Closures and variable binding with use in PHP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a closure that uses the variable $x.

PHP

<?php
$x = 5;
$func = function() use ([1]) {
    return $x * 2;
};
echo $func();
?>

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

A$x

B$y

C$z

D$a

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name not defined outside the closure.

Forgetting to use the use keyword.

The use keyword imports the variable $x into the closure's scope.

2fill in blank

medium

Complete the code to modify the external variable $count inside the closure.

PHP

<?php
$count = 0;
$increment = function() use (&[1]) {
    $count++;
};
$increment();
echo $count;
?>

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

A$value

B$count

C$num

D$total

Attempts:

3 left

💡 Hint

Common Mistakes

Not using & to pass by reference.

Using a different variable name inside use.

Using &$count imports the variable by reference, allowing the closure to modify it.

3fill in blank

hard

Fix the error in the closure by correctly importing the variable $name.

PHP

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

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

A$this->name

B$_name

C$name

D$greet

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to access external variables directly inside the closure.

Using incorrect variable names.

The closure must import $name using use ($name) to access it inside.

4fill in blank

hard

Fill both blanks to create a closure that captures $a and $b by value and returns their sum.

PHP

<?php
$a = 3;
$b = 4;
$sum = function() use ([1], [2]) {
    return [1] + [2];
};
echo $sum();
?>

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

A$a

B$b

C$c

D$d

Attempts:

3 left

💡 Hint

Common Mistakes

Using variables not defined outside the closure.

Not importing variables with use.

The closure imports $a and $b by value and returns their sum.

5fill in blank

hard

Fill all three blanks to create a closure that captures $x by reference, increments it, and returns the new value.

PHP

<?php
$x = 10;
$increment = function() use ([1]) {
    [2]++;
    return [3];
};
echo $increment();
?>

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

A&$x

B$x

D$y

Attempts:

3 left

💡 Hint

Common Mistakes

Not using reference in use.

Using different variable names inside the closure.

The closure imports $x by reference using &$x, increments it, and returns the updated value.

Practice

1. What does the use keyword do in a PHP closure?

easy

A. It defines a new variable inside the closure.

B. It creates a global variable accessible everywhere.

C. It imports variables from the parent scope into the closure.

D. It calls the closure immediately.

Solution

1. Step 1: Understand closure scope

   A closure is a function inside another function that can access variables from the outside only if explicitly told.

2. Step 2: Role of use

   The use keyword brings variables from the parent scope into the closure's scope so the closure can use them.

3. Final Answer:

   It imports variables from the parent scope into the closure. -> Option C

4. Quick Check:

   use = import variables [OK]

Hint: Remember: use imports outside variables into closure [OK]

Common Mistakes:

• Thinking use creates new variables inside closure
• Confusing use with global keyword
• Assuming use calls the closure

2. Which of the following is the correct syntax to pass a variable $x by reference into a closure using use?

easy

A. function() use ($x) { }

B. function() use (&x) { }

C. function() use ($&x) { }

D. function() use (&$x) { }

Solution

1. Step 1: Understand passing by reference

   To allow a closure to modify an outside variable, it must be passed by reference using &.

2. Step 2: Correct syntax with use

   The correct syntax is use (&$x) where & is before the variable name with the dollar sign.

3. Final Answer:

   function() use (&$x) { } -> Option D

4. Quick Check:

   Reference = & before $variable [OK]

Hint: Use & before $var inside use to pass by reference [OK]

Common Mistakes:

• Placing & after $ or before variable name without $
• Omitting & when intending to modify variable
• Using invalid syntax like $&x

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

$count = 0;
$func = function() use (&$count) { $count++; echo $count; };
$func();
$func();
echo $count;

medium

A. 111

B. 122

C. 120

D. Error

Solution

1. Step 1: Analyze closure calls and reference

   The closure uses &$count by reference, so changes inside affect the outside variable.

2. Step 2: Trace each call and output

   First call: increments count to 1 and echoes '1'. Second call: increments count to 2 and echoes '2'. Finally, outside echo prints 2.

3. Final Answer:

   122 -> Option B

4. Quick Check:

   Reference increments count = 122 [OK]

Hint: Reference in use means closure changes outside variable [OK]

Common Mistakes:

• Assuming closure copies variable, so count stays 0
• Expecting output '111' by mistake
• Thinking code causes error

4. Identify the error in this PHP code snippet:

$x = 5;
$func = function() use ($x) { $x++; echo $x; };
$func();
echo $x;

medium

A. Output will be '65' because $x inside closure is a copy.

B. Runtime error because $x is undefined inside closure.

C. Syntax error because $x is not passed by reference.

D. Output will be '55' because closure does not run.

Solution

1. Step 1: Understand variable binding without reference

   Using use ($x) passes $x by value, so closure gets a copy.

2. Step 2: Trace output of closure and outside variable

   Closure increments its copy to 6 and echoes it. Outside $x remains 5, so echo prints 5.

3. Final Answer:

   Output will be '65' because $x inside closure is a copy. -> Option A

4. Quick Check:

   Value copy in use means outside unchanged [OK]

Hint: Without &, closure changes copy, not original variable [OK]

Common Mistakes:

• Expecting outside $x to change without &
• Thinking code causes syntax or runtime error
• Confusing output order

5. Given this code, what will be the output?

$values = [1, 2, 3];
$sum = 0;
$funcs = [];
foreach ($values as $v) {
    $funcs[] = function() use (&$sum, $v) { $sum += $v; };
}
foreach ($funcs as $f) { $f(); }
echo $sum;

Why does this output occur?

hard

A. 6, because each closure adds its own $v to $sum by value and $sum is by reference.

B. 3, because $v is shared and only last value is added multiple times.

C. 0, because closures do not run correctly.

D. Error, because $v cannot be used without reference.

Solution

1. Step 1: Understand variable binding in loop closures

   Each closure captures $v by value (copy) and $sum by reference.

2. Step 2: Trace closure execution

   Each closure adds its own $v (1, 2, 3) to the shared $sum. After all run, $sum = 6.

3. Final Answer:

   6, because each closure adds its own $v to $sum by value and $sum is by reference. -> Option A

4. Quick Check:

   Value copy for $v + reference $sum = 6 [OK]

Hint: Use value for loop var, reference for accumulator [OK]

Common Mistakes:

• Assuming $v is shared reference causing wrong sum
• Expecting error due to $v usage
• Thinking closures do not run