PHP variables and scope

In php variables have scope (don't all languages)...

A variable declared outside of a function will not be available inside of a function, unless be declare it inside the function with the global command.



Anyway, the below serves as a personal reminder.

If we take $loggedIn for example, this would not work when the function is called if we didn't have the global $loggedIn statement, unlike the Python and Java examples for reference below.


<?php
$loggedIn=True;

function setLoginIcon()
 {
  global $loggedIn;

  if ($loggedIn)
    {
    echo 'True';
}
else
{
echo 'False';
}
 }
?>

Also note that declaring a variable static inside a function will make it available outside the function.

As an example of how, say Python is different, see below;

a = True
def myFunction():    
   if (a):        
      print("True")    
   else:        
      print("False")
      
myFunction()

Similarly, see below for a Java example;

public class JavaApplication20
{
    boolean a = true;
    public void myFunction()
    {
        if (a)
        {
            System.out.println("True");  
        }
        else
        {
            System.out.println("False"); 
        }
        
    }


    public static void main(String[] args)
    {
        JavaApplication20 a = new JavaApplication20();
        a.myFunction();
    }    
}

On another note, scope for local variables should generally be considered to be within the bounds of which is was created/declared, see the following example in C;

#include <stdio.h>

int main()
{
    int a=5; // local var a set to 5
    printf("I'm a local: %d addr: %d\n", a, &a);

    if (1)
    {
        // We'll always run...
        int a=10; // local var a set to 10
        printf("I'm also a local: %d addr: %d\n", a, &a);
    }
   
    // which local var will we print here...
    printf("finally, %d addr: %d", a, &a);
    return 0;
}

In the above we see that the address of the variable inside the if statement is different from the other.