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