Named arguments in PHP - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of function call with named arguments

What is the output of this PHP code using named arguments?

PHP

<?php
function greet($name, $greeting = 'Hello') {
    return "$greeting, $name!";
}
echo greet(greeting: 'Hi', name: 'Alice');
?>

A Hi, Alice!

B SyntaxError

C Hello, Alice!

D Hi, greeting!

💡 Hint

Named arguments allow you to specify parameters by name in any order.

The function greet is called with named arguments greeting: 'Hi' and name: 'Alice'. This overrides the default greeting and prints 'Hi, Alice!'.

❓ Predict Output

intermediate

Named arguments with default values

What will this PHP code output when using named arguments?

PHP

<?php
function calculate($x, $y = 10, $z = 5) {
    return $x + $y * $z;
}
echo calculate(z: 3, x: 2);
?>

A 17

B SyntaxError

C 32

D 15

💡 Hint

Remember the order of operations and default values.

Named arguments assign x=2 and z=3. y uses default 10. So calculation is 2 + 10 * 3 = 2 + 30 = 32.

❓ Predict Output

advanced

Named arguments with variadic function

What is the output of this PHP code using named arguments with a variadic function?

PHP

<?php
function sum($a, ...$numbers) {
    return $a + array_sum($numbers);
}
echo sum(numbers: [2, 3, 4], a: 1);
?>

A Fatal error

B 1

C Warning: Missing argument for variadic parameter

D 10

💡 Hint

Variadic parameters cannot be passed as named arguments as arrays directly.

Passing an array to a variadic parameter using named arguments causes a fatal error because PHP expects separate arguments, not an array.

❓ Predict Output

advanced

Named arguments with type mismatch

What error does this PHP code produce when calling a function with named arguments?

PHP

<?php
function multiply(int $a, int $b) {
    return $a * $b;
}
echo multiply(a: 5, b: 'three');
?>

A 15

B TypeError

C Warning: Invalid argument

D 5three

💡 Hint

PHP 7+ enforces type declarations strictly by default.

Passing a string 'three' to an int parameter causes a TypeError in PHP 7+ with strict typing.

🧠 Conceptual

expert

Behavior of named arguments with inheritance

Consider this PHP code with inheritance and named arguments. What is the output?

PHP

<?php
class Base {
    public function display($x, $y = 10) {
        echo $x + $y;
    }
}
class Child extends Base {
    public function display($x, $y = 20) {
        echo $x * $y;
    }
}
$obj = new Child();
$obj->display(x: 3, y: 4);
?>

A 12

B 7

C Fatal error

D 60

💡 Hint

Check method signatures and parameter order in inheritance.

Child class method display has parameters in the same order as Base but with different default values, which is allowed. So output is 3 * 4 = 12.