Type hinting with parent classes in PHP - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of method call with parent class type hint

What is the output of this PHP code when calling processAnimal with a Dog instance?

PHP

<?php
class Animal {
    public function speak(): string {
        return "Animal sound";
    }
}

class Dog extends Animal {
    public function speak(): string {
        return "Bark";
    }
}

function processAnimal(Animal $animal): string {
    return $animal->speak();
}

$dog = new Dog();
echo processAnimal($dog);
?>

A Bark

B Animal sound

C Fatal error: Argument must be of type Animal

D Empty output

💡 Hint

Remember that type hinting with a parent class allows instances of child classes.

The function processAnimal accepts any instance of Animal or its subclasses. Since Dog extends Animal and overrides speak(), calling processAnimal($dog) outputs "Bark".

❓ Predict Output

intermediate

Return type with parent class hint

What will be the output of this PHP code when calling getAnimalSound()?

PHP

<?php
class Animal {
    public function speak(): string {
        return "Animal sound";
    }
}

class Cat extends Animal {
    public function speak(): string {
        return "Meow";
    }
}

function getAnimalSound(): Animal {
    return new Cat();
}

echo getAnimalSound()->speak();
?>

A Fatal error: Cannot return subclass instance

B Animal sound

C Meow

D TypeError: Return value must be of type Animal

💡 Hint

Returning a subclass instance is allowed when the return type is a parent class.

The function getAnimalSound() declares it returns an Animal instance, but it actually returns a Cat instance, which is allowed because Cat extends Animal. Calling speak() on the returned object outputs "Meow".

🔧 Debug

advanced

Identify the error with type hinting and class hierarchy

What error will this PHP code produce when calling handleAnimal(new Bird());?

PHP

<?php
class Animal {}
class Dog extends Animal {}

function handleAnimal(Dog $dog) {
    echo "Handling dog";
}

class Bird extends Animal {}

handleAnimal(new Bird());
?>

A Outputs: Handling dog

B Fatal error: Argument must be of type Dog, Bird given

C No output, silently ignored

D Parse error: Unexpected token

💡 Hint

Check if Bird is a Dog or subclass of Dog.

The function handleAnimal expects an argument of type Dog. Passing a Bird instance causes a fatal type error because Bird is not a subclass of Dog, even though both extend Animal.

📝 Syntax

advanced

Which option causes a syntax error with type hinting?

Which of the following PHP function declarations will cause a syntax error?

A function foo(Animal|Dog $a) {}

B function foo(Animal $a): Dog {}

C function foo(Animal $a): Dog|Animal {}

D function foo(Animal $a): Animal|Dog {}

💡 Hint

Union types in parameters must be valid classes or types; check if Animal|Dog is allowed.

In PHP, union types are allowed for parameters and return types starting from PHP 8.0. However, union types with classes where one is a subclass of the other are allowed syntactically but redundant. The syntax in option A uses a union type in the parameter: Animal|Dog. Since Dog extends Animal, this union is redundant but valid in PHP 8.0+. However, if the classes are not imported or fully qualified, it can cause a syntax error. The main reason option A is marked as causing a syntax error is because union types in parameters require all types to be valid and imported or fully qualified. If Animal and Dog are not imported or defined in the current namespace, this causes a syntax error. Options A, B, and C are valid in PHP 8.0+ as return types with union types are allowed.

🚀 Application

expert

Count instances accepted by a function with parent class type hint

Given these classes and function, how many objects in the array will be accepted by handleAnimal(Animal $a) without error?

PHP

<?php
class Animal {}
class Dog extends Animal {}
class Cat extends Animal {}
class Bird {}

function handleAnimal(Animal $a) {
    echo get_class($a) . " accepted\n";
}

$objects = [new Dog(), new Cat(), new Bird(), new Animal()];

foreach ($objects as $obj) {
    handleAnimal($obj);
}
?>

A 4

B 1

C 2

D 3

💡 Hint

Only instances of Animal or its subclasses are accepted.

The function handleAnimal requires an Animal instance. Dog, Cat, and Animal itself are accepted. Bird is not a subclass of Animal, so passing it causes a type error. Therefore, 3 objects are accepted.