Move to 2.13.3 (from 2.13.tar.gz - sigh).

1 files changed, 120 insertions, 0 deletions

diff --git a/missing/strtol.c b/missing/strtol.c
new file mode 100644
index 00000000..e102ae34
--- /dev/null
+++ b/missing/strtol.c
@@ -0,0 +1,120 @@
+/*
+Article 4291 of comp.lang.c:
+From: chris@mimsy.umd.edu (Chris Torek)
+Newsgroups: comp.lang.c
+Subject: Re: error checking strtol
+Message-ID: <24445@mimsy.umd.edu>
+Date: 17 May 90 09:31:17 GMT
+Organization: U of Maryland, Dept. of Computer Science, Coll. Pk., MD 20742
+
+The following is a working strtol.  It depends only on the existence of
+correct header files (including <limits.h>) and on ASCII (IBM programmers
+will have to use strchr()).  It does not support locales other than `C'.
+System V programmers should be able to replace their current strtol with
+this one.  (After writing this, I checked the SVR2 source; it did not
+handle several cases correctly.)
+*/
+
+#ifdef __STDC__
+#include <limits.h>
+#else
+#define	LONG_MIN	(-0x80000000)	/* for 32-bit 2s-complement at least */
+#define	LONG_MAX	0x7fffffff
+#endif
+
+#if 0
+#include <limits.h>
+#include <ctype.h>
+#include <errno.h>
+#endif
+
+#ifndef _MSC_VER
+int	errno;
+#endif
+
+/*
+ * Convert a string to a long integer.
+ *
+ * Ignores `locale' stuff.  Assumes that the upper and lower case
+ * alphabets and digits are each contiguous.
+ */
+long
+strtol(nptr, endptr, base)
+	const char *nptr;
+	char **endptr;
+	register int base;
+{
+	register const char *s = nptr;
+	register unsigned long acc;
+	register int c;
+	register unsigned long cutoff;
+	register int neg = 0, any, cutlim;
+
+	/*
+	 * Skip white space and pick up leading +/- sign if any.
+	 * If base is 0, allow 0x for hex and 0 for octal, else
+	 * assume decimal; if base is already 16, allow 0x.
+	 */
+	do {
+		c = *s++;
+	} while (isspace(c));
+	if (c == '-') {
+		neg = 1;
+		c = *s++;
+	} else if (c == '+')
+		c = *s++;
+	if ((base == 0 || base == 16) &&
+	    c == '0' && (*s == 'x' || *s == 'X')) {
+		c = s[1];
+		s += 2;
+		base = 16;
+	}
+	if (base == 0)
+		base = c == '0' ? 8 : 10;
+
+	/*
+	 * Compute the cutoff value between legal numbers and illegal
+	 * numbers.  That is the largest legal value, divided by the
+	 * base.  An input number that is greater than this value, if
+	 * followed by a legal input character, is too big.  One that
+	 * is equal to this value may be valid or not; the limit
+	 * between valid and invalid numbers is then based on the last
+	 * digit.  For instance, if the range for longs is
+	 * [-2147483648..2147483647] and the input base is 10,
+	 * cutoff will be set to 214748364 and cutlim to either
+	 * 7 (neg==0) or 8 (neg==1), meaning that if we have accumulated
+	 * a value > 214748364, or equal but the next digit is > 7 (or 8),
+	 * the number is too big, and we will return a range error.
+	 *
+	 * Set any if any `digits' consumed; make it negative to indicate
+	 * overflow.
+	 */
+	cutoff = neg ? -(unsigned long)LONG_MIN : LONG_MAX;
+	cutlim = cutoff % (unsigned long)base;
+	cutoff /= (unsigned long)base;
+	for (acc = 0, any = 0;; c = *s++) {
+		if (isdigit(c))
+			c -= '0';
+		else if (isalpha(c))
+			c -= isupper(c) ? 'A' - 10 : 'a' - 10;
+		else
+			break;
+		if (c >= base)
+			break;
+		if (any < 0 || acc > cutoff || acc == cutoff && c > cutlim)
+			any = -1;
+		else {
+			any = 1;
+			acc *= base;
+			acc += c;
+		}
+	}
+	if (any < 0) {
+		acc = neg ? LONG_MIN : LONG_MAX;
+		errno = ERANGE;
+	} else if (neg)
+		acc = -acc;
+	if (endptr != 0)
+		*endptr = (char *) (any ? s - 1 : nptr);
+	return (acc);
+}