[Solved] What is the issue with the following PHP code snippet?class Animal {} class Cat extends Animal {} function petAnimal(Animal $a) { echo "Petting animal"; } petAnimal('cat'); | PHP

PHP - Inheritance and Polymorphism

What is the issue with the following PHP code snippet?

class Animal {}
class Cat extends Animal {}

function petAnimal(Animal $a) {
    echo "Petting animal";
}

petAnimal('cat');

APassing a string instead of an object causes a type error.

BThe function should type hint with Cat instead of Animal.

CThe class Cat must implement an interface to be accepted.

DThere is no error; the code runs fine.

Step-by-Step Solution

Solution:

Step 1: Check parameter type hint
The function expects an object of class Animal or subclass.
Step 2: Analyze the argument
The argument passed is a string 'cat', not an object.
Step 3: Result
PHP will throw a TypeError because the argument is not an object of Animal or subclass.
Final Answer:
Passing a string instead of an object causes a type error. -> Option A
Quick Check:
Type hint requires object, not string [OK]

Quick Trick: Type hint expects object, not string [OK]

Common Mistakes:

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