Jim Gibson wrote:
On 4/17/09 Fri Apr 17, 2009 1:50 PM, "Brian" <brian5432...@yahoo.co.uk> scribbled:Brian wrote: oops, should read...... $Year_out = $Year_in; while ($Year_out > 100) {$Year_out -= 100;} if (($Year_out > 00) && ($Year_out <= 25)) {$string = $string1;} if (($Year_out > 25) && ($Year_out <= 50)) {$Year_out -= 25;$string = $string2;} if (($Year_out > 50) && ($Year_out <= 75)) {$Year_out -= 50;$string = $string3;} if (($Year_out > 75) && ($Year_out <= 100)) {$Year_out -= 75;$string = $string4;}You are doing quite a few redundant tests here. Note that after your while loop, $Year_out cannot be greater than 100. Also, if $Year_out is greater than 25, it must be greater than 0. In addition, when you are done, $Year_out will be between 0 and 25. Therefore, you can take the value modulo 25 and rearrange the tests to be a little more efficient (I am assuming that $Year_out is always >= zero): while ( $Year_out > 100 ) { $Year_out -= 100; } if ( $Year_out > 75 ) { $string = $string4; }elsif ( $Year_out > 50 { $string = $string3; }elsif ( $Year_out > 25 ) { $string = $string2; }elsif ( $Year_out > 0 ) { $string = $string1; } $Year_out = $Year_out % 25;
Thanks for that, a better structure, but I don't understand the use of % 25 in this instance. $year is actually always going to be >= zero , even if someone enters a negative. I haven't factored for the Julian/Gregorian changeover, so this prog will calc year 1 as it would have been had todays calendar been in use 2000+ years ago. A point I should make here is that I am actually working over 400 year cycles and not 100, so...
100 = 400 75 = 300 50 = 200 25 = 100 but I suppose the results will be this same in your code above. (I changed certain parameters to stop any robots from stealing my code) ;-) -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
