Re^9: How to call a sub reff from a hash

I'm not understanding the syntax

This Can be written more clearly as

${$aref}[$i] $aref->[$i]

${$href}{$k} $href->{$k}

&{$cref}($arg) $cref->($arg)

This	Can be written more clearly as
`${$aref}[$i]`	`$aref->[$i]`
`${$href}{$k}`	`$href->{$k}`
`&{$cref}($arg)`	`$cref->($arg)`

In this case, the code ref is stored in a hash, so $cref is a hash lookup.

my $cref = $hash{$k}; $cref->($arg)

    |
    v

( $hash{$k} )->($arg);

    |
    v

$hash{$k}->($arg);
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Re^10: How to call a sub reff from a hash by NateTut (Deacon) on Dec 16, 2008 at 19:12 UTC
Thank you that helped clear up my misunderstanding.	[reply] [d/l]