Re: Re: I'm looking for some 'heavy magic'

Howdy!

Not quite.

The actual examples in perldelta are:

@x->[2]
scalar(@x)->[2]
[download]

"@$ref->[0]" can be parsed as @{$ref->[0]}, where $ref refers to an array whose first element is an arrayref.

yours,
Michael

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Re: Re: Re: I'm looking for some 'heavy magic' by Anonymous Monk on Jan 25, 2002 at 10:56 UTC
Nuh-uh! `@$ref->[0]` can be parsed as `@{$ref->[0]}` That's not what's happening. From perlref: Because of being able to omit the curlies for the simple case of `$$x`, people often make the mistake of viewing the dereferencing symbols as proper operators, and wonder about their precedence. Example code: `use strict; my $r = [qw/foo bar/]; print $r->[0]; # 'foo' print @$r->[0]; # 'foo', due to bug. print @{$r->[0]}; # Can't use string ("foo") as an ARRAY ref...` [download] I bet that you've several times have heard "use references as variable names". That is a good way to describe references, and it works here too. Having `$r = \@x`, `@$r->[0]` will be the same as `@x->[0]`. So it is indeed the same issue as in perldelta.	[reply] [d/l] [select]

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Re: Re: Re: I'm looking for some 'heavy magic'
by Anonymous Monk on Jan 25, 2002 at 10:56 UTC

@$ref->[0] can be parsed as @{$ref->[0]}

Because of being able to omit the curlies for the simple case of $$x, people often make the mistake of viewing the dereferencing symbols as proper operators, and wonder about their precedence.

use strict;
my $r = [qw/foo bar/];
print $r->[0];    # 'foo'
print @$r->[0];   # 'foo', due to bug.
print @{$r->[0]}; # Can't use string ("foo") as an ARRAY ref...
[download]

$r = \@x

@$r->[0]

@x->[0]

perldelta

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