In recent versions of Perl, you can do:
$var1 = ($var =~ s/oo//gir);
In pretty much any version of Perl, you can do:
($var1 = $var) =~ s/oo//gi;
Update: the second method is about 35% faster according to my benchmarking. The first one is sometimes more elegant - especially when the variable you're operating on (and don't want to be changed) is $_ such as this map block:
my @without_oo = map { s/oo//rig } @with_oo;
# traditional version:
my @without_oo = map { (my $x = $_) =~ s/oo//ig; $x } @with_oo;
But in your original example, neither of the two solutions I presented above seems more elegant than the other, so I'd prefer the faster one.
perl -E'sub Monkey::do{say$_,for@_,do{($monkey=[caller(0)]->[3])=~s{::}{ }and$monkey}}"Monkey say"->Monkey::do'
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