Can't pass number as OOP Parameter

[unemployment]

Why doesn't the second option work?

 

This works...

$number = 2;
$data = new user_info();
$data->info_data($number);

 

 

This doesn't...

$data = new user_info();
$data->info_data(2);

 

[xyph]

Yes, it does.

[AyKay47]

Why doesn't the second option work?

 

This works...

$number = 2;
$data = new user_info();
$data->info_data($number);

 

 

This doesn't...

$data = new user_info();
$data->info_data(2);

I don't understand how that wouldn't work...setting a variable to the int 2 is the same as passing the actual int 2...maybe something in your class itself?

[KevinM1]

More to the point, what does "doesn't work" actually mean?

[unemployment]

My misunderstanding I was trying to pass it by reference.  I was just trying to learn what by reference actually does.  Can anyone enlighten me.

 

class user_info
{
public function info_data(&$number){

	$number *= 3;

	echo $number;
}
}

$data = new user_info();
$data->info_data(2);

[AyKay47]

http://php.net/manual/en/language.references.pass.php

 

[xyph]

<?php

$value = 5;

double( $value );

echo $value; // outputs 10

function double( &$num ) {
$num = $num * 2;
}

?>