Re: In place replace, ignoring between quotes

Brute force...

use warnings;
use strict;

my $str = q(cd / ; /path/to/R/R_latest --vanilla --args "fName='rGSDPl
+an';jobCode=682718;jobId=6827181;" < job682718.R > job_6827181.txt);
my $new;
my $out = 1;
for (split //, $str) {
    $out = ! $out if /"/;
    $new .= (/;/ and $out) ? '&&' :  $_;
}
[download]

Comment on Re: In place replace, ignoring between quotes Download Code

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Re^2: In place replace, ignoring between quotes by trippledubs (Deacon) on Oct 25, 2013 at 19:09 UTC
This is neat. This is an implementation of a state machine, right? In the debugger it looks like $out just undefines and redefines itself for every double quote? I don't understand how you can just say, equal not yourself. How does that work?	[reply]
Re^3: In place replace, ignoring between quotes by toolic (Bishop) on Oct 25, 2013 at 19:23 UTC
Yes, this is a state machine. I spend a lot of time coding in Verilog, where toggling a bit is a natural part of the language. I guess this would be cleaner in Perl: `$out = $out ? 0 : 1 if $_ eq '"';` [download] I do not know why Perl sets $out to undef.	[reply] [d/l]
Re^3: In place replace, ignoring between quotes by SuicideJunkie (Vicar) on Oct 25, 2013 at 19:20 UTC
If $var is true, then !$var is false (''). If $var is false, then !$var is true (1). That new value is then assigned to $var Alternatively, you could say `$var ^= 1` to do the same sort of thing, just with 1 and 0 instead of 1 and ''.	[reply] [d/l]
Re^4: In place replace, ignoring between quotes by trippledubs (Deacon) on Oct 25, 2013 at 19:47 UTC
Nice. ^=1 makes more sense to me. I guess I have a hard time understanding because if $var = 0 (or "") then you say $var = ! $var, why 1? Why not, 50? I mean practically speaking it works and that is what is important. I guess it doesn't have to exactly translate to English. Thanks for sharing!	[reply]