in reply to Re: Different way to use while loop
in thread Different way to use while loop

repeats the substitution till it fails ...

What kind of substitution is it?

Its a global substitution, so even if it was  s/FOO/BAR/g there would be no need to repeat that, the global substitution took care of it in one pass

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Re^3: Different way to use while loop ( updated )
by LanX (Saint) on Jan 09, 2015 at 03:52 UTC
    > the global substitution took care of it in one pass

    not in scalar context which is enforced by the boolean context of the while condition.

    see s///g in perlop, perlre , perlretut

    Cheers Rolf

    PS: Je suis Charlie!

    update
    Global matching

    In scalar context, successive invocations against a string will have //g jump from match to match, keeping track of position in the string as it goes along. You can get or set the position with the pos() function.

    update

    I was wrong the scalar thing is only valid for m//g , s///g just returns the number of all substitutions

      DB<112> $_ = "a a a"
 => "a a a"

  DB<113> scalar s/a/x/g
 => 3

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    update

    See this example in perlop for a similar use case

    1 while s/(\d)(\d\d\d)(?!\d)/$1,$2/g;

    But without knowing Foo and Bar we can only speculate.

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Re^3: Different way to use while loop
by dsheroh (Monsignor) on Jan 09, 2015 at 12:40 UTC
    Its a global substitution, so even if it was s/FOO/BAR/g there would be no need to repeat that, the global substitution took care of it in one pass
    Not necessarily: 
    $ perl -E '$_ = "caattt"; s/at//g; say'
catt
$ perl -E '$_ = "caattt"; 1 while s/at//g; say'
ct

    [download]
    Without knowing FOO and BAR, we can't say whether new instances of FOO might be created in the string when the substitution is performed.
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