Re: Different way to use while loop

This

1 while s/^FOO/BAR/g;

repeats the substitution till it failsš ...

...which doesn't make much sense because of the ^ anchoring at the start, i.e. it can't substitute more than once.

This should lead to the same results

while (<FH>) {
  s/^FOO/BAR/;
  print "$_";
}
[download]

The rest (iterating over lines from filehandle with $_) should be obvious, right?

If not have a look at perlintro

Cheers Rolf

PS: Je suis Charlie!

update

š) it's a postfix equivalent of

while (s/^FOO/BAR/g) {
    1 ; # no op
}
[download]

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Re^2: Different way to use while loop by Anonymous Monk on Jan 09, 2015 at 03:48 UTC
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	[reply] [d/l]
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` [download] update See this example in perlop for a similar use case `1 while s/(\d)(\d\d\d)(?!\d)/$1,$2/g;` But without knowing Foo and Bar we can only speculate.	[reply] [d/l] [select]
Re^3: Different way to use while loop by dsheroh (Monsignor) on Jan 09, 2015 at 12:40 UTC
Its a global substitution, so even if it was s/FOO/BAR/g there would be no need to repeat that, the global substitution took care of it in one pass Not necessarily: `$ 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.	[reply] [d/l] [select]