Re^2: Arrays in arrays, how to access them

subcall( \@{$M[0]} );

Which passes a reference to the first array in @M.

So it does, but if the array @M is initialized per something like
my @M = (\@a1, \@a2, \@a3);
then the elements of @M are all array references to begin with, so why not just
subcall($M[0]);
and avoid taking a reference to a de-referenced array reference?

(And see also perllol "Manipulating Arrays of Arrays in Perl".) (Update: And also perlreftut.)

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Re^3: Arrays in arrays, how to access them by viored (Novice) on Oct 31, 2013 at 02:27 UTC
When I use `subcall( $M[0] );` [download] I get ARRAY(0xa07b5d8), not a the output of the function I'm interested in. Same thing for `subcall( \@{M[0]} )` [download] What am I missing?	[reply] [d/l] [select]
Re^3: Arrays in arrays, how to access them by viored (Novice) on Oct 31, 2013 at 02:46 UTC
Ah, but if I use `subcall ( @{\@{$M[0]}} )` [download] It works. Thank you so much, both of of you.	[reply] [d/l]
Re^4: Arrays in arrays, how to access them by AnomalousMonk (Archbishop) on Oct 31, 2013 at 07:06 UTC
`subcall( $M[0] );` Assuming that `@M` is initialized as shown in Re^2: Arrays in arrays, how to access them, you get, in this instance, an array reference passed to the `subcall()` function because that's what element 0 of the `@M` array is. `subcall( \@{M[0]} )` In this instance (assuming it's fixed to `subcall( \@{$M[0]} )` by adding the missing `$` sigil), you have, working from the inside out: an array reference: `$M[0]` de-referenced to an array: `@{$M[0]}` to which a reference is taken: `\@{$M[0]}` this reference then being passed to a function: `subcall( \@{$M[0]} )` In this instance, you're simply wrapping the access to `$M[0]` in a redundant pair of complementary operations — and maybe in this case the compiler would even be smart enough to optimize away the redundancy! It's as if you had asked "Why do the two function calls `func($x);` and `func($x + 1729 - 137 - 1729 + 137);` both pass the same value to the function?" Ah, but if I use `subcall ( @{\@{$M[0]}} )` ... If you use `subcall ( @{\@{$M[0]}} )` you extend the process of the second instance above to de-reference the array reference created (redundantly) by `\@{$M[0]}` and pass an array (in the example, the `@a1` array) to the `subcall()` function, which is apparently what it takes to make that function happy. `>perl -wMstrict -le "my @a1 = qw(a b c); my @a2 = qw(p q r); my @a3 = qw(x y z); my @M = (\@a1, \@a2, \@a3); ;; print $M[0]; print \@{$M[0]}; ;; print 'array references are the same' if $M[0] == \@{$M[0]}; ;; print qq{-@{\@{$M[0]}}- -@{$M[0]}-}; " ARRAY(0x43281c) ARRAY(0x43281c) array references are the same -a b c- -a b c-` [download] (Also see perlreftut.)	[reply] [d/l] [select]