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Author	Message
kings_on_steeds Senior Member Joined: 05 Apr 2004 Posts: 290 Location: Bookham, Surrey, UK	Posted: Mon Apr 18, 2005 3:42 pm    Post subject: dose anyone know how to take 1 away from a variabul? i am a little stuck. Quote: $var1 == 6; -1 print $var1; prints 5
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Gavin God Like Joined: 15 Mar 2004 Posts: 661 Location: Manchester, UK	Posted: Mon Apr 18, 2005 4:21 pm    Post subject: Do you mean just: Code: my $var1 = 6;<br />my $var1 = $var1 - 1; _________________ MSN: gavin [at] gavinbrittain [dot] co [dot] uk E-Portfolio: www.gavinbrittain.co.uk (New version comming soon)
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Cer Upgraded Agent Joined: 03 Feb 2004 Posts: 3776 Location: Michigan votes: 4	Posted: Mon Apr 18, 2005 6:50 pm    Post subject: If you're subtracting (or adding) just 1, you can just use -- and ++ Code: $var = 6;<br />$var--; # subtract 1<br />$var++; # add 1 _________________ Current Site (2008) http://www.cuvou.com/
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Siebe God Like Joined: 06 Jan 2004 Posts: 562 Location: Netherlands	Posted: Mon Apr 18, 2005 7:54 pm    Post subject: Note you can also use quick operators, so instead of doing Code: my $a = 5;<br />$a = $a + 10;<br />$a = $a - 10;<br />$a = $a / 10; You can also do... Code: my $a = 5;<br />$a += 10;<br />$a -= 10;<br />$a /= 10; It works on pretty much every operator (though I'm not sure about **, exponent). Either way, this is usually a tad bit faster (although you won't notice it that much.. but hey, any speed profit is welcome) because perl does not have to create a symbolic copy of $a before doing something with it (instead it can use the original value, which saves one "stack copy"). Technical details aside, it's pretty nifty (and usually a lot cleaner).
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kings_on_steeds Senior Member Joined: 05 Apr 2004 Posts: 290 Location: Bookham, Surrey, UK	Posted: Tue Apr 19, 2005 9:15 pm    Post subject: it worked. yahh, thanks guys.
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