Re: [opensuse] Quick Perl Question #2 - Why does chomp($float); = "1"?

On Saturday 03 May 2008 13:22, Randall R Schulz wrote:

...

Chomp is not a function in the sense you're using it. It alters the variable you pass in-place, rather than returning the "comped" value.

Change that line to this:

chomp($TEMPC); # Works! Removes New Line from Number

What "chomp" returns, by the way, is 1 / true to tell you it changed its argument or 0 / false to tell you it did not alter it. Randall Schulz -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org For additional commands, e-mail: opensuse+help@opensuse.org