in reply to Re^2: Shortcut operator for $a>{'b'}=$b if $b; in thread Shortcut operator for $a>{'b'}=$b if $b;
You could write the operator yourself:
sub assign_if_defined {
$_[0] = $_[1] if defined $_[1];
}
assign_if_defined($a>{'b'}, $b);
By the way, I'm using defined since you said you didn't want empty keys. Checking for truthfullness (as you have been doing) removes both empty and false keys. Just remove the word defined for a truth test.
Re^4: Shortcut operator for $a>{'b'}=$b if $b; by leriksen (Curate) on Sep 21, 2005 at 03:24 UTC 
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
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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.
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