S.S.R. GOST 28149-89, "Cryptographic Transformation
* Algorithm", effective 1 July 1990. ([email protected])
*
* That is a draft, and may contain errors, which will be faithfully
* reflected here, along with possible exciting new bugs.
*
* Some details have been cleared up by the paper "Soviet Encryption
* Algorithm" by Josef Pieprzyk and Leonid Tombak of the University
* of Wollongong, New South Wales. (josef/[email protected])
*
* The standard is written by A. Zabotin (project leader), G.P. Glazkov,
* and V.B. Isaeva. It was accepted and introduced into use by the
* action of the State Standards Committee of the USSR on 2 June 89 as
* No. 1409. It was to be reviewed in 1993, but whether anyone wishes
* to take on this obligation from the USSR is questionable.
*
* This code is placed in the public domain.
*/
/*
* If you read the standard, it belabors the point of copying corresponding
* bits from point A to point B quite a bit. It helps to understand that
* the standard is uniformly little-endian, although it numbers bits from
* 1 rather than 0, so bit n has value 2^(n-1). The least significant bit
* of the 32-bit words that are manipulated in the algorithm is the first,
* lowest-numbered, in the bit string.
*/
/* A 32-bit data type */
#ifdef __alpha /* Any other 64-bit machines? */
typedef unsigned int word32;
#else
typedef unsigned long word32;
#endif
/*
* The standard does not specify the contents of the 8 4 bit->4 bit
* substitution boxes, saying they're a parameter of the network
* being set up. For illustration purposes here, I have used
* the first rows of the 8 S-boxes from the DES. (Note that the
* DES S-boxes are numbered starting from 1 at the msb. In keeping
* with the rest of the GOST, I have used little-endian numbering.
* Thus, k8 is S-box 1.
*
* Obviously, a careful look at the cryptographic properties of the cipher
* must be undertaken before "production" substitution boxes are defined.
*
* The standard also does not specify a standard bit-string representation
* for the contents of these blocks.
*/
static unsigned char const k8[16] = {
14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7 };
static unsigned char const k7[16] = {
15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10 };
static unsigned char const k6[16] = {
10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8 };
static unsigned char const k5[16] = {
7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15 };
static unsigned char const k4[16] = {
2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9 };
static unsigned char const k3[16] = {
12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11 };
static unsigned char const k2[16] = {
4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1 };
static unsigned char const k1[16] = {
13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7 };
/* Byte-at-a-time substitution boxes */
static unsigned char k87[256];
static unsigned char k65[256];
static unsigned char k43[256];
static unsigned char k21[256];
/*
* Build byte-at-a-time subtitution tables.
* This must be called once for global setup.
*/
void
kboxinit(void)
{
int i;
for (i = 0; i < 256; i++) {
k87[i] = k8[i >> 4] << 4 | k7[i & 15];
k65[i] = k6[i >> 4] << 4 | k5[i & 15];
k43[i] = k4[i >> 4] << 4 | k3[i & 15];
k21[i] = k2[i >> 4] << 4 | k1[i & 15];
}
}
/*
* Do the substitution and rotation that are the core of the operation,
* like the expansion, substitution and permutation of the DES.
* It would be possible to perform DES-like optimisations and store
* the table entries as 32-bit words, already rotated, but the
* efficiency gain is questionable.
*
* This should be inlined for maximum speed
*/
#if __GNUC__
__inline__
#endif
static word32
f(word32 x)
{
/* Do substitutions */
#if 0
/* This is annoyingly slow */
x = k8[x>>28 & 15] << 28 | k7[x>>24 & 15] << 24 |
k6[x>>20 & 15] << 20 | k5[x>>16 & 15] << 16 |
k4[x>>12 & 15] << 12 | k3[x>> 8 & 15] << 8 |
k2[x>> 4 & 15] << 4 | k1[x & 15];
#else
/* This is faster */
x = k87[x>>24 & 255] << 24 | k65[x>>16 & 255] << 16 |
k43[x>> 8 & 255] << 8 | k21[x & 255];
#endif
/* Rotate left 11 bits */
return x<<11 | x>>(32-11);
}
/*
* The GOST standard defines the input in terms of bits 1..64, with
* bit 1 being the lsb of in[0] and bit 64 being the msb of in[1].
*
* The keys are defined similarly, with bit 256 being the msb of key[7].
