Array walk function in PHP - Step-by-Step Execution

Concept Flow - Array walk function

Start with array

↓

Pick first element

↓

Apply callback function

↓

Move to next element

↓

More elements?

Yes→Repeat callback

No↓

End

The array walk function goes through each element of an array and applies a callback function to it one by one.

Execution Sample

PHP

<?php
$arr = [1, 2, 3];
array_walk($arr, function(&$v) { $v *= 2; });
print_r($arr);
?>

This code doubles each element in the array by walking through it and applying a function.

Execution Table

Step	Current Element (value)	Callback Action	Array State After Action
1	1	Multiply by 2 → 2	[2, 2, 3]
2	2	Multiply by 2 → 4	[2, 4, 3]
3	3	Multiply by 2 → 6	[2, 4, 6]
4	No more elements	Stop	[2, 4, 6]

💡 All elements processed, array walk ends.

Variable Tracker

Variable	Start	After 1	After 2	After 3	Final
arr	[1, 2, 3]	[2, 2, 3]	[2, 4, 3]	[2, 4, 6]	[2, 4, 6]
v (current element)	1	2	4	6	6

Key Moments - 2 Insights

Why does the array change even though the callback uses a parameter?

Because the callback parameter is passed by reference (&$v), changes inside the callback affect the original array elements, as shown in steps 1-3 of the execution_table.

What happens if the callback does not modify the element?

The array stays the same because array_walk applies the callback but does not change elements unless the callback modifies them, as seen in the exit step where no changes occur.

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after step 2?

A [2, 2, 3]

B [2, 4, 3]

C [1, 2, 3]

D [2, 4, 6]

💡 Hint

Check the 'Array State After Action' column for step 2 in the execution_table.

At which step does the array element 3 become 6?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'Current Element' and 'Callback Action' columns in the execution_table.

If the callback function did not use & (reference), what would happen to the array?

A Array elements would remain unchanged

B Array elements would still be doubled

C Array would be emptied

D Callback would cause an error

💡 Hint

Refer to variable_tracker and key_moments about passing by reference.

Concept Snapshot

array_walk(array, callback) applies callback to each element.
Callback can modify elements if parameter is passed by reference (&).
Array changes happen inside callback if references used.
No return value; original array is modified directly.
Useful for simple element-wise operations.

Full Transcript

The array_walk function in PHP takes an array and a callback function. It goes through each element of the array one by one. For each element, it runs the callback function. If the callback changes the element by reference, the original array changes. In the example, each number in the array is doubled. The execution table shows each step: starting with 1, doubling it to 2, then 2 to 4, and 3 to 6. After all elements are processed, the function ends. This helps understand how array_walk modifies arrays by applying a function to each element.

Practice

1. What does the PHP function array_walk do?

easy

A. It applies a user-defined function to every element of an array.

B. It sorts the array in ascending order.

C. It merges two arrays into one.

D. It removes duplicate values from an array.

Solution

1. Step 1: Understand the purpose of array_walk

   The function array_walk is designed to apply a callback function to each element of an array.

2. Step 2: Compare with other array functions

   Sorting, merging, or removing duplicates are done by other functions like sort, array_merge, or array_unique.

3. Final Answer:

   It applies a user-defined function to every element of an array. -> Option A

4. Quick Check:

   array_walk = apply function to each element [OK]

Hint: array_walk runs a function on each array item [OK]

Common Mistakes:

• Confusing array_walk with sorting functions
• Thinking it merges arrays
• Assuming it removes duplicates

2. Which of the following is the correct syntax to use array_walk with a callback function named modify on an array $arr?

easy

A. array_walk('modify', $arr);

B. array_walk(modify, $arr);

C. array_walk($arr, modify);

D. array_walk($arr, 'modify');

Solution

1. Step 1: Recall array_walk syntax

   The correct syntax is array_walk(array, callback), where callback is a string with the function name.

2. Step 2: Check the options

   Only array_walk($arr, 'modify'); matches the correct order and uses quotes around the callback function name.

