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
unicode(Rune * k)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
latin1(Rune * k,int n)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 }
77