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4.39: What is the difference between a list and an array?

(contributed by brian d foy)

A list is a fixed collection of scalars. An array is a variable that
holds a variable collection of scalars. An array can supply its
collection for list operations, so list operations also work on arrays:

# slices
( 'dog', 'cat', 'bird' )[2,3];

# iteration
foreach ( qw( dog cat bird ) ) { ... }
foreach ( @animals ) { ... }

my @three = grep { length == 3 } qw( dog cat bird );
my @three = grep { length == 3 } @animals;

# supply an argument list
wash_animals( qw( dog cat bird ) );
wash_animals( @animals );

Array operations, which change the scalars, reaaranges them, or adds or
subtracts some scalars, only work on arrays. These can't work on a list,
which is fixed. Array operations include "shift", "unshift", "push",
"pop", and "splice".

An array can also change its length:

$#animals = 1; # truncate to two elements
$#animals = 10000; # pre-extend to 10,001 elements

You can change an array element, but you can't change a list element:

$animals[0] = 'Rottweiler';
qw( dog cat bird )[0] = 'Rottweiler'; # syntax error!

foreach ( @animals ) {
s/^d/fr/; # works fine

foreach ( qw( dog cat bird ) ) {
s/^d/fr/; # Error! Modification of read only value!

However, if the list element is itself a variable, it appears that you
can change a list element. However, the list element is the variable,
not the data. You're not changing the list element, but something the
list element refers to. The list element itself doesn't change: it's
still the same variable.

You also have to be careful about context. You can assign an array to a
scalar to get the number of elements in the array. This only works for
arrays, though:

my $count = @animals; # only works with arrays

If you try to do the same thing with what you think is a list, you get a
quite different result. Although it looks like you have a list on the
righthand side, Perl actually sees a bunch of scalars separated by a

my $scalar = ( 'dog', 'cat', 'bird' ); # $scalar gets bird

Since you're assigning to a scalar, the righthand side is in scalar
context. The comma operator (yes, it's an operator!) in scalar context
evaluates its lefthand side, throws away the result, and evaluates it's
righthand side and returns the result. In effect, that list-lookalike
assigns to $scalar it's rightmost value. Many people mess this up
becuase they choose a list-lookalike whose last element is also the
count they expect:

my $scalar = ( 1, 2, 3 ); # $scalar gets 3, accidentally


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