/* * Copyright 2004 Sun Microsystems, Inc. All rights reserved. * Use is subject to license terms. */ #pragma ident "%Z%%M% %I% %E% SMI" /* * Copyright (C) 1998 by the FundsXpress, INC. * * All rights reserved. * * Export of this software from the United States of America may require * a specific license from the United States Government. It is the * responsibility of any person or organization contemplating export to * obtain such a license before exporting. * * WITHIN THAT CONSTRAINT, permission to use, copy, modify, and * distribute this software and its documentation for any purpose and * without fee is hereby granted, provided that the above copyright * notice appear in all copies and that both that copyright notice and * this permission notice appear in supporting documentation, and that * the name of FundsXpress. not be used in advertising or publicity pertaining * to distribution of the software without specific, written prior * permission. FundsXpress makes no representations about the suitability of * this software for any purpose. It is provided "as is" without express * or implied warranty. * * THIS SOFTWARE IS PROVIDED ``AS IS'' AND WITHOUT ANY EXPRESS OR * IMPLIED WARRANTIES, INCLUDING, WITHOUT LIMITATION, THE IMPLIED * WARRANTIES OF MERCHANTIBILITY AND FITNESS FOR A PARTICULAR PURPOSE. */ #include /* * Solaris Kerberos defines memory management macros in , * which is included by , so we need not include */ /* #include */ /* n-fold(k-bits): l = lcm(n,k) r = l/k s = k-bits | k-bits rot 13 | k-bits rot 13*2 | ... | k-bits rot 13*(r-1) compute the 1's complement sum: n-fold = s[0..n-1]+s[n..2n-1]+s[2n..3n-1]+..+s[(k-1)*n..k*n-1] */ /* representation: msb first, assume n and k are multiples of 8, and that k>=16. this is the case of all the cryptosystems which are likely to be used. this function can be replaced if that assumption ever fails. */ /* input length is in bits */ void krb5_nfold(inbits, in, outbits, out) int inbits; krb5_const unsigned char *in; int outbits; unsigned char *out; { int a,b,c,lcm; int byte, i, msbit; /* the code below is more readable if I make these bytes instead of bits */ inbits >>= 3; outbits >>= 3; /* first compute lcm(n,k) */ a = outbits; b = inbits; while(b != 0) { c = b; b = a%b; a = c; } lcm = outbits*inbits/a; /* now do the real work */ (void) memset(out, 0, outbits); byte = 0; /* this will end up cycling through k lcm(k,n)/k times, which is correct */ for (i=lcm-1; i>=0; i--) { /* compute the msbit in k which gets added into this byte */ msbit = (/* first, start with the msbit in the first, unrotated byte */ ((inbits<<3)-1) /* then, for each byte, shift to the right for each repetition */ +(((inbits<<3)+13)*(i/inbits)) /* last, pick out the correct byte within that shifted repetition */ +((inbits-(i%inbits))<<3) )%(inbits<<3); /* pull out the byte value itself */ byte += (((in[((inbits-1)-(msbit>>3))%inbits]<<8)| (in[((inbits)-(msbit>>3))%inbits])) >>((msbit&7)+1))&0xff; /* do the addition */ byte += out[i%outbits]; out[i%outbits] = byte&0xff; #if 0 printf("msbit[%d] = %d\tbyte = %02x\tsum = %03x\n", i, msbit, (((in[((inbits-1)-(msbit>>3))%inbits]<<8)| (in[((inbits)-(msbit>>3))%inbits])) >>((msbit&7)+1))&0xff, byte); #endif /* keep around the carry bit, if any */ byte >>= 8; #if 0 printf("carry=%d\n", byte); #endif } /* if there's a carry bit left over, add it back in */ if (byte) { for (i=outbits-1; i>=0; i--) { /* do the addition */ byte += out[i]; out[i] = byte&0xff; /* keep around the carry bit, if any */ byte >>= 8; } } }