1 #include <u.h>
2 #include <libc.h>
3 #include <draw.h>
4 
5 /*
6  * The code makes two assumptions: strlen(ld) is 1 or 2; latintab[i].ld can be a
7  * prefix of latintab[j].ld only when j<i.
8  */
9 static struct cvlist
10 {
11 	char	*ld;		/* must be seen before using this conversion */
12 	char	*si;		/* options for last input characters */
13 	Rune	so[60];		/* the corresponding Rune for each si entry */
14 } latintab[] = {
15 #include "latin1.h"
16 	0, 0, { 0 }
17 };
18 
19 /*
20  * Given 5 characters k[0]..k[n], find the rune or return -1 for failure.
21  */
22 static long
23 unicode(Rune *k, int n)
24 {
25 	long i, c;
26 
27 	c = 0;
28 	for(i=0; i<n; i++,k++){
29 		c <<= 4;
30 		if('0'<=*k && *k<='9')
31 			c += *k-'0';
32 		else if('a'<=*k && *k<='f')
33 			c += 10 + *k-'a';
34 		else if('A'<=*k && *k<='F')
35 			c += 10 + *k-'A';
36 		else
37 			return -1;
38 		if(c > Runemax)
39 			return -1;
40 	}
41 	return c;
42 }
43 
44 /*
45  * Given n characters k[0]..k[n-1], find the corresponding rune or return -1 for
46  * failure, or something < -1 if n is too small.  In the latter case, the result
47  * is minus the required n.
48  */
49 int
50 latin1(Rune *k, int n)
51 {
52 	struct cvlist *l;
53 	int c;
54 	char* p;
55 
56 	if(k[0] == 'X'){
57 		if(n < 2)
58 			return -2;
59 		if(k[1] == 'X') {
60 			if(n < 3)
61 				return -3;
62 			if(k[2] == 'X') {
63 				if(n < 9) {
64 					if(unicode(k+3, n-3) < 0)
65 						return -1;
66 					return -(n+1);
67 				}
68 				return unicode(k+3, 6);
69 			}
70 			if(n < 7) {
71 				if(unicode(k+2, n-2) < 0)
72 					return -1;
73 				return -(n+1);
74 			}
75 			return unicode(k+2, 5);
76 		}
77 		if(n < 5) {
78 			if(unicode(k+1, n-1) < 0)
79 				return -1;
80 			return -(n+1);
81 		}
82 		return unicode(k+1, 4);
83 	}
84 
85 	for(l=latintab; l->ld!=0; l++)
86 		if(k[0] == l->ld[0]){
87 			if(n == 1)
88 				return -2;
89 			if(l->ld[1] == 0)
90 				c = k[1];
91 			else if(l->ld[1] != k[1])
92 				continue;
93 			else if(n == 2)
94 				return -3;
95 			else
96 				c = k[2];
97 			for(p=l->si; *p!=0; p++)
98 				if(*p == c)
99 					return l->so[p - l->si];
100 			return -1;
101 		}
102 	return -1;
103 }