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

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

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

Comment on Re^4: Shortcut operator for $a->{'b'}=$b if $b; Select or Download Code

Replies are listed 'Best First'.
Re^5: Shortcut operator for $a->{'b'}=$b if $b; by ikegami (Patriarch) on Sep 21, 2005 at 05:28 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]

In Section Seekers of Perl Wisdom