Re^3: Shortcut operator for $a->{'b'}=$b if $b;

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You could write the operator yourself:

sub assign_if_defined {
   $_[0] = $_[1] if defined $_[1];
}

assign_if_defined($a->{'b'}, $b);
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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.

Comment on Re^3: Shortcut operator for $a->{'b'}=$b if $b; Select or Download Code

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Re^4: Shortcut operator for $a->{'b'}=$b if $b; by leriksen (Curate) on Sep 20, 2005 at 23:24 UTC
Maybe someone can suggest how to define a '?=' operator `$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	[reply] [d/l] [select]
Re^5: Shortcut operator for $a->{'b'}=$b if $b; by ikegami (Patriarch) on Sep 21, 2005 at 01:28 UTC
`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.	[reply] [d/l] [select]

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