Why Global Variable Shows Undefined When Using The return Function ?

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 ?

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