Re: implicit numeric conversion isn't

Are you sure $step is really a string (and not some kind of object with overloaded operators)? What does ref($step) return?

Also what's in $lupdate?

Normal string -> number conversion works:

perl -e'my $step = "300"; print 10 % $step'
10
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Re^2: implicit numeric conversion isn't by stargrazer (Novice) on Nov 25, 2006 at 00:47 UTC
Thanx for your response. If I print( "'" . ref($step) . "'\n" ); I get nothing. I made another post where I did a Dump. Does it make any sense?	[reply]
Re^3: implicit numeric conversion isn't by NetWallah (Canon) on Nov 25, 2006 at 05:48 UTC
Yes - that does make sense - $step is NOT a reference, so you get nothing. Here are a couple of somewhat lame suggestions - but if you are desparate enough, you may want to try these: Use Data::Dumper, and print Dumper $hash; Try using RRDs::fetch instead of info - This will probably give you more info than you want, but if info isnt working for you .... "A closed mouth gathers no feet." --Unknown	[reply]