04-11-2002
I have just one simple questio, WHY ARE YOU WHINING ABOUT 3 SECONDS DIFFERENCE ?!?!?! I am stuck on a 2.8KBPS dialup connection , im lucky if a site will load in 1 minute let alone 5 seconds. Be happy that you have DSL at all, I live exactly one mile away from where the firbe optics end. So be happy with 5 seconds.
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LEARN ABOUT DEBIAN
print_time_table
PARSE_TIME(3) BSD Library Functions Manual PARSE_TIME(3)
NAME
parse_time, print_time_table, unparse_time, unparse_time_approx, -- parse and unparse time intervals
LIBRARY
The roken library (libroken, -lroken)
SYNOPSIS
#include <parse_time.h>
int
parse_time(const char *timespec, const char *def_unit);
void
print_time_table(FILE *f);
size_t
unparse_time(int seconds, char *buf, size_t len);
size_t
unparse_time_approx(int seconds, char *buf, size_t len);
DESCRIPTION
The parse_time() function converts a the period of time specified in into a number of seconds. The timespec can be any number of <number
unit> pairs separated by comma and whitespace. The number can be negative. Number without explicit units are taken as being def_unit.
The unparse_time() and unparse_time_approx() does the opposite of parse_time(), that is they take a number of seconds and express that as
human readable string. unparse_time produces an exact time, while unparse_time_approx restricts the result to only include one units.
print_time_table() prints a descriptive list of available units on the passed file descriptor.
The possible units include:
second, s
minute, m
hour, h
day
week seven days
month 30 days
year 365 days
Units names can be arbitrarily abbreviated (as long as they are unique).
RETURN VALUES
parse_time() returns the number of seconds that represents the expression in timespec or -1 on error. unparse_time() and
unparse_time_approx() return the number of characters written to buf. if the return value is greater than or equal to the len argument, the
string was too short and some of the printed characters were discarded.
EXAMPLES
#include <stdio.h>
#include <parse_time.h>
int
main(int argc, char **argv)
{
int i;
int result;
char buf[128];
print_time_table(stdout);
for (i = 1; i < argc; i++) {
result = parse_time(argv[i], "second");
if(result == -1) {
fprintf(stderr, "%s: parse error
", argv[i]);
continue;
}
printf("--
");
printf("parse_time = %d
", result);
unparse_time(result, buf, sizeof(buf));
printf("unparse_time = %s
", buf);
unparse_time_approx(result, buf, sizeof(buf));
printf("unparse_time_approx = %s
", buf);
}
return 0;
}
$ ./a.out "1 minute 30 seconds" "90 s" "1 y -1 s"
1 year = 365 days
1 month = 30 days
1 week = 7 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
1 second
--
parse_time = 90
unparse_time = 1 minute 30 seconds
unparse_time_approx = 1 minute
--
parse_time = 90
unparse_time = 1 minute 30 seconds
unparse_time_approx = 1 minute
--
parse_time = 31535999
unparse_time = 12 months 4 days 23 hours 59 minutes 59 seconds
unparse_time_approx = 12 months
BUGS
Since parse_time() returns -1 on error there is no way to parse "minus one second". Currently "s" at the end of units is ignored. This is a
hack for English plural forms. If these functions are ever localised, this scheme will have to change.
HEIMDAL
October 31, 2004 HEIMDAL