Re: problem with perl 5

The /r modifier for the tr/// and s/// operators has been added in version 5.14, see the perl 5.14 delta

Edit: you don't need it though, without /r $string =~ s/search/rep/; will apply its modifications directly on $string. The /r modifier is used to keep the original string intact and output the result to another one.

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Re^2: problem with perl 5 by praveenchappa (Acolyte) on Mar 09, 2015 at 08:55 UTC
ya got it thank you	[reply]
Re^2: problem with perl 5 by praveenchappa (Acolyte) on Mar 09, 2015 at 08:52 UTC
ok then what could be the solution in 5.10 ,how can replicate the same behaviour in 5.10,please help me out	[reply]
Re^3: problem with perl 5 by Anonymous Monk on Mar 09, 2015 at 09:03 UTC
It sounds like you've already solved the problem, but for the benefit of others: The `/r` modifier causes the expression `$string=~s/\Q$oldstring\E/$newstring/gsr` to return the modified version of `$string` without changing the content of `$string`. However, since then you're just assigning that value back to `$string`, the expression `if ($string = $string=~s/\Q$oldstring\E/$newstring/gsr)` is equivalent to `if ($string=~s/\Q$oldstring\E/$newstring/gs)`, since without `/r`, the `$string=~s///;` expression modifies `$string` directly.	[reply] [d/l] [select]
Re^3: problem with perl 5 by Anonymous Monk on Mar 09, 2015 at 09:41 UTC
If you want to have the result in a second string you have to copy the input first. `my $output = $input; $output = s/\Q$oldstring\E/$newstring/gs;` [download]	[reply] [d/l]
Re^4: problem with perl 5 by Anonymous Monk on Mar 09, 2015 at 09:50 UTC
Not quite, that second `=` should be a `=~`, and the shorter (more idiomatic, I'd say) way to write that is: `(my $output = $string) =~ s/\Q$oldstring\E/$newstring/gs;` [download]	[reply] [d/l] [select]