Yes:

my %hash;
$hash{key1} = 'longstring';
$hash{key2} = alter($hash{key1});
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Admittedly, you have to change your initialization syntax somewhat. The reason is that list on the right hand side of the = is built before the assignment even starts. (See perlop for operator precedence.) Some people seem to think %h = (a=>b, c=>d); is a hash constructor/initializer, but it's not. It's simply a list which gets assigned to a hash.