Global keyword behavior in PHP - Step-by-Step Execution

Concept Flow - Global keyword behavior

Start

↓

Define global variable

↓

Enter function

↓

Use 'global' keyword

↓

Access/modify global variable

↓

Exit function

↓

Global variable updated

↓

End

This flow shows how a global variable is defined outside a function, then accessed and modified inside a function using the 'global' keyword.

Execution Sample

PHP

<?php
$x = 5;
function test() {
  global $x;
  $x = 10;
}
test();
echo $x;
?>

This code defines a global variable $x, modifies it inside a function using 'global', then prints the updated value.

Execution Table

Step	Action	Variable	Value	Notes
1	Define global variable	$x	5	Global scope
2	Call function test()	-	-	Enter function scope
3	Declare 'global $x' inside function	$x	5	Link local $x to global $x
4	Assign $x = 10 inside function	$x	10	Modifies global $x
5	Exit function test()	-	-	Return to global scope
6	Echo $x	$x	10	Prints updated global variable

💡 Function ends and global variable $x is updated to 10

Variable Tracker

Variable	Start	After function call	Final
$x	5	10	10

Key Moments - 2 Insights

Why does $x inside the function change the global $x?

Because of the 'global $x;' declaration inside the function (see step 3 in execution_table), the local $x refers to the global $x, so modifying it changes the global variable.

What happens if we don't use 'global' inside the function?

Without 'global', $x inside the function is a new local variable, so changes won't affect the global $x (not shown in table but implied).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of $x after step 4 inside the function?

A 10

B 5

C Undefined

D 0

💡 Hint

Check the 'Value' column at step 4 where $x is assigned 10 inside the function.

At which step does the global variable $x get updated?

A Step 2

B Step 3

C Step 4

D Step 6

💡 Hint

Look at the 'Action' and 'Notes' columns around step 4 where $x is assigned 10.

If we remove 'global $x;' from the function, what would be the output at step 6?

A 10

B 5

C Error

D 0

💡 Hint

Without 'global', the function modifies a local $x, so global $x remains unchanged as shown in variable_tracker.

Concept Snapshot

PHP 'global' keyword:
- Declares variables inside functions as global.
- Allows access and modification of global variables.
- Without 'global', variables inside functions are local.
- Changes to global variables persist after function ends.

Full Transcript

This example shows how PHP's 'global' keyword works. We start by defining a global variable $x with value 5. When we enter the function test(), we declare 'global $x;' which links the local $x to the global one. Then we assign 10 to $x inside the function, which updates the global variable. After the function ends, printing $x shows 10, confirming the global variable was changed. Without 'global', the function would only change a local copy, leaving the global $x unchanged.

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