Array References in PHP

In PHP, unlike other programming languages, arrays are also copied, not passed by reference.

Let's verify this:

php<?php
  $arr1 = [1, 2, 3, 4, 5];
  $arr2 = $arr1;
  $arr2[0] = '!';
  
  var_dump($arr1); // won't change
?>

Now let's force the array to be passed by reference:

php<?php
  $arr1 = [1, 2, 3, 4, 5];
  $arr2 = &$arr1;
  $arr2[0] = '!';
  
  var_dump($arr1); // will change
?>

Задание 1

Explain what the result of executing the code will be:

php<?php
  $arr1 = [1, 2, 3, 4, 5];
  
  $arr2 = $arr1;
  $arr2[0] = '!';
  
  echo $arr1[0];
  echo $arr2[0];
?>

Задание 2

Explain what the result of executing the code will be:

php<?php
  $arr1 = [1, 2, 3, 4, 5];
  
  $arr2 = &$arr1;
  $arr1[0]++;
  
  echo $arr1[0];
  echo $arr2[0];
?>

Задание 3

Explain what the result of executing the code will be:

php<?php
  $arr1 = [1, 2, 3, 4, 5];
  
  $arr2 = &$arr1;
  
  $arr1[0]++;
  $arr2[0]++;
  
  echo $arr1[0];
  echo $arr2[0];
?>