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diff --git a/src/lib/crypto/krb/nfold.c b/src/lib/crypto/krb/nfold.c
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+/*
+ * 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 "k5-int.h"
+#ifdef HAVE_MEMORY_H
+#include <memory.h>
+#endif
+
+/*
+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(unsigned int inbits, const unsigned char *in, unsigned 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 */
+
+ 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;
+ }
+ }
+}
+