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- #include "u.h"
- #include "../port/lib.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 "../port/latin1.h"
- 0, 0, 0
- };
- /*
- * Given n characters k[0]..k[n-1], find the rune or return -1 for failure.
- */
- long
- unicode(Rune *k, int n)
- {
- long c;
- Rune *r;
- c = 0;
- for(r = &k[1]; r<&k[n]; r++){ /* +1 to skip [Xx] */
- c <<= 4;
- if('0'<=*r && *r<='9')
- c += *r-'0';
- else if('a'<=*r && *r<='f')
- c += 10 + *r-'a';
- else if('A'<=*r && *r<='F')
- c += 10 + *r-'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, 5);
- else
- return -5;
- if(k[0] == 'x')
- if(n>=UTFmax*2+1)
- return unicode(k, UTFmax*2+1);
- else
- return -(UTFmax+1);
- 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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