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Maybe someone can suggest how to define a '?=' operator
$a->{'b'} ?= $b; assigns $b to $a->{'b'} if $b is defined, does not autovivify $a->{'b'} if $b not defined is that possible? - I'm thinking of use overload but I dimly recall that there is a finite set of operators you can overload and ?= isnt one of them. ...reality must take precedence over public relations, for nature cannot be fooled. - R P Feynmann In reply to Re^4: Shortcut operator for $a->{'b'}=$b if $b;
by leriksen
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