Re: day of year calculation

[mlh2003]

One way to speed things up a little is to get rid of the for loop by replacing the array containing the number of days in each month with the cumulative number of days. If the year is a leap year, add 1 to the result if the month is >2.

UPDATE (for explanation): In the code I check for the month > 1 since I subtract 1 from the month in the previous line.

#!/usr/local/bin/perl

use strict;

sub dayofyear {
    my ($day1,$month,$year)=@_;
    my @cumul_d_in_m=(0,31,59,90,120,151,181,212,243,273,304,334,365);
    my $doy=$cumul_d_in_m[--$month]+$day1;
    $doy++ if (&leap($year) && $month>1);
    return $doy;
}
sub leap {
    my $y = shift;
    return 0 unless $y % 4 == 0;
    return 1 unless $y % 100 == 0;
    return 0 unless $y % 400 == 0;
    return 1;
}
print dayofyear(10,3,2005)."\n";
print dayofyear(10,3,2004)."\n";
print dayofyear(10,2,2005)."\n";
print dayofyear(10,2,2004)."\n";

This gives the following output:
69
70
41
41

[fglock]

1 less function call:

#!/usr/local/bin/perl

use strict;

sub dayofyear {
    my ($day1,$month,$year)=@_;
    my @cumul_d_in_m =
(0,31,59,90,120,151,181,212,243,273,304,334,365);
    my $doy=$cumul_d_in_m[--$month]+$day1;
    return $doy if $month < 2;
    return $doy unless $year % 4 == 0;
    return ++$doy unless $year % 100 == 0;
    return $doy unless $year % 400 == 0;
    return ++$doy;
}

print dayofyear(10,3,2005)."\n";
print dayofyear(10,3,2004)."\n";
print dayofyear(10,2,2005)."\n";
print dayofyear(10,2,2004)."\n";