Re^4: Shortcut operator for $a->{'b'}=$b if $b;

[leriksen]

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.

[ikegami]

overload doesn't define operators; it redefines them. ?= would have to be added to perl by patch.

There is an alternative to a new or overloaded operator. One could use a tied hash. The only difference would be in the setter.

By the way, my function does not autovivify. At least, not in 5.6.x or 5.8.x.