sys/src/cmd/vnc/latin1.c

   1 #include <u.h>
   2
   3 /*
   4  * The code makes two assumptions: strlen(ld) is 1 or 2; latintab[i].ld can be a
   5  * prefix of latintab[j].ld only when j<i.
   6  */
   7 struct cvlist
   8 {
   9         char    *ld;            /* must be seen before using this conversion */
  10         char    *si;            /* options for last input characters */
  11         Rune    *so;            /* the corresponding Rune for each si entry */
  12 } latintab[] = {
  13 #include "latin1.h"
  14         0,      0,              0
  15 };
  16
  17 /*
  18  * Given 5 characters k[0]..k[4], find the rune or return -1 for failure.
  19  */
  20 long
  21 unicode(Rune *k)
  22 {
  23         long i, c;
  24
  25         k++;    /* skip 'X' */
  26         c = 0;
  27         for(i=0; i<4; i++,k++){
  28                 c <<= 4;
  29                 if('0'<=*k && *k<='9')
  30                         c += *k-'0';
  31                 else if('a'<=*k && *k<='f')
  32                         c += 10 + *k-'a';
  33                 else if('A'<=*k && *k<='F')
  34                         c += 10 + *k-'A';
  35                 else
  36                         return -1;
  37         }
  38         return c;
  39 }
  40
  41 /*
  42  * Given n characters k[0]..k[n-1], find the corresponding rune or return -1 for
  43  * failure, or something < -1 if n is too small.  In the latter case, the result
  44  * is minus the required n.
  45  */
  46 long
  47 latin1(Rune *k, int n)
  48 {
  49         struct cvlist *l;
  50         int c;
  51         char* p;
  52
  53         if(k[0] == 'X')
  54                 if(n>=5)
  55                         return unicode(k);
  56                 else
  57                         return -5;
  58         for(l=latintab; l->ld!=0; l++)
  59                 if(k[0] == l->ld[0]){
  60                         if(n == 1)
  61                                 return -2;
  62                         if(l->ld[1] == 0)
  63                                 c = k[1];
  64                         else if(l->ld[1] != k[1])
  65                                 continue;
  66                         else if(n == 2)
  67                                 return -3;
  68                         else
  69                                 c = k[2];
  70                         for(p=l->si; *p!=0; p++)
  71                                 if(*p == c)
  72                                         return l->so[p - l->si];
  73                         return -1;
  74                 }
  75         return -1;
  76 }