des_locl.h

Netscape公司提供的安全套接字层

/* des_locl.h */
/* Copyright (C) 1993 Eric Young - see README for more details */
#include <stdio.h>
#include "deseay.h"

#if defined(__STDC__) || defined(VMS) || defined(M_XENIX) || defined(MSDOS)
#include <string.h>
#define bcopy(b1,b2,len) memcpy(b2, b1, (size_t)(len))
#define bzero(b,len) memset(b, 0, (size_t)(len))
#define bcmp(b1,b2,len) memcmp(b1, b2, (size_t)(len))
#define index(s1,char) strchr(s1,char)
#endif

#ifdef MSDOS
#define getpid() 2
#define RAND
extern int errno;
#define PROTO
#endif

#ifdef __STDC__
#define PROTO
#endif

#ifdef RAND
#define random() rand()
#define srandom(s) srand(s)
#endif

#define ITERATIONS 16
#define HALF_ITERATIONS 8

/* used in des_read and des_write */
#define MAXWRITE    (1024*16)
#define BSIZE       (MAXWRITE+4)

#define c2l(c,l)    (l =((unsigned long)(*((c)++)))    , \
             l|=((unsigned long)(*((c)++)))<< 8, \
             l|=((unsigned long)(*((c)++)))<<16, \
             l|=((unsigned long)(*((c)++)))<<24)

/* NOTE - c is not incremented as per c2l */
#define c2ln(c,l1,l2,n) { \
            c+=n; \
            l1=l2=0; \
            switch (n) { \
            case 8: l2|=((unsigned long)(*(--(c))))<<24; \
            case 7: l2|=((unsigned long)(*(--(c))))<<16; \
            case 6: l2|=((unsigned long)(*(--(c))))<< 8; \
            case 5: l2|=((unsigned long)(*(--(c))));     \
            case 4: l1|=((unsigned long)(*(--(c))))<<24; \
            case 3: l1|=((unsigned long)(*(--(c))))<<16; \
            case 2: l1|=((unsigned long)(*(--(c))))<< 8; \
            case 1: l1|=((unsigned long)(*(--(c))));     \
                } \
            }

#define l2c(l,c)    (*((c)++)=(unsigned char)(((l)    )&0xff), \
             *((c)++)=(unsigned char)(((l)>> 8)&0xff), \
             *((c)++)=(unsigned char)(((l)>>16)&0xff), \
             *((c)++)=(unsigned char)(((l)>>24)&0xff))

/* replacements for htonl and ntohl since I have no idea what to do
 * when faced with machines with 8 byte longs. */
#define HDRSIZE 4

#define n2l(c,l)    (l =((unsigned long)(*((c)++)))<<24, \
             l|=((unsigned long)(*((c)++)))<<16, \
             l|=((unsigned long)(*((c)++)))<< 8, \
             l|=((unsigned long)(*((c)++))))

#define l2n(l,c)    (*((c)++)=(unsigned char)(((l)>>24)&0xff), \
             *((c)++)=(unsigned char)(((l)>>16)&0xff), \
             *((c)++)=(unsigned char)(((l)>> 8)&0xff), \
             *((c)++)=(unsigned char)(((l)    )&0xff))

/* NOTE - c is not incremented as per l2c */
#define l2cn(l1,l2,c,n) { \
            c+=n; \
            switch (n) { \
            case 8: *(--(c))=(unsigned char)(((l2)>>24)&0xff); \
            case 7: *(--(c))=(unsigned char)(((l2)>>16)&0xff); \
            case 6: *(--(c))=(unsigned char)(((l2)>> 8)&0xff); \
            case 5: *(--(c))=(unsigned char)(((l2)    )&0xff); \
            case 4: *(--(c))=(unsigned char)(((l1)>>24)&0xff); \
            case 3: *(--(c))=(unsigned char)(((l1)>>16)&0xff); \
            case 2: *(--(c))=(unsigned char)(((l1)>> 8)&0xff); \
            case 1: *(--(c))=(unsigned char)(((l1)    )&0xff); \
                } \
            }

/* The changes to this macro may help or hinder, depending on the
 * compiler and the achitecture.  gcc2 always seems to do well :-).
 * Inspired by Dana How <[e-mail]>
 * DO NOT use the alternative version on machines with 8 byte longs. */
#ifdef ALT_ECB
#define D_ENCRYPT(L,R,S) \
    u=((R^s[S  ])<<2);  \
    t= R^s[S+1]; \
    t=((t>>2)+(t<<30)); \
    L^= \
    *(unsigned long *)(des_SP+0x0100+((t    )&0xfc))+ \
    *(unsigned long *)(des_SP+0x0300+((t>> 8)&0xfc))+ \
    *(unsigned long *)(des_SP+0x0500+((t>>16)&0xfc))+ \
    *(unsigned long *)(des_SP+0x0700+((t>>24)&0xfc))+ \
    *(unsigned long *)(des_SP+       ((u    )&0xfc))+ \
    *(unsigned long *)(des_SP+0x0200+((u>> 8)&0xfc))+ \
    *(unsigned long *)(des_SP+0x0400+((u>>16)&0xfc))+ \
    *(unsigned long *)(des_SP+0x0600+((u>>24)&0xfc));
#else /* original version */
#ifdef MSDOS
#define D_ENCRYPT(L,R,S)    \
    U.l=R^s[S+1]; \
    T.s[0]=((U.s[0]>>4)|(U.s[1]<<12))&0x3f3f; \
    T.s[1]=((U.s[1]>>4)|(U.s[0]<<12))&0x3f3f; \
    U.l=(R^s[S  ])&0x3f3f3f3f; \
    L^= des_SPtrans[1][(T.c[0])]| \
        des_SPtrans[3][(T.c[1])]| \
        des_SPtrans[5][(T.c[2])]| \
        des_SPtrans[7][(T.c[3])]| \
        des_SPtrans[0][(U.c[0])]| \
        des_SPtrans[2][(U.c[1])]| \
        des_SPtrans[4][(U.c[2])]| \
        des_SPtrans[6][(U.c[3])];
#else
#define D_ENCRYPT(L,R,S)    \
    u=(R^s[S  ]); \
    t=R^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];
#endif
#endif

    /* 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)))