Closures and variable binding with use in PHP - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of closure with variable binding by value

What is the output of this PHP code?

PHP

<?php
$number = 10;
$func = function() use ($number) {
    echo $number;
};
$number = 20;
$func();

A 10

B 0

C 20

D Error: Undefined variable

💡 Hint

Remember that variables passed with use are copied by value unless passed by reference.

The closure captures $number by value at the time of definition, so it prints 10 even though $number changes later.

❓ Predict Output

intermediate

Output of closure with variable binding by reference

What will this PHP code output?

PHP

<?php
$value = 5;
$func = function() use (&$value) {
    $value += 10;
    echo $value;
};
$func();
echo ' ' . $value;

A 5 15

B Error: Cannot use variable by reference

C 15 5

D 15 15

💡 Hint

Binding by reference means changes inside the closure affect the original variable.

The closure modifies $value by adding 10. Since it is bound by reference, both echo statements print 15.

🔧 Debug

advanced

Why does this closure not update the variable?

Consider this PHP code snippet. Why does the variable $count not increment outside the closure?

PHP

<?php
$count = 0;
$increment = function() use ($count) {
    $count++;
};
$increment();
echo $count;

A Because $count is not initialized before use.

B Because $count is passed by value, so increment inside closure doesn't affect outside variable.

C Because closures cannot modify variables from outer scope.

D Because $count is passed by reference but not incremented properly.

💡 Hint

Think about how variables are passed into closures with use().

The closure captures $count by value, so incrementing it inside does not change the original $count outside.

❓ Predict Output

advanced

Output of nested closures with variable binding

What is the output of this PHP code?

PHP

<?php
$a = 1;
$outer = function() use ($a) {
    $b = 2;
    $inner = function() use ($a, &$b) {
        $a++;
        $b++;
        echo $a . ' ' . $b;
    };
    $inner();
    echo ' ' . $b;
};
$outer();

A 1 2 2

B 2 3 3

C 1 3 3

D 2 2 2

💡 Hint

Remember that $a is passed by value and $b by reference in inner closure.

Inside inner closure, $a++ affects only local copy (1->2 but not reflected outside), $b++ affects original $b (2->3). So echo prints '2 3' and then outer echo prints ' 3'.

🧠 Conceptual

expert

Why does binding by reference in closure cause unexpected behavior?

In PHP, binding a variable by reference in a closure can sometimes cause unexpected results when the variable changes after closure creation. Why does this happen?

A Because the closure holds a reference to the original variable, so changes outside affect closure behavior dynamically.

B Because PHP copies the variable value at closure creation, so changes outside do not affect closure.

C Because closures cannot access variables by reference in PHP.

D Because binding by reference causes a syntax error if variable changes later.

💡 Hint

Think about how references work in PHP and how closures keep references.

Binding by reference means the closure uses the original variable's memory. If the variable changes outside, the closure sees the updated value, which can cause unexpected dynamic behavior.

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