3. Final Answer:

   array_walk($arr, 'modify'); -> Option D

4. Quick Check:

   array_walk(array, 'function') syntax [OK]

Hint: array_walk takes array first, then callback as string [OK]

Common Mistakes:

• Swapping the order of parameters
• Omitting quotes around callback name
• Passing callback without quotes

3. What will be the output of the following PHP code?

$numbers = [1, 2, 3];
array_walk($numbers, function(&$value) {
    $value *= 2;
});
print_r($numbers);

medium

A. Array ( [0] => 2 [1] => 4 [2] => 6 )

B. Array ( [0] => 1 [1] => 2 [2] => 3 )

C. Array ( [0] => 0 [1] => 0 [2] => 0 )

D. Error: Cannot modify array elements

Solution

1. Step 1: Understand the callback function

   The callback multiplies each value by 2, and the value is passed by reference (&$value), so changes affect the original array.

2. Step 2: Apply changes to each element

   1*2=2, 2*2=4, 3*2=6, so the array becomes [2, 4, 6].

3. Final Answer:

   Array ( [0] => 2 [1] => 4 [2] => 6 ) -> Option A

4. Quick Check:

   Reference callback doubles values [OK]

Hint: Pass value by reference to modify array items [OK]

Common Mistakes:

• Not using & to modify original array
• Expecting original array unchanged
• Confusing output format

4. Identify the error in this code snippet:

$arr = [1, 2, 3];
array_walk($arr, function($val) {
    $val += 5;
});
print_r($arr);

medium

A. The array must be passed by reference to array_walk.

B. array_walk requires a named function, not an anonymous function.

C. Callback function parameter should be passed by reference to modify array.

D. print_r cannot print arrays modified by array_walk.

Solution

1. Step 1: Check how callback modifies array

   The callback parameter $val is passed by value, so changes inside do not affect original array.

2. Step 2: Correct usage for modification

   Passing &$val by reference allows modifying the original array elements.

3. Final Answer:

   Callback function parameter should be passed by reference to modify array. -> Option C

4. Quick Check:

   Use & in callback to change array items [OK]

Hint: Use & in callback parameter to modify array elements [OK]

Common Mistakes:

• Ignoring reference in callback parameter
• Thinking array_walk needs array by reference
• Believing anonymous functions are not allowed

5. Given the array $fruits = ['apple' => 2, 'banana' => 3, 'cherry' => 5];, how can you use array_walk to add 10 to each value only if the value is less than 4?

hard

A. array_walk($fruits, function($val) { if ($val < 4) $val += 10; });

B. array_walk($fruits, function(&$val) { if ($val < 4) $val += 10; });

C. array_walk($fruits, function(&$val, $key) { if ($key < 4) $val += 10; });

D. array_walk($fruits, function(&$val, $key) { if ($val > 4) $val += 10; });

Solution

1. Step 1: Identify condition and modification

   We want to add 10 to values less than 4, so check $val < 4 inside the callback.

2. Step 2: Ensure modification affects original array

   Pass $val by reference (&$val) to change original array values.

3. Step 3: Verify options

   array_walk($fruits, function(&$val) { if ($val < 4) $val += 10; }); correctly uses reference and condition. array_walk($fruits, function($val) { if ($val < 4) $val += 10; }); lacks reference, so no change. array_walk($fruits, function(&$val, $key) { if ($key < 4) $val += 10; }); checks key instead of value. array_walk($fruits, function(&$val, $key) { if ($val > 4) $val += 10; }); adds 10 when value is greater than 4, which is wrong.

4. Final Answer:

   array_walk($fruits, function(&$val) { if ($val < 4) $val += 10; }); -> Option B

5. Quick Check:

   Reference + condition on value < 4 [OK]

Hint: Use & and if condition inside callback to selectively modify [OK]

Common Mistakes:

• Not passing value by reference
• Checking key instead of value
• Using wrong comparison operator