ref: d21de87db1e6068d35f66da7411a98231c361a70
dir: /emu/port/latin1.c/
#include "dat.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.
*/
static
struct cvlist
{
char *ld; /* must be seen before using this conversion */
char *si; /* options for last input characters */
char *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.
*/
static long
unicode(uchar *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(uchar *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) {
Rune r;
int i = p - l->si;
p = l->so;
for(; i >= 0; i--)
p += chartorune(&r, p);
return r;
}
return -1;
}
return -1;
}