James Board has asked for the wisdom of the Perl Monks concerning the following question:
However, if I include that in another perl file as a library, it no longer works:#!/usr/bin/perl -P use strict; #define C 10 my $var = C; print $var . "\n";
The problem is that the preprocessor (notice the -P switch) isn't processing included library files. Is there any way to make the preprocessor process included files? Jim Update: This is a contrived example. I simplified it for this forum to isolate the problem, which is the fact that the preprocessor ignores included library files. In reality, I have several perl library files which I created, and I want subroutines in those files to be able to call the same subroutine messageLog() and have it behave differently based on which file it was called from (so I can turn off/on message logging from a specific file). I don't know if I can do that with Perl syntax, but I know I can do it with macros: just define the macro for messageLog() at the beginning of each file.#!/usr/bin/perl -P require "progA.pl";
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Re: Preprocessor, library files
by Jenda (Abbot) on Dec 26, 2008 at 20:00 UTC | |
Re: Preprocessor, library files
by gwadej (Chaplain) on Dec 26, 2008 at 19:43 UTC | |
Re: Preprocessor, library files
by webfiend (Vicar) on Dec 26, 2008 at 19:45 UTC | |
Re: Preprocessor, library files
by webfiend (Vicar) on Dec 27, 2008 at 06:28 UTC | |
Re: Preprocessor, library files
by spmlingam (Scribe) on Dec 27, 2008 at 06:19 UTC |