20060303

1 files changed, 76 insertions, 0 deletions

diff --git a/os/port/latin1.c b/os/port/latin1.c
new file mode 100644
index 00000000..8dcb0d7e
--- /dev/null
+++ b/os/port/latin1.c
@@ -0,0 +1,76 @@
+#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 "../port/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;
+}