Type declarations for parameters in PHP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the parameter type as integer.

PHP

<?php
function addOne([1] $num) {
    return $num + 1;
}
?>

Drag options to blanks, or click blank then click option'

Abool

Bint

Cfloat

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'string' instead of 'int' for a number parameter.

Forgetting to declare the type at all.

The parameter type int declares that the function expects an integer.

2fill in blank

medium

Complete the code to declare the parameter type as string.

PHP

<?php
function greet([1] $name) {
    return "Hello, " . $name;
}
?>

Drag options to blanks, or click blank then click option'

Astring

Bbool

Carray

Dfloat

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'array' instead of 'string' for a text parameter.

Not declaring any type, which allows any type.

The parameter type string declares that the function expects a text value.

3fill in blank

hard

Fix the error in the function parameter type declaration.

PHP

<?php
function multiply([1] $value1, int $value2) {
    return $value1 * $value2;
}
?>

Drag options to blanks, or click blank then click option'

Anumber

Bfloat

Cinteger

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'integer' instead of 'int' causes a syntax error.

Using 'number' which is not a valid PHP type.

The correct type declaration for integers in PHP is int, not 'integer' or 'number'.

4fill in blank

hard

Fill both blanks to declare parameter types for a function that takes a float and a boolean.

PHP

<?php
function checkValues([1] $score, [2] $passed) {
    if ($passed) {
        return $score;
    }
    return 0;
}
?>

Drag options to blanks, or click blank then click option'

Afloat

Bbool

Cint

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of 'float' for decimal numbers.

Using 'string' instead of 'bool' for true/false.

The first parameter is a floating-point number, so use float. The second is a true/false value, so use bool.

5fill in blank

hard

Fill all three blanks to declare parameter types for a function that takes a string, an integer, and a boolean.

PHP

<?php
function processData([1] $name, [2] $age, [3] $isMember) {
    if ($isMember) {
        return "Welcome " . $name . ", age " . $age;
    }
    return "Guest";
}
?>

Drag options to blanks, or click blank then click option'

Astring

Bint

Cbool

Dfloat

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up types like using 'float' for age.

Not using 'bool' for true/false parameters.

The parameters are a text name (string), an age number (int), and a true/false membership status (bool).

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