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ltoa.c
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1994-03-02
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/*
* ltoa(num, base) - return string form of the given number in the
* given base.
*
* If the base has magnitude greater than 10, the sign of the base
* determines which case of letters will be used to represent digits
* larger than 9. A positive base will use upper-case letters, while
* a negative base will use lower-case letters.
*
* If the base has magnitude 10, the sign of the base determines whether
* the number is to be considered signed or unsigned. A base of +10
* is considered to be signed, while a base of -10 is considered to
* be unsigned. Odd, huh?
*
* Only base 10 is potentially signed; all other bases are unsigned.
*
* For performance reasons there is no check for an invalid base (a base
* of magnitude less than 1 or greater than 36).
*
* The return value is a pointer to a static buffer that is overwritten
* with every call.
*
* DaviD W. Sanderson
*/
char *
ltoa(num, obase)
unsigned long num;
int obase;
{
unsigned long base;
int neg = 0;/* 1 if num is negative */
char a = 'A';
/*
* buf[] is big enough for the longest binary string together
* with terminal NUL and possible minus sign.
*
* Note that there does not need to be a separate byte allocated
* for the minus sign. The minus sign is present only for
* numbers to be represented in decimal. Since the base 10
* representation of numbers larger than 1 is shorter than
* the base 2 representation, there will always be room for
* the minus sign in the base 10 string.
*/
static char buf[sizeof(unsigned long) * 8 + 1];
/* p points to the last char in buf */
char *p = buf + (sizeof buf - 1);
*p = '\0'; /* terminate the string */
/*
* The only signed-ness occurs when obase == +10.
*
* If obase == -10, the decimal number is unsigned.
*/
if (obase == 10 && (long) num < 0)
{
neg = 1;
num = -(long) num;
}
if (obase < 0)
{
a = 'a';
obase = -obase;
}
/*
* now that obase is nonnegative, assign it to base
*/
base = (unsigned long) obase;
/*
* subtract 10 from the code for the letter 'a' since rem
* will be 10 more than the offset from a
*/
a -= 10;
do
{
long rem;
rem = num % base; /* obtain value of digit */
*--p = ((rem < 10) ? '0' : a) + rem;
}
while ((num /= base) != 0);
if (neg)
*--p = '-';
return p;
}