62 lines
1.7 KiB
C
62 lines
1.7 KiB
C
/*
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** Calculate a percentage without resorting to floating point
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** and return a pointer to a string
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**
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** "digits" is the number of digits past the decimal place you want
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** (zero being the straight percentage with no decimals)
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**
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** Erik E. Fair <fair@clock.org>, May 8, 1997
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*/
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#include <stdio.h>
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#include <machine/limits.h>
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#include <sys/types.h>
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extern char * strpct(u_long num, u_long denom, u_int digits);
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char *
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strpct(numerator, denominator, digits)
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u_long numerator, denominator;
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u_int digits;
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{
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register int i;
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u_long result, factor = 100L;
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static char percent[32];
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/* I should check for digit overflow here, too XXX */
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for(i = 0; i < digits; i++) {
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factor *= 10;
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}
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/* watch out for overflow! */
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if (numerator < (ULONG_MAX / factor)) {
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numerator *= factor;
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} else {
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/* toss some of the bits of lesser significance */
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denominator /= factor;
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}
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/* divide by zero is just plain bad */
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if (denominator == 0L) {
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denominator = 1L;
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}
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result = numerator / denominator;
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if (digits == 0) {
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(void) snprintf(percent, sizeof(percent), "%lu%%", result);
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} else {
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char fmt[32];
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/* indirection to produce the right output format */
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(void) snprintf(fmt, sizeof(fmt), "%%lu.%%0%ulu%%%%", digits);
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factor /= 100L; /* undo initialization */
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(void) snprintf(percent, sizeof(percent),
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fmt, result / factor, result % factor);
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}
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return(percent);
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}
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