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
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Re^5: Shortcut operator for $a->{'b'}=$b if $b;
by ikegami (Pope) on Sep 21, 2005 at 05:28 UTC |

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