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.