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

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

update

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.