[Solved] Identify the error in this PHP code snippet:$arr = [1, 2, 3, 4]; $filtered = array_filter($arr, function($n) { if ($n % 2 == 0) { return true; } }); | PHP

PHP - Array Functions

Identify the error in this PHP code snippet:

$arr = [1, 2, 3, 4];
$filtered = array_filter($arr, function($n) {
  if ($n % 2 == 0) {
    return true;
  }
});

AThe callback function does not return false for odd numbers

Barray_filter requires a third parameter

CThe array must be associative

DThe callback function should not use if statements

Step-by-Step Solution

Solution:

Step 1: Analyze callback return values
The callback returns true for even numbers but returns nothing (null) for odd numbers.
Step 2: Understand array_filter behavior
Null is treated as false, so odd numbers are filtered out correctly, but explicit false return is better practice.
Final Answer:
The callback function does not return false for odd numbers -> Option A
Quick Check:
Callback should return true or false explicitly [OK]

Quick Trick: Always return true or false explicitly in callback [OK]

Common Mistakes:

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