Type declarations for parameters in PHP - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a type declaration for a parameter in PHP?

A type declaration for a parameter in PHP specifies the expected data type of the argument passed to a function or method. It helps catch errors early by ensuring the correct type is used.

beginner

How do you declare a parameter to accept only integers in PHP?

You add the type before the parameter name, like this: <code>function example(int $number) { }</code>. This means the function expects an integer argument.

intermediate

What happens if you pass a wrong type to a function with a declared parameter type in PHP?

PHP will throw a TypeError at runtime, stopping the program unless caught. This helps prevent bugs caused by wrong data types.

intermediate

Can you declare a parameter to accept multiple types in PHP?

Yes, since PHP 8.0, you can use union types like <code>function example(int|string $value) { }</code> to accept either an integer or a string.

beginner

What is the difference between type declarations and type hints in PHP?

They mean the same thing: specifying the expected type of a parameter. The term 'type hint' was used before PHP 7, now 'type declaration' is more common.

How do you declare a function parameter to accept only strings in PHP?

A function example(string $text) {}

B function example($text: string) {}

C function example(text string) {}

D function example($text) : string {}

The correct syntax is to put the type before the parameter name: string $text.

What error does PHP throw if a wrong type is passed to a typed parameter?

A TypeError

B SyntaxError

C ValueError

D RuntimeException

PHP throws a TypeError when a parameter type declaration is violated.

Which PHP version introduced union types for parameters?

A PHP 7.4

B PHP 5.6

C PHP 7.0

D PHP 8.0

Union types were introduced in PHP 8.0, allowing multiple types for a parameter.

What does this function declaration mean? function test(array $data) {}

A The function accepts any type of argument.

B The function accepts only arrays as argument.

C The function returns an array.

D The function accepts only strings.

The parameter $data must be an array due to the type declaration.

Can you omit the type declaration for a parameter in PHP?

A Only in PHP 8 and above.

B No, all parameters must have a type declaration.

C Yes, parameters can be untyped and accept any value.

D Only for class methods.

Type declarations are optional; if omitted, the parameter accepts any type.

Explain how type declarations for parameters improve PHP code quality.

Think about how telling PHP what type to expect helps avoid mistakes.

Describe how to declare a parameter that accepts either an integer or a string in PHP.

Use the vertical bar | to separate types.

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