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  • Perl

Idiom #312 Test for list equality

Set b to true if the lists p and q have the same size and the same elements, false otherwise.

use v5.10.1;
no warnings 'experimental::smartmatch';
$b = @array1 ~~ @array2;

The smartmatch operator will recursively compare two arrays (lists actually) and return true if it finds the same element values in both. See the documentation for specifics.

Unfortunately, smartmatch remains experimental because of its byzantine complexity. For this application it's fine.
use v5.10;
use smart::match;
my $b = \@array1 |M| \@array2;

Uses the CPAN smart::match module to enable smart match operator |M|. Unlike perl's experimental smartmatch operator ~~, array arguments must be references (hence the backslashes).
use v5.10;
use strict;
sub lcomp {
    my ($a, $b) = @_;

    return 0 if ( @$a != @$b ); # lengths different

    my $matched = 1;
    for (my $i=0; $i < @$a; $i++) {
        return 0 unless $a->[$i] == $b->[$i];
    }

    return 1;
}

This does a numeric comparison of two flat lists, with minimal error checking. Perl's dynamic typing means lists can contain arbitrary mixing and nesting of strings and numbers or lists or hashes; For that you should use a CPAN module like Data::Compare.
use Data::Compare;
$b = Compare( \@p, \@q );

CPAN module Data:Compare imports function Compare which can deeply compare two perl structures (lists, hashes, arbitrarily mixed and nested). If they are the same, true is returned.
#include <algorithm>
bool b = ::std::ranges::equal(p, q);
New implementation...
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programming-idioms.org 
$b = @array1 ~~ @array2;
my $b = \@array1 |M| \@array2;
sub lcomp {
    my ($a, $b) = @_;

    return 0 if ( @$a != @$b ); # lengths different

    my $matched = 1;
    for (my $i=0; $i < @$a; $i++) {
        return 0 unless $a->[$i] == $b->[$i];
    }

    return 1;
}

$b = Compare( \@p, \@q );
bool b = ::std::ranges::equal(p, q);
  if (size(p,1) /= size(q,1)) then
    b = false
  else
    b = all (p == q)
  end if

b := slices.Equal(p, q)
b = p.equals(q);
  b := (p.count = q.count);
  if b then for i := 0 to p.count-1 do 
    if (p.items[i] <> q.items[i]) then
    begin
      b := false;
      break;
    end;
b = p == q
b = p == q
b = p == q;