Type hinting with parent classes in PHP - Mini Project: Build & Apply

Type hinting with parent classes

📖 Scenario: You are building a simple system to manage different types of vehicles. Each vehicle has a method to display its type. You want to use type hinting with a parent class to ensure your functions accept any vehicle type.

🎯 Goal: Create a parent class Vehicle and two child classes Car and Bike. Then write a function that accepts a Vehicle type and calls its method to display the vehicle type.

📋 What You'll Learn

Create a parent class called Vehicle with a method getType().

Create two child classes Car and Bike that extend Vehicle and override getType().

Write a function printVehicleType that accepts a parameter type hinted as Vehicle.

Call getType() inside printVehicleType and print the result.

💡 Why This Matters

🌍 Real World

Type hinting with parent classes helps write flexible code that works with many related objects, like different vehicle types in a transport app.

💼 Career

Understanding inheritance and type hinting is essential for writing clean, maintainable PHP code in many software development jobs.

1

Create the parent class Vehicle

Create a class called Vehicle with a public method getType() that returns the string 'Generic Vehicle'.

PHP

<?php

// Create class Vehicle with method getType()
// Your code here

Hint

Use class Vehicle {} and inside it define public function getType(): string that returns 'Generic Vehicle'.

2

Create child classes Car and Bike

Create two classes called Car and Bike that extend Vehicle. Override the getType() method in each to return 'Car' and 'Bike' respectively.

PHP

<?php

class Vehicle {
    public function getType(): string {
        return 'Generic Vehicle';
    }
}

// Create class Car extending Vehicle
// Create class Bike extending Vehicle
// Your code here

Hint

Use class Car extends Vehicle and override getType() to return 'Car'. Do the same for Bike returning 'Bike'.

3

Write function with type hinting for Vehicle

Write a function called printVehicleType that accepts one parameter $vehicle type hinted as Vehicle. Inside the function, call $vehicle->getType() and print the returned string.

PHP

<?php

class Vehicle {
    public function getType(): string {
        return 'Generic Vehicle';
    }
}

class Car extends Vehicle {
    public function getType(): string {
        return 'Car';
    }
}

class Bike extends Vehicle {
    public function getType(): string {
        return 'Bike';
    }
}

// Write function printVehicleType with parameter type hinted as Vehicle
// Your code here

Hint

Define function printVehicleType(Vehicle $vehicle): void and inside use echo $vehicle->getType() to print the type.

4

Call function with Car and Bike objects and print output

Create an object $car of class Car and an object $bike of class Bike. Call printVehicleType with $car and then with $bike to print their types.

PHP

getType() . "\n";
}

// Create $car as Car object
// Create $bike as Bike object
// Call printVehicleType with $car and $bike
// Your code here

Hint

Create objects with new Car() and new Bike(). Then call printVehicleType($car) and printVehicleType($bike).

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