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Hi, I understand passing by reference eg..

 

function f(&$param)

{

...

}

 

But I can't properly grasp returning by reference when the & is before the function.

 

Does it have anything to do with this:

$a = 1;

$b = &$a;

So now $a and $b basically reference the same value.

 

PS, I have not got to classes yet, so please don't use them in any explanations.

Thanks for any help.

 

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if you give a function an &$variable, then whatever is done to that variable within the function is relegated back to the variable you put as the argument.  For instance

function addOne(&$a) {
     $a += 1;
}
$b = 1;
echo $b;   // Prints 1
addOne($b);
echo $b;  // Prints 2

Thanks, I get that, but thats not what I'm asking...

 

If I have a function:

 

function &func($a)

{

....

}

 

Whats going on here? I think I have to return something, I'm not sure.

Is it somehow related to this..

 

##############################

$a = 1;

$b = &$a;

// so now $a and $b reference the same variable

##############################

 

Thanks again for any help

Thanks, I get that, but thats not what I'm asking...

 

If I have a function:

 

function &func($a)

{

....

}

 

Whats going on here? I think I have to return something, I'm not sure.

Is it somehow related to this..

 

##############################

$a = 1;

$b = &$a;

// so now $a and $b reference the same variable

##############################

 

Thanks again for any help

 

See: http://php.net/manual/en/language.references.return.php

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