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- /* ecb_encrypt.c */
- /* Copyright (C) 1992 Eric Young - see COPYING for more details */
- #include "des_local.h"
- #include "sp.h"
-
- int des_ecb_encrypt(input,output,ks,encrypt)
- des_cblock *input;
- des_cblock *output;
- des_key_schedule ks;
- int encrypt;
- {
- register ulong l0,l1;
- register uchar *in,*out;
- ulong ll[2];
-
- in=(uchar *)input;
- out=(uchar *)output;
- c2l(in,l0);
- c2l(in,l1);
- ll[0]=l0;
- ll[1]=l1;
- des_encrypt(ll,ll,ks,encrypt);
- l0=ll[0];
- l1=ll[1];
- l2c(l0,out);
- l2c(l1,out);
- return(0);
- }
-
- #define D_ENCRYPT(L,R,S) \
- t=(R<<1)|(R>>31); \
- u=(t^s[S ]); \
- t=(t^s[S+1]); \
- t=(t>>4)|(t<<28); \
- L^= des_SPtrans[1][(t )&0x3f]| \
- des_SPtrans[3][(t>> 8)&0x3f]| \
- des_SPtrans[5][(t>>16)&0x3f]| \
- des_SPtrans[7][(t>>24)&0x3f]| \
- des_SPtrans[0][(u )&0x3f]| \
- des_SPtrans[2][(u>> 8)&0x3f]| \
- des_SPtrans[4][(u>>16)&0x3f]| \
- des_SPtrans[6][(u>>24)&0x3f];
-
- /* IP and FP
- * The problem is more of a geometric problem that random bit fiddling.
- 0 1 2 3 4 5 6 7 62 54 46 38 30 22 14 6
- 8 9 10 11 12 13 14 15 60 52 44 36 28 20 12 4
- 16 17 18 19 20 21 22 23 58 50 42 34 26 18 10 2
- 24 25 26 27 28 29 30 31 to 56 48 40 32 24 16 8 0
-
- 32 33 34 35 36 37 38 39 63 55 47 39 31 23 15 7
- 40 41 42 43 44 45 46 47 61 53 45 37 29 21 13 5
- 48 49 50 51 52 53 54 55 59 51 43 35 27 19 11 3
- 56 57 58 59 60 61 62 63 57 49 41 33 25 17 9 1
-
- The output has been subject to swaps of the form
- 0 1 -> 3 1 but the odd and even bits have been put into
- 2 3 2 0
- different words. The main trick is to remember that
- t=((l>>size)^r)&(mask);
- r^=t;
- l^=(t<<size);
- can be used to swap and move bits between words.
-
- So l = 0 1 2 3 r = 16 17 18 19
- 4 5 6 7 20 21 22 23
- 8 9 10 11 24 25 26 27
- 12 13 14 15 28 29 30 31
- becomes (for size == 2 and mask == 0x3333)
- t = 2^16 3^17 -- -- l = 0 1 16 17 r = 2 3 18 19
- 6^20 7^21 -- -- 4 5 20 21 6 7 22 23
- 10^24 11^25 -- -- 8 9 24 25 10 11 24 25
- 14^28 15^29 -- -- 12 13 28 29 14 15 28 29
-
- Thanks for hints from Richard Outerbridge - he told me IP&FP
- could be done in 15 xor, 10 shifts and 5 ands.
- When I finally started to think of the problem in 2D
- I first got ~42 operations without xors. When I remembered
- how to use xors :-) I got it to its final state.
- */
- #define PERM_OP(a,b,t,n,m) ((t)=((((a)>>(n))^(b))&(m)),\
- (b)^=(t),\
- (a)^=((t)<<(n)))
-
- int des_encrypt(input,output,ks,encrypt)
- ulong *input;
- ulong *output;
- des_key_schedule ks;
- int encrypt;
- {
- register ulong l,r,t,u;
- register int i;
- register ulong *s;
-
- l=input[0];
- r=input[1];
-
- /* do IP */
- PERM_OP(r,l,t, 4,0x0f0f0f0f);
- PERM_OP(l,r,t,16,0x0000ffff);
- PERM_OP(r,l,t, 2,0x33333333);
- PERM_OP(l,r,t, 8,0x00ff00ff);
- PERM_OP(r,l,t, 1,0x55555555);
- /* r and l are reversed - remember that :-) */
- t=l;
- l=r;
- r=t;
-
- s=(ulong *)ks;
-
- /* I don't know if it is worth the effort of loop unrolling the
- * inner loop */
- if (encrypt)
- {
- for (i=0; i<32; i+=4)
- {
- D_ENCRYPT(l,r,i+0); /* 1 */
- D_ENCRYPT(r,l,i+2); /* 2 */
- }
- }
- else
- {
- for (i=30; i>0; i-=4)
- {
- D_ENCRYPT(l,r,i-0); /* 16 */
- D_ENCRYPT(r,l,i-2); /* 15 */
- }
- }
-
- /* swap l and r
- * we will not do the swap so just remember they are
- * reversed for the rest of the subroutine
- * luckily FP fixes this problem :-) */
-
- PERM_OP(r,l,t, 1,0x55555555);
- PERM_OP(l,r,t, 8,0x00ff00ff);
- PERM_OP(r,l,t, 2,0x33333333);
- PERM_OP(l,r,t,16,0x0000ffff);
- PERM_OP(r,l,t, 4,0x0f0f0f0f);
-
- output[0]=l;
- output[1]=r;
- return(0);
- }
-
-