1 while s/^FOO/BAR/g;

repeats the substitution till it fails¹ ...

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

If not have a look at perlintro

update

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

> 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.

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
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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.

The following code should illustrate the difference:

#!/usr/bin/perl

$_ = "Foollll\nFoollll\n";
1 while s/^Fool/Foo/g;
print "1:\n\n$_";

print "\n\n";

$_ = "Foollll\nFoollll\n";
s/^Fool/Foo/g;
print "2:\n\n$_";
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Can you see what the output will be...? ;)

Howdy!

Thanks for the info and examples! I read perlop and perlre years ago, but don't recall seeing the 1 while example(s).

Thanks Again!

1 while condition;
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while(condition){};
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The repetition could make a difference, if BAR sometimes could sometimes contain FOO (what I'd call a recursive substitution, even if in fact it's an iteration ;-)

I just had a co-worker ask what this code did and I didn't know. ... It seems to work just like: ... Any thoughts?

Use more words to explain the program, for example start with <c> $_ contains "foo" , then loop for blah blah and on blah iteration blah happen s...

Do that for 2/3 of the loops you posted

Think about what you're saying

Explain in great detail what happens to the variable at each step, what operator or control structure does

Say it out loud, write it down, it shouldn't take more than two minutes to write it up ... I'll wait

Here are some vocabulary hints (magic word is "Condition") courtesy of ppi_dumper
Read more... (3 kB)