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[4], find the rune or return -1 for failure.
21  */
22 static long
23 unicode(Rune *k)
24 {
25 	long i, c;
26 
27 	k++;	/* skip 'X' */
28 	c = 0;
29 	for(i=0; i<4; i++,k++){
30 		c <<= 4;
31 		if('0'<=*k && *k<='9')
32 			c += *k-'0';
33 		else if('a'<=*k && *k<='f')
34 			c += 10 + *k-'a';
35 		else if('A'<=*k && *k<='F')
36 			c += 10 + *k-'A';
37 		else
38 			return -1;
39 	}
40 	return c;
41 }
42 
43 /*
44  * Given n characters k[0]..k[n-1], find the corresponding rune or return -1 for
45  * failure, or something < -1 if n is too small.  In the latter case, the result
46  * is minus the required n.
47  */
48 int
49 _latin1(Rune *k, int n)
50 {
51 	struct cvlist *l;
52 	int c;
53 	char* p;
54 
55 	if(k[0] == 'X'){
56 		if(n>=5)
57 			return unicode(k);
58 		else
59 			return -5;
60 	}
61 	
62 	for(l=latintab; l->ld!=0; l++)
63 		if(k[0] == l->ld[0]){
64 			if(n == 1)
65 				return -2;
66 			if(l->ld[1] == 0)
67 				c = k[1];
68 			else if(l->ld[1] != k[1])
69 				continue;
70 			else if(n == 2)
71 				return -3;
72 			else
73 				c = k[2];
74 			for(p=l->si; *p!=0; p++)
75 				if(*p == c)
76 					return l->so[p - l->si];
77 			return -1;
78 		}
79 	return -1;
80 }