Array access and modification in PHP - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this PHP code?

Consider the following PHP code snippet:

$arr = [1, 2, 3, 4];
$arr[2] = $arr[0] + $arr[1];
echo $arr[2];

What will be printed?

PHP

$arr = [1, 2, 3, 4];
$arr[2] = $arr[0] + $arr[1];
echo $arr[2];

A 5

B 3

C 6

D Undefined offset error

💡 Hint

Remember that array indices start at 0 and you can assign new values to existing keys.

The original value at index 2 is 3, but it is replaced by the sum of values at indices 0 and 1, which are 1 and 2. So, 1 + 2 = 3, but since the original value was 3, the new value is 3. However, the code actually sums 1 + 2 = 3, so the output is 3.

❓ Predict Output

intermediate

What is the output after modifying a nested array?

Look at this PHP code:

$arr = ["a" => [1, 2], "b" => [3, 4]];
$arr["a"][1] = 5;
echo $arr["a"][1];

What will it print?

PHP

$arr = ["a" => [1, 2], "b" => [3, 4]];
$arr["a"][1] = 5;
echo $arr["a"][1];

A 5

B 2

C 1

D Undefined index error

💡 Hint

Think about how to access and change values inside nested arrays.

The code changes the second element (index 1) of the array under key "a" from 2 to 5. So echo prints 5.

🔧 Debug

advanced

Identify the error in this array modification code

What error does this PHP code produce?

$arr = [10, 20, 30];
$arr[3] += 5;
echo $arr[3];

PHP

$arr = [10, 20, 30];
$arr[3] += 5;
echo $arr[3];

A Outputs 5

B Notice: Undefined offset: 3

C Fatal error: Cannot use undefined offset

D Outputs 0

💡 Hint

Think about what happens when you try to add to an array element that does not exist yet.

Accessing $arr[3] before it is set triggers a notice about undefined offset. Then PHP treats it as 0 and adds 5, so echo prints 5, but the notice is shown.

📝 Syntax

advanced

Which option causes a syntax error in array modification?

Which of the following PHP code snippets will cause a syntax error?

A

$arr = [1, 2, 3];
$arr[0] = $arr[0] + 1;

B

$arr = [1, 2, 3];
$arr[] = 4;

C

$arr = [1, 2, 3];
$arr[2] +=;

D

$arr = [1, 2, 3];
$arr[1] = 5;

💡 Hint

Look carefully at the assignment operators and their usage.

Option C has an incomplete assignment operator '+=' without a value, causing a syntax error.

🚀 Application

expert

How many elements are in the array after these modifications?

Given this PHP code:

$arr = [];
$arr[2] = 'a';
$arr[] = 'b';
$arr[5] = 'c';
$arr[] = 'd';

How many elements does $arr have after these lines run?

PHP

$arr = [];
$arr[2] = 'a';
$arr[] = 'b';
$arr[5] = 'c';
$arr[] = 'd';

A 6

B 3

C 5

D 4

💡 Hint

Remember how PHP assigns keys when using empty brackets and how explicit keys affect the array.

After the first assignment, $arr has key 2. Then $arr[] adds 'b' at the next highest integer key, which is 3. Then key 5 is set to 'c'. Then $arr[] adds 'd' at key 6. So keys are 2,3,5,6 → 4 elements total.

Practice

1. What is the correct way to access the first element of an array $arr in PHP?

easy

A. $arr{0}

B. $arr[1]

C. $arr[0]

D. $arr->0

Solution

1. Step 1: Understand PHP array indexing

   PHP arrays start at index 0, so the first element is at index 0.

2. Step 2: Identify correct syntax for array access

   Use square brackets with the index number: $arr[0].

3. Final Answer:

   $arr[0] -> Option C

4. Quick Check:

   First element index = 0 [OK]

Hint: First element is always at index 0 in PHP arrays [OK]

Common Mistakes:

• Using index 1 for first element
• Using curly braces for array access
• Using object property syntax

2. Which of the following is the correct way to change the value of the third element in array $arr to 10?

easy

A. $arr(3) = 10;

B. $arr[2] = 10;

C. $arr->2 = 10;

D. $arr{3} = 10;

Solution

1. Step 1: Identify the index of the third element

   Arrays start at 0, so the third element is at index 2.

2. Step 2: Use correct syntax to assign value

   Use square brackets with index 2: $arr[2] = 10;.

3. Final Answer:

   $arr[2] = 10; -> Option B

4. Quick Check:

   Index 2 for third element assignment [OK]

Hint: Use square brackets and zero-based index to modify array [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Using object property syntax
• Using wrong index number

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

$arr = [5, 10, 15];
$arr[1] = $arr[1] + 5;
echo $arr[1];

medium

A. 15

B. 10

C. 20

D. 5

Solution

1. Step 1: Understand initial array values

   $arr = [5, 10, 15]; so $arr[1] is 10.

2. Step 2: Calculate new value of $arr[1]

   $arr[1] = $arr[1] + 5 means 10 + 5 = 15.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   10 + 5 = 15 [OK]

Hint: Add to existing value using $arr[index] + number [OK]

Common Mistakes:

• Printing old value without update
• Adding index instead of value
• Confusing array indexes

4. Identify the error in the following PHP code snippet:

$arr = [1, 2, 3];
$arr[3] = 4;
echo $arr[3];

medium

A. Accessing an undefined index causes an error

B. Cannot echo array elements directly

C. Syntax error in array declaration

D. Assigning to index 3 adds a new element without error

Solution

1. Step 1: Check array indexes and assignment

   Array has indexes 0,1,2 initially. Assigning $arr[3] = 4 adds a new element at index 3.

2. Step 2: Confirm if this causes error

   PHP allows adding new elements by assigning to new indexes without error.

3. Final Answer:

   Assigning to index 3 adds a new element without error -> Option D

4. Quick Check:

   PHP arrays auto-expand on new index assignment [OK]

Hint: Assigning new index adds element; no error in PHP [OK]

Common Mistakes:

• Thinking accessing undefined index causes error
• Confusing syntax error with runtime behavior
• Believing echo can't print array elements

5. Given the array $arr = [2, 4, 6, 8];, which code snippet correctly doubles each element's value by modifying the array in place?

hard

A. for ($i = 0; $i < count($arr); $i++) { $arr[$i] *= 2; }

B. $arr = array_fill(0, count($arr), 2 * $arr[0]);

C. array_map(fn($v) => $v * 2, $arr);

D. foreach ($arr as $value) { $value = $value * 2; }

Solution

1. Step 1: Understand how to modify array elements in place

   Using a for loop with index allows direct modification of each element.

2. Step 2: Analyze each option

   A: foreach copies values, does not modify original array.
   B: array_fill replaces all elements with same value, incorrect.
   C: array_map returns new array, original unchanged.
   D: for loop modifies $arr[$i] directly.

3. Final Answer:

   for ($i = 0; $i < count($arr); $i++) { $arr[$i] *= 2; } -> Option A

4. Quick Check:

   Use indexed loop to modify array elements [OK]

Hint: Use for loop with indexes to change array elements directly [OK]

Common Mistakes:

• Modifying foreach loop variable instead of array
• Expecting array_map to change original array
• Using array_fill incorrectly