*/
void
gostcrypt(word32 const in[2], word32 out[2], word32 const key[8])
{
register word32 n1, n2; /* As named in the GOST */
n1 = in[0];
n2 = in[1];
/* Instead of swapping halves, swap names each round */
n2 ^= f(n1+key[0]);
n1 ^= f(n2+key[1]);
n2 ^= f(n1+key[2]);
n1 ^= f(n2+key[3]);
n2 ^= f(n1+key[4]);
n1 ^= f(n2+key[5]);
n2 ^= f(n1+key[6]);
n1 ^= f(n2+key[7]);
n2 ^= f(n1+key[0]);
n1 ^= f(n2+key[1]);
n2 ^= f(n1+key[2]);
n1 ^= f(n2+key[3]);
n2 ^= f(n1+key[4]);
n1 ^= f(n2+key[5]);
n2 ^= f(n1+key[6]);
n1 ^= f(n2+key[7]);
n2 ^= f(n1+key[0]);
n1 ^= f(n2+key[1]);
n2 ^= f(n1+key[2]);
n1 ^= f(n2+key[3]);
n2 ^= f(n1+key[4]);
n1 ^= f(n2+key[5]);
n2 ^= f(n1+key[6]);
n1 ^= f(n2+key[7]);
n2 ^= f(n1+key[7]);
n1 ^= f(n2+key[6]);
n2 ^= f(n1+key[5]);
n1 ^= f(n2+key[4]);
n2 ^= f(n1+key[3]);
n1 ^= f(n2+key[2]);
n2 ^= f(n1+key[1]);
n1 ^= f(n2+key[0]);
/* There is no swap after the last round */
out[0] = n2;
out[1] = n1;
}
/*
* The key schedule is somewhat different for decryption.
* (The key table is used once forward and three times backward.)
* You could define an expanded key, or just write the code twice,
* as done here.
*/
void
gostdecrypt(word32 const in[2], word32 out[2], word32 const key[8])
{
register word32 n1, n2; /* As named in the GOST */
n1 = in[0];
n2 = in[1];
n2 ^= f(n1+key[0]);
n1 ^= f(n2+key[1]);
n2 ^= f(n1+key[2]);
n1 ^= f(n2+key[3]);
n2 ^= f(n1+key[4]);
n1 ^= f(n2+key[5]);
n2 ^= f(n1+key[6]);
n1 ^= f(n2+key[7]);
n2 ^= f(n1+key[7]);
n1 ^= f(n2+key[6]);
n2 ^= f(n1+key[5]);
n1 ^= f(n2+key[4]);
n2 ^= f(n1+key[3]);
n1 ^= f(n2+key[2]);
n2 ^= f(n1+key[1]);
n1 ^= f(n2+key[0]);
n2 ^= f(n1+key[7]);
n1 ^= f(n2+key[6]);
n2 ^= f(n1+key[5]);
n1 ^= f(n2+key[4]);
n2 ^= f(n1+key[3]);
n1 ^= f(n2+key[2]);
n2 ^= f(n1+key[1]);
n1 ^= f(n2+key[0]);
n2 ^= f(n1+key[7]);
n1 ^= f(n2+key[6]);
n2 ^= f(n1+key[5]);
n1 ^= f(n2+key[4]);
n2 ^= f(n1+key[3]);
n1 ^= f(n2+key[2]);
n2 ^= f(n1+key[1]);
n1 ^= f(n2+key[0]);
out[0] = n2;
out[1] = n1;
}
/*
* The GOST "Output feedback" standard. It seems closer morally
* to the counter feedback mode some people have proposed for DES.
* The avoidance of the short cycles that are possible in OFB seems
* like a Good Thing.
*
* Calling it the stream mode makes more sense.
*
* The IV is encrypted with the key to produce the initial counter value.
* Then, for each output block, a constant is added, modulo 2^32-1
* (0 is represented as all-ones, not all-zeros), to each half of
* the counter, and the counter is encrypted to produce the value
* to XOR with the output.
*
* Len is the number of blocks. Sub-block encryption is
* left as an exercise for the user. Remember that the
* standard defines everything in a little-endian manner,
* so you want to use the low bit of gamma[0] first.
*
* OFB is, of course, self-inverse, so there is only one function.
