/*
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);
}