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
in thread Shortcut operator for $a->{'b'}=$b if $b; by rokadave

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