Type hinting with parent classes in PHP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to type hint the parameter as the parent class.

PHP

<?php
class Animal {}
function feed([1] $animal) {
    echo "Feeding animal";
}
?>

Drag options to blanks, or click blank then click option'

ABird

BDog

CCat

DAnimal

Attempts:

3 left

💡 Hint

Common Mistakes

Using a child class name instead of the parent class for type hinting.

The parameter should be type hinted as the parent class Animal to accept any subclass instance.

2fill in blank

medium

Complete the method signature to accept any subclass of Vehicle using type hinting.

PHP

<?php
class Vehicle {}
class Car extends Vehicle {}
function startEngine([1] $vehicle) {
    echo "Engine started";
}
?>

Drag options to blanks, or click blank then click option'

ACar

BVehicle

CEngine

DMotor

Attempts:

3 left

💡 Hint

Common Mistakes

Type hinting with a subclass instead of the parent class.

The method should accept the parent class Vehicle to allow any subclass like Car.

3fill in blank

hard

Fix the error by correctly type hinting the parameter to accept any subclass of Fruit.

PHP

<?php
class Fruit {}
class Apple extends Fruit {}
function eat([1] $fruit) {
    echo "Eating fruit";
}
?>

Drag options to blanks, or click blank then click option'

AFruit

BApple

CFood

DVegetable

Attempts:

3 left

💡 Hint

Common Mistakes

Using a subclass name which restricts accepted types.

Type hinting with the parent class Fruit allows the function to accept any subclass like Apple.

4fill in blank

hard

Fill both blanks to create a function that accepts any subclass of Device and calls its method.

PHP

<?php
class Device {
    public function powerOn() {
        echo "Device powered on";
    }
}
class Phone extends Device {}
function activate([1] $device) {
    $device->[2]();
}
?>

Drag options to blanks, or click blank then click option'

ADevice

BpowerOn

Cstart

DPhone

Attempts:

3 left

💡 Hint

Common Mistakes

Type hinting with a subclass or calling a method that does not exist.

The parameter should be type hinted as Device to accept any subclass. The method called is powerOn defined in the parent class.

5fill in blank

hard

Fill all three blanks to define a function that accepts any subclass of Employee, checks the role, and prints a message.

PHP

<?php
class Employee {
    public function getRole() {
        return "Employee";
    }
}
class Manager extends Employee {
    public function getRole() {
        return "Manager";
    }
}
function printRole([1] $emp) {
    if ($emp->getRole() === [2]) {
        echo [3];
    }
}
?>

Drag options to blanks, or click blank then click option'

AEmployee

B"Manager"

C"Manager role detected"

DManager

Attempts:

3 left

💡 Hint

Common Mistakes

Using subclass name as type hint or incorrect string literals.

The function parameter is type hinted as Employee to accept any subclass. It compares the role string to "Manager" and prints the message "Manager role detected".

Practice

1.

What does type hinting with a parent class in PHP allow?

For example, using function foo(ParentClass $obj).

easy

A. It disables type checking for the function parameter.

B. It restricts the function to accept only the exact parent class, not child classes.

C. It only accepts objects of child classes, not the parent class itself.

D. It allows passing any object that is an instance of the parent class or its child classes.

Solution

1. Step 1: Understand type hinting with parent classes

   Type hinting with a parent class means the function accepts objects of that class or any subclass.

2. Step 2: Apply to the example

   Using ParentClass $obj allows passing instances of ParentClass or any class that extends it.

3. Final Answer:

   It allows passing any object that is an instance of the parent class or its child classes. -> Option D

4. Quick Check:

   Parent class type hint = accepts parent or child objects [OK]

Hint: Parent class hint accepts parent and child objects [OK]

Common Mistakes:

• Thinking only exact parent class objects are accepted
• Believing child classes are excluded
• Confusing type hinting with disabling checks

2.

Which of the following is the correct syntax to type hint a function parameter with a parent class Animal in PHP?

function feed(____ $pet) { /* ... */ }

easy

A. Animal $pet

B. $pet Animal

C. class Animal $pet

D. type Animal $pet

Solution

1. Step 1: Recall PHP type hint syntax

   In PHP, the type hint comes before the variable name: Type $variable.

2. Step 2: Apply to the example

   For class Animal, the correct syntax is Animal $pet.

3. Final Answer:

   Animal $pet -> Option A

4. Quick Check:

   Type hint before variable name = Animal $pet [OK]

Hint: Type hint is class name before variable with $ [OK]

Common Mistakes:

• Placing variable before type hint
• Adding extra keywords like 'class' or 'type'
• Omitting the $ sign on variable

3.

Given the code below, what will be the output?

class Vehicle {}
class Car extends Vehicle {}

function startEngine(Vehicle $v) {
    echo get_class($v);
}

$car = new Car();
startEngine($car);

medium

A. Error: Type mismatch

B. Vehicle

C. Car

D. No output

Solution

1. Step 1: Understand type hinting with inheritance

   The function expects a Vehicle object or subclass. Car extends Vehicle, so passing Car is valid.

2. Step 2: Analyze output of get_class()

   get_class($v) returns the actual class name of the object, which is 'Car'.

3. Final Answer:

   Car -> Option C

4. Quick Check:

   Type hint accepts subclass; get_class returns actual class [OK]

Hint: get_class shows actual object class, not type hint [OK]

Common Mistakes:

• Assuming output is parent class name
• Expecting a type error for subclass
• Thinking no output is produced

4.

What is the error in the following PHP code?

class Fruit {}
class Apple extends Fruit {}

function eat(Apple $fruit) {
    echo "Eating apple";
}

$fruit = new Fruit();
eat($fruit);

medium

A. Syntax error: missing semicolon

B. Fatal error: Argument must be of type Apple, Fruit given

C. No error, outputs 'Eating apple'

D. Warning: Undefined variable $fruit

Solution

1. Step 1: Check type hint and argument

   The function expects an Apple object, but the argument is a Fruit object (parent class).

2. Step 2: Understand PHP type hint rules

   PHP requires the argument to be the hinted class or subclass, not a parent class. Passing Fruit causes a fatal error.

3. Final Answer:

   Fatal error: Argument must be of type Apple, Fruit given -> Option B

4. Quick Check:

   Type hint mismatch with parent class argument = fatal error [OK]

Hint: Child class hint rejects parent class objects [OK]

Common Mistakes:

• Thinking parent class can replace child hint
• Expecting no error on type mismatch
• Confusing syntax error with type error

5.

Consider the following classes and function:

class Employee {}
class Manager extends Employee {}
class Developer extends Employee {}

function assignTask(Employee $e) {
    if ($e instanceof Manager) {
        return "Assigning management task";
    } elseif ($e instanceof Developer) {
        return "Assigning coding task";
    } else {
        return "Assigning general task";
    }
}

$worker = new Manager();
echo assignTask($worker);

What will be the output and why?

hard

A. Assigning management task

B. Assigning coding task

C. Assigning general task

D. Type error: argument must be Employee

Solution

1. Step 1: Check type hint and argument

   The function accepts any Employee or subclass. Manager is a subclass, so passing $worker is valid.

2. Step 2: Evaluate instanceof conditions

   The code checks if $e is a Manager, which is true, so it returns "Assigning management task".

3. Final Answer:

   Assigning management task -> Option A

4. Quick Check:

   Type hint accepts subclass; instanceof detects Manager [OK]

Hint: Type hint parent + instanceof child checks for behavior [OK]

Common Mistakes:

• Expecting type error for subclass argument
• Ignoring instanceof checks order
• Assuming general task is assigned