Type declarations for parameters in PHP - Mini Project: Build & Apply

Type Declarations for Parameters in PHP

📖 Scenario: You are building a simple calculator that adds two numbers. To make sure the calculator works correctly, you want to tell PHP exactly what kind of values the calculator should accept.

🎯 Goal: Create a PHP function that adds two numbers using type declarations for the parameters to ensure only numbers are accepted.

📋 What You'll Learn

Create a function named addNumbers with two parameters named $a and $b.

Use type declarations to specify that both $a and $b must be integers.

Call the function with two integer values.

Print the result of the function call.

💡 Why This Matters

🌍 Real World

Type declarations help prevent errors by making sure functions get the right kind of data, which is important in real-world applications like calculators, data processing, and web forms.

💼 Career

Understanding type declarations is useful for writing clear, reliable PHP code, a skill valued in web development jobs.

1

Create the function skeleton

Write a function named addNumbers that takes two parameters named $a and $b without type declarations yet.

PHP

<?php

// Define the function addNumbers with parameters $a and $b
function addNumbers($a, $b) {
    // Your code here
}

Hint

Use the function keyword followed by the function name and parentheses with parameters.

2

Add type declarations to parameters

Modify the addNumbers function to add type declarations so that both parameters $a and $b are declared as int.

PHP

<?php

// Add type declarations to parameters $a and $b
function addNumbers(int $a, int $b) {
    // Your code here
}

Hint

Put int before each parameter name inside the parentheses.

3

Add the function logic

Inside the addNumbers function, write code to return the sum of $a and $b.

PHP

<?php

function addNumbers(int $a, int $b) {
    // Return the sum of $a and $b
    // Your code here
}

Hint

Use the return keyword followed by $a + $b.

4

Call the function and print the result

Call the addNumbers function with the integers 5 and 7, store the result in a variable named $result, and print $result.

PHP

<?php

function addNumbers(int $a, int $b) {
    return $a + $b;
}

// Call addNumbers with 5 and 7, store in $result
// Print $result
// Your code here

Hint

Use $result = addNumbers(5, 7); and then print($result);.

Practice

1. What does a type declaration for a function parameter in PHP do?

easy

A. It tells PHP what kind of data the function expects for that parameter.

B. It changes the function's return type automatically.

C. It makes the function run faster by optimizing memory.

D. It hides the parameter from being accessed outside the function.

Solution

1. Step 1: Understand the purpose of type declarations

   Type declarations specify the expected data type for function parameters, helping PHP check inputs.

2. Step 2: Identify what type declarations do not do

   They do not change return types, optimize speed, or hide parameters.

3. Final Answer:

   It tells PHP what kind of data the function expects for that parameter. -> Option A

4. Quick Check:

   Type declarations = specify expected parameter type [OK]

Hint: Type declarations specify expected input types for functions [OK]

Common Mistakes:

• Thinking type declarations change return types
• Believing they improve speed directly
• Confusing parameter visibility with type declarations

2. Which of the following is the correct syntax to declare a function parameter as an integer in PHP?

easy

A. function add(integer $a, integer $b) {}

B. function add($int a, $int b) {}

C. function add(int: $a, int: $b) {}

D. function add(int $a, int $b) {}

Solution

1. Step 1: Recall PHP type declaration syntax

   PHP uses the type name followed by the parameter with a $ sign, like int $a.

2. Step 2: Check each option's syntax

   function add(int $a, int $b) {} matches correct syntax. Options A, B, and D use invalid syntax.

3. Final Answer:

   function add(int $a, int $b) {} -> Option D

4. Quick Check:

   Correct syntax = type before $parameter [OK]

Hint: Type before $parameter name, no extra symbols [OK]

Common Mistakes:

• Putting $ before type name
• Using 'integer' instead of 'int'
• Adding colons or extra symbols

3. What will be the output of this PHP code?

function multiply(float $x, float $y) {
    return $x * $y;
}
echo multiply(3, 4.5);

medium

A. 13.5

B. 12

C. TypeError

D. 7.5

Solution

1. Step 1: Understand parameter types and argument passing

   The function expects floats. PHP will convert integer 3 to float 3.0 automatically.

2. Step 2: Calculate the multiplication

   3.0 * 4.5 = 13.5, so the function returns 13.5.

3. Final Answer:

   13.5 -> Option A

4. Quick Check:

   3 * 4.5 = 13.5 [OK]

Hint: PHP auto-converts int to float if declared [OK]

Common Mistakes:

• Expecting TypeError for int passed to float
• Calculating wrong multiplication
• Confusing output with integer result

4. Identify the error in this PHP function declaration:

function greet(string name) {
    return "Hello, $name!";
}

medium

A. Wrong type declaration keyword

B. Function must return int

C. Missing $ before parameter name

D. No error, code is correct

Solution

1. Step 1: Check parameter syntax

   In PHP, parameter names must start with a $ sign, so string name is incorrect.

2. Step 2: Identify the correct syntax

   The correct syntax is string $name with $ before the parameter name.

3. Final Answer:

   Missing $ before parameter name -> Option C

4. Quick Check:

   Parameter names need $ [OK]

Hint: Parameter names always start with $ in PHP [OK]

Common Mistakes:

• Omitting $ before parameter name
• Confusing type declaration with variable name
• Assuming no $ needed in function parameters

5. Given this function:

function process(array $data, int $limit = 10) {
    return array_slice($data, 0, $limit);
}

Which call will cause a TypeError?

hard

A. process([1, 2, 3], 2)

B. process('not an array', 5)

C. process([1, 2, 3])

D. process([1, 2, 3], 0)

Solution

1. Step 1: Analyze parameter types and defaults

   The first parameter must be an array; the second is an int with default 10.

2. Step 2: Check each call

   process([1, 2, 3], 2) passes array and int correctly. process('not an array', 5) passes a string instead of array, causing TypeError. Options A, B, and C pass valid types.

3. Final Answer:

   process('not an array', 5) causes TypeError -> Option B

4. Quick Check:

   First param must be array, string causes error [OK]

Hint: Check argument types match declared parameter types [OK]

Common Mistakes:

• Passing wrong type for array parameter
• Assuming default values fix type errors
• Ignoring parameter order and types