Global keyword behavior in PHP

Introduction

The global keyword lets you use variables from outside a function inside that function. It helps share data between different parts of your code.

When you want to change a variable inside a function that was created outside it.

When you need to read or update a value that is shared across multiple functions.

When you want to avoid passing many variables as function parameters.

When you have a setting or counter that many functions should access and modify.

Syntax

PHP

function example() {
    global $variableName;
    // Now $variableName refers to the global variable
}

You must declare each global variable inside the function with the global keyword.

It is used inside functions and class methods to access global variables (not class properties).

Examples

This example shows how the function changes the global variable $number by adding 5.

PHP

<?php
$number = 10;
function addFive() {
    global $number;
    $number += 5;
}
addFive();
echo $number;

The function uses the global $message variable to print a greeting.

PHP

<?php
$message = "Hello";
function greet() {
    global $message;
    echo $message . " World!";
}
greet();

Sample Program

This program uses the global keyword to increase the $count variable inside the function. It calls the function twice, so the count becomes 2.

PHP

<?php
$count = 0;
function increment() {
    global $count;
    $count++;
}
increment();
increment();
echo "Count is: " . $count;

Important Notes

Using many global variables can make your code harder to understand and debug.

Consider passing variables as function arguments when possible for clearer code.

Global variables keep their values between function calls unless changed.

Summary

The global keyword lets functions access variables defined outside them.

You must declare each global variable inside the function to use it.

Use global variables carefully to keep your code clean and easy to follow.

Practice

1. What does the global keyword do in PHP?

easy

A. It deletes a variable from the global scope.

B. It creates a new local variable inside the function.

C. It makes a variable constant and unchangeable.

D. It allows a function to access variables defined outside its scope.

Solution

1. Step 1: Understand variable scopes in PHP

   Variables defined outside functions are in the global scope and not accessible inside functions by default.

2. Step 2: Role of the global keyword

   The global keyword inside a function tells PHP to use the variable from the global scope instead of creating a new local one.

3. Final Answer:

   It allows a function to access variables defined outside its scope. -> Option D

4. Quick Check:

   global keyword = access outside variables [OK]

Hint: Global lets functions use outside variables directly [OK]

Common Mistakes:

• Thinking global creates a new variable inside the function
• Assuming global deletes variables
• Confusing global with constants

2. Which of the following is the correct way to use the global keyword inside a function to access a variable $count defined outside?

easy

A. global $count;

B. use $count;

C. global($count);

D. extern $count;

Solution

1. Step 1: Recall PHP syntax for global variables

   In PHP, to access a global variable inside a function, you write global $variableName; without parentheses.

2. Step 2: Check each option

   global $count; uses correct syntax. use $count; is invalid in PHP. global($count); uses parentheses which is incorrect. extern $count; uses 'extern' which is not a PHP keyword.

3. Final Answer:

   global $count; -> Option A

4. Quick Check:

   Correct syntax = global $var; [OK]

Hint: Use 'global $var;' without parentheses inside functions [OK]

Common Mistakes:

• Adding parentheses after global
• Using 'use' or 'extern' keywords incorrectly
• Forgetting the $ sign before variable name

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

$x = 5;
function test() {
  global $x;
  $x = $x + 10;
}
test();
echo $x;

medium

A. 15

B. 5

C. 10

D. Error

Solution

1. Step 1: Understand the initial value of $x

   $x is set to 5 in the global scope before the function call.

2. Step 2: Effect of the function test()

   Inside test(), global $x; makes $x inside the function refer to the global $x. Then $x is increased by 10, so $x becomes 15 globally.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   global $x modified to 15 [OK]

Hint: global lets function change outside variable value [OK]

Common Mistakes:

• Thinking $x inside function is local and unchanged
• Expecting output 5 or 10 instead of 15
• Assuming error due to global keyword

4. Identify the error in this PHP code snippet:

$count = 10;
function increment() {
  $count = $count + 1;
}
increment();
echo $count;

medium

A. Variable $count is undefined.

B. Syntax error in function declaration.

C. Missing global keyword inside function to access $count.

D. Cannot echo variable outside function.

Solution

1. Step 1: Check variable scope inside function

   Inside increment(), $count is treated as a new local variable, not the global one.

2. Step 2: Missing global keyword

   To modify the global $count, the function must declare global $count; before using it.

3. Final Answer:

   Missing global keyword inside function to access $count. -> Option C

4. Quick Check:

   global needed to access outside variable [OK]

Hint: Declare global inside function to modify outside variable [OK]

Common Mistakes:

• Assuming $count inside function is global by default
• Thinking echo outside function causes error
• Confusing syntax errors with scope errors

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

$a = 3;
$b = 4;
function sum() {
  global $a, $b;
  $b = $a + $b;
}
sum();
echo $b;

What if global $a, $b; is replaced with global $a; only?

hard

A. Output is 7 in both cases.

B. Output is 7 with global $a, $b; and 4 with global $a; only.

C. Output is 4 with global $a, $b; and 7 with global $a; only.

D. Code causes error without global $b;.

Solution

1. Step 1: When using global $a, $b;

   Both $a and $b inside the function refer to global variables. $b is updated to 3 + 4 = 7 globally.

2. Step 2: When using global $a; only

   $a inside function is global, but $b is local and uninitialized. So $b = $a + $b uses local $b (null or 0), $b becomes 3 + 0 = 3 locally, global $b remains 4. echo $b outputs global $b which is 4.

3. Final Answer:

   Output is 7 with global $a, $b; and 4 with global $a; only. -> Option B

4. Quick Check:

   global vars must be declared to modify them [OK]

Hint: Declare all globals you want to use inside function [OK]

Common Mistakes:

• Assuming global $a; also makes $b global
• Expecting output 7 without declaring global $b
• Thinking missing global causes syntax error