Re: Re: Re: Re: Backslash Interpolation

Ah this gives interesting results

my $string = '\\this is a test on \ escapes . ';
$string =~ s/\\/\\\\/g;
print $string;
[download]

I get the following output:
\\this is a test on \\ escapes .
for reference this is on perl version 5.006001

Comment on Re: Re: Re: Re: Backslash Interpolation Download Code

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Re: Re: Re: Re: Re: Backslash Interpolation by Graham (Deacon) on Jun 14, 2001 at 18:33 UTC
This is because your string has taken the single escape char and escaped it for you If you run perl in debug mode and look at the following it will show what I mean: $ perl -de0 Loading DB routines from perl5db.pl version 1.0402 Emacs support available. Enter h or `h h' for help. main::(-e:1): 0 DB<1> my $string = '\\this is a test on \ escapes . '; DB<2> print $string \this is a test on \ escapes . DB<3> x $string 0 '\\this is a test on \\ escapes . ' DB<4> $string =~ s/\\/\\\\/g; DB<5> print $string \\this is a test on \\ escapes . [download] As you can see the evaluation of $string at step 3 shows it to now contain two backslashes where only one was entered. This is because it has evaluated correctly a single backslash escape but doesn't evaluate \\ correctly. I assume that it assumes you are escaping and it need do nothing. I think you're going to have to write a neat little subroutine accounting for all the above problems. A few smart substitution regex's combined with quotemeta should eventually cover all the bases....	[reply] [d/l]

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Re: Re: Re: Re: Re: Backslash Interpolation
by Graham (Deacon) on Jun 14, 2001 at 18:33 UTC

$ perl -de0

Loading DB routines from perl5db.pl version 1.0402
Emacs support available.

Enter h or `h h' for help.

main::(-e:1):   0
  DB<1> my $string = '\\this is a test on \ escapes . ';

  DB<2> print $string
\this is a test on \ escapes . 
  DB<3> x $string
0  '\\this is a test on \\ escapes . '
  DB<4> $string =~ s/\\/\\\\/g;

  DB<5> print $string
\\this is a test on \\ escapes .
[download]

As you can see the evaluation of $string at step 3 shows it to now contain two backslashes where only one was entered.
This is because it has evaluated correctly a single backslash escape but doesn't evaluate \\ correctly. I assume that it assumes you are escaping and it need do nothing.
I think you're going to have to write a neat little subroutine accounting for all the above problems. A few smart substitution regex's combined with quotemeta should eventually cover all the bases....

[reply]
[d/l]