*/
/* The constants for addition */
#define C1 0x01010104
#define C2 0x01010101
void
gostofb(word32 const *in, word32 *out, int len,
word32 const iv[2], word32 const key[8])
{
word32 temp[2]; /* Counter */
word32 gamma[2]; /* Output XOR value */
/* Compute starting value for counter */
gostcrypt(iv, temp, key);
while (len--) {
temp[0] += C2;
if (temp[0] < C2) /* Wrap modulo 2^32? */
temp[0]++; /* Make it modulo 2^32-1 */
temp[1] += C1;
if (temp[1] < C1) /* Wrap modulo 2^32? */
temp[1]++; /* Make it modulo 2^32-1 */
gostcrypt(temp, gamma, key);
*out++ = *in++^gamma[0];
*out++ = *in++^gamma[1];
}
}
/*
* The CFB mode is just what you'd expect. Each block of ciphertext y[] is
* derived from the input x[] by the following pseudocode:
* y[i] = x[i]^gostcrypt(y[i-1])
* x[i] = y[i]^gostcrypt(y[i-1])
* Where y[-1] is the IV.
*
* The IV is modified in place. Again, len is in *blocks*.
*/
void
gostcfbencrypt(word32 const *in, word32 *out, int len,
word32 iv[2], word32 const key[8])
{
while (len--) {
gostcrypt(iv, iv, key);
iv[0] = *out++ ^= iv[0];
iv[1] = *out++ ^= iv[1];
}
}
void
gostcfbdecrypt(word32 const *in, word32 *out, int len,
word32 iv[2], word32 const key[8])
{
word32 t;
while (len--) {
gostcrypt(iv, iv, key);
t = *out;
*out++ ^= iv[0];
iv[0] = t;
t = *out;
*out++ ^= iv[1];
iv[1] = t;
}
}
/*
* The message suthetication code uses only 16 of the 32 rounds.
* There *is* a swap after the 16th round.
* The last block should be padded to 64 bits with zeros.
* len is the number of *blocks* in the input.
*/
void
gostmac(word32 const *in, int len, word32 out[2], word32 const key[8])
{
register word32 n1, n2; /* As named in the GOST */
n1 = 0;
n2 = 0;
while (len--) {
n1 ^= *in++;
n2 = *in++;
/* Instead of swapping halves, swap names each round */
n2 ^= f(n1+key[0]);
n1 ^= f(n2+key[1]);
n2 ^= f(n1+key[2]);
n1 ^= f(n2+key[3]);
n2 ^= f(n1+key[4]);
n1 ^= f(n2+key[5]);
n2 ^= f(n1+key[6]);
n1 ^= f(n2+key[7]);
n2 ^= f(n1+key[0]);
n1 ^= f(n2+key[1]);
n2 ^= f(n1+key[2]);
n1 ^= f(n2+key[3]);
n2 ^= f(n1+key[4]);
n1 ^= f(n2+key[5]);
n2 ^= f(n1+key[6]);
n1 ^= f(n2+key[7]);
}
out[0] = n1;
out[1] = n2;
}
#ifdef TEST
#include <stdio.h>
#include <stdlib.h>
/* Designed to cope with 15-bit rand() implementations */
#define RAND32 ((word32)rand() << 17^(word32)rand() << 9^rand())
int
main(void)
{
word32 key[8];
word32 plain[2];
word32 cipher[2];
int i, j;
kboxinit();
printf("GOST 21847-89 test driver.\n");
for (i = 0; i < 1000; i++) {
for (j = 0; j < 8; j++)
key[j] = RAND32;
plain[0] = RAND32;
plain[1] = RAND32;
printf("%3d\r", i);
fflush(stdout);
gostcrypt(plain, cipher, key);
for (j = 0; j < 99; j++)
gostcrypt(cipher, cipher, key);
for (j = 0; j < 100; j++)
gostdecrypt(cipher, cipher, key);
if (plain[0] != cipher[0] || plain[1] != cipher[1]) {
fprintf(stderr, "\nError! i = %d\n", i);
return 1;
}
}
printf("All tests passed.\n");
return 0;
}
#endif /* TEST */
我该怎么做才能让这段代码运行？
在mingw32？用'-mno-windows'链接它。 –
@Kerrek的评论给出了不正确和愚蠢的建议。他没有看过代码。 –
我们在1993年有什么Windows版本？自那以后会有什么改变吗？ –