ref: 7975eec3e602fd5bc3e93c78f4166259190f2fe9
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; }