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- #include <u.h>
- /*
- * The code makes two assumptions: strlen(ld) is 1 or 2; latintab[i].ld can be a
- * prefix of latintab[j].ld only when j<i.
- */
- struct cvlist
- {
- char *ld; /* must be seen before using this conversion */
- char *si; /* options for last input characters */
- Rune *so; /* the corresponding Rune for each si entry */
- } latintab[] = {
- #include "latin1.h"
- 0, 0, 0
- };
- /*
- * Given 5 characters k[0]..k[4], find the rune or return -1 for failure.
- */
- long
- unicode(Rune *k)
- {
- long i, c;
- k++; /* skip 'X' */
- c = 0;
- for(i=0; i<4; i++,k++){
- c <<= 4;
- if('0'<=*k && *k<='9')
- c += *k-'0';
- else if('a'<=*k && *k<='f')
- c += 10 + *k-'a';
- else if('A'<=*k && *k<='F')
- c += 10 + *k-'A';
- else
- return -1;
- }
- return c;
- }
- /*
- * Given n characters k[0]..k[n-1], find the corresponding rune or return -1 for
- * failure, or something < -1 if n is too small. In the latter case, the result
- * is minus the required n.
- */
- long
- latin1(Rune *k, int n)
- {
- struct cvlist *l;
- int c;
- char* p;
- if(k[0] == 'X')
- if(n>=5)
- return unicode(k);
- else
- return -5;
- for(l=latintab; l->ld!=0; l++)
- if(k[0] == l->ld[0]){
- if(n == 1)
- return -2;
- if(l->ld[1] == 0)
- c = k[1];
- else if(l->ld[1] != k[1])
- continue;
- else if(n == 2)
- return -3;
- else
- c = k[2];
- for(p=l->si; *p!=0; p++)
- if(*p == c)
- return l->so[p - l->si];
- return -1;
- }
- return -1;
- }
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