In recent versions of Perl, you can do:

$var1 = ($var =~ s/oo//gir);
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In pretty much any version of Perl, you can do:

($var1 = $var) =~ s/oo//gi;
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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;
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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.