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[求助]请教一个TrueCrypt源码中serpent算法的一个简单问题。
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2012-2-21 11:39 3048
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请教一下下面代码函数中,ks 是做什么用的?
void serpent_set_key(const unsigned __int8 userKey[], int keylen, unsigned __int8 *ks)
{
unsigned __int32 a,b,c,d,e;
unsigned __int32 *k = (unsigned __int32 *)ks;
unsigned __int32 t;
int i;
for (i = 0; i < keylen / (int)sizeof(__int32); i++)
k[i] = LE32(((unsigned __int32*)userKey)[i]);
if (keylen < 32)
k[keylen/4] |= (unsigned __int32)1 << ((keylen%4)*8);
k += 8;
t = k[-1];
for (i = 0; i < 132; ++i)
k[i] = t = rotlFixed(k[i-8] ^ k[i-5] ^ k[i-3] ^ t ^ 0x9e3779b9 ^ i, 11);
k -= 20;
#define LK(r, a, b, c, d, e) {\
a = k[(8-r)*4 + 0]; \
b = k[(8-r)*4 + 1]; \
c = k[(8-r)*4 + 2]; \
d = k[(8-r)*4 + 3];}
#define SK(r, a, b, c, d, e) {\
k[(8-r)*4 + 4] = a; \
k[(8-r)*4 + 5] = b; \
k[(8-r)*4 + 6] = c; \
k[(8-r)*4 + 7] = d;} \
for (i=0; i<4; i++)
{
afterS2(LK); afterS2(S3); afterS3(SK);
afterS1(LK); afterS1(S2); afterS2(SK);
afterS0(LK); afterS0(S1); afterS1(SK);
beforeS0(LK); beforeS0(S0); afterS0(SK);
k += 8*4;
afterS6(LK); afterS6(S7); afterS7(SK);
afterS5(LK); afterS5(S6); afterS6(SK);
afterS4(LK); afterS4(S5); afterS5(SK);
afterS3(LK); afterS3(S4); afterS4(SK);
}
afterS2(LK); afterS2(S3); afterS3(SK);
}
void serpent_set_key(const unsigned __int8 userKey[], int keylen, unsigned __int8 *ks)
{
unsigned __int32 a,b,c,d,e;
unsigned __int32 *k = (unsigned __int32 *)ks;
unsigned __int32 t;
int i;
for (i = 0; i < keylen / (int)sizeof(__int32); i++)
k[i] = LE32(((unsigned __int32*)userKey)[i]);
if (keylen < 32)
k[keylen/4] |= (unsigned __int32)1 << ((keylen%4)*8);
k += 8;
t = k[-1];
for (i = 0; i < 132; ++i)
k[i] = t = rotlFixed(k[i-8] ^ k[i-5] ^ k[i-3] ^ t ^ 0x9e3779b9 ^ i, 11);
k -= 20;
#define LK(r, a, b, c, d, e) {\
a = k[(8-r)*4 + 0]; \
b = k[(8-r)*4 + 1]; \
c = k[(8-r)*4 + 2]; \
d = k[(8-r)*4 + 3];}
#define SK(r, a, b, c, d, e) {\
k[(8-r)*4 + 4] = a; \
k[(8-r)*4 + 5] = b; \
k[(8-r)*4 + 6] = c; \
k[(8-r)*4 + 7] = d;} \
for (i=0; i<4; i++)
{
afterS2(LK); afterS2(S3); afterS3(SK);
afterS1(LK); afterS1(S2); afterS2(SK);
afterS0(LK); afterS0(S1); afterS1(SK);
beforeS0(LK); beforeS0(S0); afterS0(SK);
k += 8*4;
afterS6(LK); afterS6(S7); afterS7(SK);
afterS5(LK); afterS5(S6); afterS6(SK);
afterS4(LK); afterS4(S5); afterS5(SK);
afterS3(LK); afterS3(S4); afterS4(SK);
}
afterS2(LK); afterS2(S3); afterS3(SK);
}
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