Interpolation question

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

I have a line like this...

print DESTINATION "old: $_new: $line\n";

Where $_ and $line are strings with a \n on the end that I want to write to the file.

Obviously $_new won't work, but neither will $_\new because it treats \n as a newline. Does anybody know of a better way to do it, "old: $_"."new: $line\n" seems bodgy.

Ta.

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Re: Interpolation question by Ovid (Cardinal) on May 24, 2001 at 20:13 UTC
When interpolating a variable in quotes, Perl cannot always tell what the variable is. A variable named `$_new` is perfectly legal. Wrap the actual variable name in curly braces '`{}`' to let Perl know what variable you are referring to. `print DESTINATION "old: ${_}new: $line\n";` [download] Otherwise, you could wind up with problems like this: `my $a = 1; my $ab = 1; my $abc = 1; my $abcd = 1; # what do we want to print? print "$abcd";` [download] Cheers, Ovid Join the Perlmonks Setiathome Group or just click on the the link and check out our stats.	[reply] [d/l] [select]
Re: Interpolation question by virtualsue (Vicar) on May 24, 2001 at 21:01 UTC
I'd probably do: `print DESTINATION "old: $_","new: $line\n";` [download] I assume you'd consider the above "bodgy" as well. I'm now curious about whether I've missed something (else) important here. Would using braces around the variable name provide a significant win over breaking the output into two quoted strings? Outside of golf, that is. :)	[reply] [d/l]
(tye)Re: Interpolation question by tye (Sage) on May 24, 2001 at 23:08 UTC
`print DESTINATION "old: $_$,new: $line\n";` Will work for that unless the value of $, has been changed (which is likely to cause lots of other problems, so I wouldn't worry about that case). - tye (but my friends call me "Tye")	[reply] [d/l]
Re: Interpolation question by srawls (Friar) on May 24, 2001 at 22:33 UTC
Well, I like the curly braces best, but since there is always more than one way to do it: `print DESTINATION "old: $_\0new: $line\n";` [download] This prints $_, then the null character (nothing), then new, thus seperating $_ from new. The 15 year old, freshman programmer, Stephen Rawls	[reply] [d/l]