Why type awareness matters in PHP - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable with an integer type.

PHP

<?php
$number = [1];
?>

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

A42

B"42"

Ctrue

D3.14

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around numbers makes them strings, not integers.

The variable $number should be assigned an integer value. The option 42 is an integer, while "42" is a string, true is a boolean, and 3.14 is a float.

2fill in blank

medium

Complete the code to declare a function that returns a string type.

PHP

<?php
function greet(): [1] {
    return "Hello!";
}
?>

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

Aint

Bfloat

Cbool

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong return type causes type errors.

The function greet returns a string, so the return type should be string. Other options represent different types.

3fill in blank

hard

Fix the error in the code by completing the type declaration for the parameter.

PHP

<?php
function doubleValue([1] $num): int {
    return $num * 2;
}
?>

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

Astring

Bbool

Cint

Dfloat

Attempts:

3 left

💡 Hint

Common Mistakes

Using string or bool causes type mismatch errors.

The function expects an integer parameter to multiply and return an integer. Declaring the parameter as int ensures type safety.

4fill in blank

hard

Fill both blanks to create an array with integer keys and string values.

PHP

<?php
$fruits = array([1] => [2]);
?>

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

A1

B"apple"

C"banana"

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Using boolean values as keys or values causes unexpected behavior.

The array key should be an integer 1, and the value should be a string like "apple". Other options are either wrong types or values.

5fill in blank

hard

Fill all three blanks to declare a typed class property and assign it a value.

PHP

<?php
class Person {
    public [1] $name = [2];
    public [3] $age = 30;
}
?>

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

Astring

B"John"

Cint

Dbool

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning a string value to an integer property or vice versa.

The property $name is a string and should be assigned a string value like "John". The property $age is an integer type.

Practice

1. Why is type awareness important in PHP programming?

easy

A. It makes the code run faster by skipping checks.

B. It helps prevent errors by knowing what kind of data is used.

C. It allows PHP to ignore variable names.

D. It removes the need for comments in code.

Solution

1. Step 1: Understand what type awareness means

   Type awareness means knowing the kind of data (like number, text) your program uses.

2. Step 2: Connect type awareness to error prevention

   Knowing data types helps avoid mistakes like adding text to numbers, which causes errors.

3. Final Answer:

   It helps prevent errors by knowing what kind of data is used. -> Option B

4. Quick Check:

   Type awareness = error prevention [OK]

Hint: Remember: types help catch mistakes early [OK]

Common Mistakes:

• Thinking type awareness speeds up code by skipping checks
• Believing variable names can be ignored
• Assuming comments replace type knowledge

2. Which of the following is the correct way to declare a function with type hints in PHP?

easy

A. function add(int $a, int $b): int { return $a + $b; }

B. function add($a, $b) { return $a + $b; }

C. function add(int $a, $b): string { return $a + $b; }

D. function add(int $a, int $b) { return "$a + $b"; }

Solution

1. Step 1: Identify correct type hint syntax

   PHP uses function name(type $param): returnType to declare types.

2. Step 2: Check each option for correct types

   function add(int $a, int $b): int { return $a + $b; } correctly types both parameters as int and return type as int.

3. Final Answer:

   function add(int $a, int $b): int { return $a + $b; } -> Option A

4. Quick Check:

   Correct type hints = function add(int $a, int $b): int { return $a + $b; } [OK]

Hint: Look for parameter and return type hints together [OK]

Common Mistakes:

• Missing return type declaration
• Mixing types in parameters and return
• Returning string when return type is int

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

function multiply(int $x, int $y): int {
    return $x * $y;
}
echo multiply(3, '4 apples');

medium

A. 12

B. TypeError

C. 3 4 apples

D. 7

Solution

1. Step 1: Understand PHP type coercion with int hints

   PHP converts string '4 apples' to integer 4 when type hinted as int.

2. Step 2: Calculate the multiplication

   3 * 4 = 12, so the function returns 12 and echoes it.

3. Final Answer:

   12 -> Option A

4. Quick Check:

   String '4 apples' coerced to int 4 = 12 output [OK]

Hint: PHP converts strings starting with numbers to int [OK]

Common Mistakes:

• Expecting a TypeError for string argument
• Thinking output is concatenated string
• Adding numbers instead of multiplying

4. Identify the error in this PHP function and how to fix it:

function divide(int $a, int $b): float {
    return $a / $b;
}
echo divide(10, 0);

medium

A. No error; code runs fine.

B. Return type should be int, not float.

C. Parameters should be float, not int.

D. Division by zero error; check if $b is zero before dividing.

Solution

1. Step 1: Analyze the division operation

   Dividing by zero causes a runtime error in PHP.

2. Step 2: Suggest fix to avoid division by zero

   Check if $b is zero before dividing to prevent error.

3. Final Answer:

   Division by zero error; check if $b is zero before dividing. -> Option D

4. Quick Check:

   Division by zero check needed = Division by zero error; check if $b is zero before dividing. [OK]

Hint: Always check divisor is not zero before dividing [OK]

Common Mistakes:

• Ignoring division by zero runtime error
• Changing return type unnecessarily
• Changing parameter types without reason

5. You want to create a PHP function that accepts an array of mixed values and returns an array with only integers doubled. Which approach best uses type awareness?

hard

A. Use no type hints and double all values assuming they are integers.

B. Declare function parameters as int to accept only integers and double them.

C. Use a function with array type hint and inside loop check if each item is int before doubling.

D. Use string type hint and convert all values to strings before doubling.

Solution

1. Step 1: Understand the input and output requirements

   Input is an array with mixed types; output is array with only integers doubled.

2. Step 2: Choose approach that uses type awareness correctly

   Using array type hint and checking each item's type inside loop ensures only integers are processed.

3. Final Answer:

   Use a function with array type hint and inside loop check if each item is int before doubling. -> Option C

4. Quick Check:

   Array type hint + type check inside loop = Use a function with array type hint and inside loop check if each item is int before doubling. [OK]

Hint: Type hint arrays, then check item types inside loop [OK]

Common Mistakes:

• Using int type hint for whole array parameter
• Skipping type checks and processing all items
• Converting all values to strings unnecessarily