Folks,
I am learning about Global Variables and Super Globals. Been experimenting. The following works. I am able to 'return $GLOBALS['z4']'.
Example 1
<?php
//Php holds all 'Global Variables' in an array. Format: $GLOBALS[index]; The 'index' holds the name of the variable.
$x4 = 4; //'Global Variable' as outside function. It now exists in "Super Global" Variable array as: $GLOBALS['x4'].
$y4 = 4; //'Global Variable' as outside function. It now exists in "Super Global" Variable array as: $GLOBALS['y4'].
function addition_4()
{
$GLOBALS['z4'] = $GLOBALS['x4'] + $GLOBALS['y4']; //A new "Global Variable" '$z4' is created inside the function.
return $GLOBALS['z4'];
}
echo addition_4();
?>
But on this following one, why am I unable to 'return $z4' ?
Example 2
<?php
//Php holds all 'Global Variables' in an array. Format: $GLOBALS[index]; The 'index' holds the name of the variable.
$x4 = 4; //'Global Variable' as outside function. It now exists in "Super Global" Variable array as: $GLOBALS['x4'].
$y4 = 4; //'Global Variable' as outside function. It now exists in "Super Global" Variable array as: $GLOBALS['y4'].
function addition_4()
{
$GLOBALS['z4'] = $GLOBALS['x4'] + $GLOBALS['y4']; //A new "Global Variable" '$z4' is created inside the function.
return $z4; //WHY I GET ERROR: "Notice: Undefined variable: z4 in ..." ?
}
echo addition_4();
?>
Folks, since this following exists "$GLOBALS['z4']" in both examples above, then I was able to "return $GLOBALS['z4'] on the 1st example.
Now, since "$GLOBALS['z4']" exists in both examples, then that also means "$z4" also exists in both examples. Correct ? So, why I am unable to "return $z4" on the 2nd example ? Why I get error that, $z4 is undefined ?
