Gettype and typeof checks in PHP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the type of the variable $value using gettype.

PHP

<?php
$value = 123;
$type = [1]($value);
echo $type;
?>

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

Atype_of

Btypeof

Cgettype

Dtype

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'typeof' which is not a PHP function.

Trying to use 'type' as a function.

The gettype() function returns the type of a variable as a string in PHP.

2fill in blank

medium

Complete the code to check if the variable $value is of type integer using gettype.

PHP

<?php
$value = 10;
if (gettype($value) === [1]) {
    echo 'Integer detected';
}
?>

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

A'integer'

B'int'

C'number'

D'digit'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of 'integer'.

Using 'number' which is not returned by gettype.

The gettype() function returns 'integer' for integer values in PHP.

3fill in blank

hard

Fix the error in the code to correctly check the type of $value using gettype.

PHP

<?php
$value = 3.14;
if ([1]($value) == 'double') {
    echo 'Float detected';
}
?>

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

Atypeof

Btype

Ctype_of

Dgettype

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'typeof' which is not a PHP function.

Using 'type' or 'type_of' which do not exist.

Only gettype() is a valid PHP function to get the type of a variable.

4fill in blank

hard

Fill both blanks to create a dictionary with words as keys and their lengths as values, only for words longer than 3 characters.

PHP

<?php
$words = ['apple', 'cat', 'banana', 'dog'];
$lengths = array();
foreach ($words as $word) {
    if (strlen($word) [1] 3) {
        $lengths[$word] = [2]($word);
    }
}
print_r($lengths);
?>

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

A>

Bstrlen

C<

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' in the condition.

Using 'count' instead of 'strlen' for string length.

We check if the length of the word is greater than 3 using strlen($word) > 3 and then assign the length using strlen($word).

5fill in blank

hard

Fill all three blanks to create an associative array with uppercase keys and values only if the value is a string.

PHP

<?php
$data = ['name' => 'John', 'age' => 30, 'city' => 'Paris'];
$result = array();
foreach ($data as [1] => [2]) {
    if (gettype([2]) === 'string') {
        $result[strtoupper([1])] = [2];
    }
}
print_r($result);
?>

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

Akey

Bvalue

C$value

D$key

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names without '$'.

Mixing up key and value variable names.

In the foreach loop, $key and $value are used to access keys and values. The condition checks if $value is a string, then adds it to $result with uppercase keys.

Practice

1. What does the PHP function gettype() return when called on a variable?

easy

A. A string describing the variable's type

B. An integer representing the type code

C. A boolean indicating if the variable is set

D. An object describing the variable

Solution

1. Step 1: Understand the purpose of gettype()

   The gettype() function returns the type of a variable as a string, such as "integer", "string", or "array".

2. Step 2: Compare options with gettype() behavior

   The option stating that it returns a string describing the variable's type is correct.

3. Final Answer:

   A string describing the variable's type -> Option A

4. Quick Check:

   gettype() = string description [OK]

Hint: Remember: gettype() always returns a string type name [OK]

Common Mistakes:

• Thinking gettype() returns a numeric code
• Confusing gettype() with isset()
• Assuming gettype() returns an object

2. Which of the following is the correct syntax to get the type of a variable $var using gettype() in PHP?

easy

A. gettype $var;

B. typeof $var;

C. typeof($var);

D. gettype($var);

Solution

1. Step 1: Recall correct function syntax in PHP

   PHP functions require parentheses around arguments, so gettype($var); is correct syntax.

2. Step 2: Check other options for syntax errors

   Syntax using typeof is invalid because it is not a PHP function. gettype $var; misses parentheses, which is invalid syntax.

3. Final Answer:

   gettype($var); -> Option D

4. Quick Check:

   Function call syntax = gettype($var) [OK]

Hint: Use parentheses with function calls in PHP [OK]

Common Mistakes:

• Using typeof instead of gettype()
• Omitting parentheses in function calls
• Confusing PHP syntax with JavaScript

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

$x = 3.14;
echo gettype($x);

medium

A. integer

B. double

C. float

D. string

Solution

1. Step 1: Identify the variable type

   The variable $x is assigned 3.14, which is a floating-point number.

2. Step 2: Understand gettype() output for floats

   In PHP, gettype() returns "double" for floating-point numbers, not "float".

3. Final Answer:

   double -> Option B

4. Quick Check:

   gettype(3.14) = double [OK]

Hint: Remember: PHP calls floats "double" in gettype() output [OK]

Common Mistakes:

• Expecting output to be "float" instead of "double"
• Confusing integer and float types
• Assuming gettype() returns "string" for numbers

4. Consider this PHP code snippet:

$value = true;
if (typeof($value) === 'boolean') {
    echo 'Boolean type';
} else {
    echo 'Not boolean';
}

What is the problem with this code?

medium

A. typeof is not a valid PHP function

B. Comparison operator === cannot be used with strings

C. Boolean values cannot be checked with typeof

D. The variable $value is not defined

Solution

1. Step 1: Check if typeof is a PHP function

   PHP does not have a built-in typeof() function; it uses gettype() instead.

2. Step 2: Analyze the impact of using typeof()

   Using typeof() causes a fatal error because the function does not exist.

3. Final Answer:

   typeof is not a valid PHP function -> Option A

4. Quick Check:

   typeof() invalid in PHP [OK]

Hint: Use gettype(), not typeof(), in PHP [OK]

Common Mistakes:

• Assuming typeof() exists in PHP like in JavaScript
• Thinking === cannot compare strings
• Ignoring variable initialization

5. In PHP 8.0+, the typeof operator returns a Type object instead of a string. Given this code:

$var = [1, 2, 3];
$type = typeof $var;
echo $type->name;

What will be the output?

hard

A. Type

B. object

C. error

D. array

Solution

1. Step 1: Verify typeof in PHP 8.0+

   PHP has no typeof operator or function in 8.0+ or any version; use gettype() instead.

2. Step 2: Predict code execution

   The undefined typeof causes a fatal error (e.g., "Call to undefined function"). No echo output.

3. Final Answer:

   error -> Option C

4. Quick Check:

   typeof invalid in PHP [OK]

Hint: PHP has no typeof; use gettype() [OK]

Common Mistakes:

• Believing typeof exists in PHP 8.0+
• Confusing PHP with JavaScript's typeof
• Expecting a Type object return