[php] How to pass an array into a function, and return the results with an array

So I'm trying to learn how to pass arrays through a function, so that I can get around PHP's inability to return multiple values. Haven't been able to get anything to work so far, but here is my best try. Can anybody point out where I'm going wrong?

function foo($array)
{
    $array[3]=$array[0]+$array[1]+$array[2];
    return $array;
}

$waffles[0]=1;
$waffles[1]=2;
$waffles[2]=3;
foo($waffles);

echo $waffles[3];

For clarification: I want to be able to pass multiple variables into a function, do something, then return multiple variables back out while keeping them seperate. This was just an example I was trying to get working as a work around for not being able to return multiple variables from an array

function foo(Array $array)
{
     return $array;
}

i know a Class is a bit the overkill

class Foo
{

 private $sum = NULL;

 public function __construct($array)
 {
   $this->sum[] = $array;
   return $this;
 }

 public function getSum()
 {
   $sum = $this->sum;
   for($i=0;$i<count($sum);$i++)
   {
      // get the last array index
      $res[$i] = $sum[$i] + $sum[count($sum)-$i];
   }
   return $res;
 }


}


$fo = new Foo($myarray)->getSum();

I always return multiple values by using a combination of list() and array()s:

function DecideStuffToReturn() {
    $IsValid = true;
    $AnswerToLife = 42;

    // Build the return array.
    return array($IsValid, $AnswerToLife);
}

// Part out the return array in to multiple variables.
list($IsValid, $AnswerToLife) = DecideStuffToReturn();

You can name them whatever you like. I chose to keep the function variables and the return variables the same for consistency but you can call them whatever you like.

See list() for more information.

Another way is:

$NAME = "John";
$EMAIL = "[email protected]";
$USERNAME = "John123";
$PASSWORD = "1234";
$array = Array ("$NAME","$EMAIL","$USERNAME","$PASSWORD");
function getAndReturn (Array $array){
    return $array;
}
print_r(getAndReturn($array));

Try

$waffles = foo($waffles);

Or pass the array by reference, like suggested in the other answers.

In addition, you can add new elements to an array without writing the index, e.g.

$waffles = array(1,2,3); // filling on initialization

or

$waffles = array();
$waffles[] = 1;
$waffles[] = 2;
$waffles[] = 3;

On a sidenote, if you want to sum all values in an array, use array_sum()

You're passing the array into the function by copy. Only objects are passed by reference in PHP, and an array is not an object. Here's what you do (note the &)

function foo(&$arr) { # note the &
  $arr[3] = $arr[0]+$arr[1]+$arr[2];
}
$waffles = array(1,2,3);
foo($waffles);
echo $waffles[3]; # prints 6

That aside, I'm not sure why you would do that particular operation like that. Why not just return the sum instead of assigning it to a new array element?

Here is how I do it. This way I can actually get a function to simulate returning multiple values;

function foo($array) 
{ 
    foreach($array as $_key => $_value) 
    {
       $str .= "{$_key}=".$_value.'&';
    }
    return $str = substr($str, 0, -1);
} 

/* Set the variables to pass to function, in an Array */

    $waffles['variable1'] = "value1"; 
    $waffles['variable2'] = "value2"; 
    $waffles['variable3'] = "value3"; 

/* Call Function */

    parse_str( foo( $waffles ));

/* Function returns multiple variable/value pairs */

    echo $variable1 ."<br>";
    echo $variable2 ."<br>";
    echo $variable3 ."<br>";

Especially usefull if you want, for example all fields in a database to be returned as variables, named the same as the database table fields. See 'db_fields( )' function below.

For example, if you have a query

select login, password, email from members_table where id = $id

Function returns multiple variables:

$login, $password and $email

Here is the function:

function db_fields($field, $filter, $filter_by,  $table = 'members_table') {

 /*
    This function will return as variable names, all fields that you request, 
    and the field values assigned to the variables as variable values.  

    $filter_by = TABLE FIELD TO FILTER RESULTS BY
    $filter =  VALUE TO FILTER BY
    $table = TABLE TO RUN QUERY AGAINST

    Returns single string value or ARRAY, based on whether user requests single
    field or multiple fields.

    We return all fields as variable names. If multiple rows
    are returned, check is_array($return_field); If > 0, it contains multiple rows.
    In that case, simply run parse_str($return_value) for each Array Item. 
*/
    $field = ($field == "*") ? "*,*" : $field;
    $fields = explode(",",$field);

    $assoc_array = ( count($fields) > 0 ) ? 1 : 0;

    if (!$assoc_array) {
        $result = mysql_fetch_assoc(mysql_query("select $field from $table where $filter_by = '$filter'"));
        return ${$field} = $result[$field];
    }
    else
    {
        $query = mysql_query("select $field from $table where $filter_by = '$filter'");
        while ($row = mysql_fetch_assoc($query)) {
            foreach($row as $_key => $_value) {
                $str .= "{$_key}=".$_value.'&';
            }
            return $str = substr($str, 0, -1);
        }
    }
}

Below is a sample call to function. So, If we need to get User Data for say $user_id = 12345, from the members table with fields ID, LOGIN, PASSWORD, EMAIL:

$filter = $user_id;
$filter_by = "ID";
$table_name = "members_table"

parse_str(db_fields('LOGIN, PASSWORD, EMAIL', $filter, $filter_by, $table_name));

/* This will return the following variables: */

echo $LOGIN ."<br>";
echo $PASSWORD ."<br>";
echo $EMAIL ."<br>";

We could also call like this:

parse_str(db_fields('*', $filter, $filter_by, $table_name));

The above call would return all fields as variable names.

You are not able to return 'multiple values' in PHP. You can return a single value, which might be an array.

function foo($test1, $test2, $test3)
{
    return array($test1, $test2, $test3);
}
$test1 = "1";
$test2 = "2";
$test3 = "3";

$arr = foo($test1, $test2, $test3);

$test1 = $arr[0];
$test2 = $arr[1];
$test3 = $arr[2];