How do i pass array as an argument for a function ?

[jd2007]

<?php
class Arrays
{
private $arr=array();
private $mystr;

function produceStr(array $arr)
{
   $this->arr=$arr;
   $this->mystr=implode("and", $this->arr);
   echo $this->mystr;
}
}
?>

 

when i instantiate an object, i do this:

<?php
$a=new Arrays();
$b=array("a", "b", "c");
$a->produceStr($b);
?>

 

this doesn't work...why ?

[trq]

There is no object of type array in php so I don't believe you can type hint using it. eg; This...

 

function produceStr(array $arr)

 

should simply be...

 

function produceStr($arr)

 

Other than that, I don't really see any problem.

[jd2007]

but i want to be array $arr but php treats it like string

[wildteen88]

Code works fine for me, using PHP5.2.x here. I noticed you are using the private keyword here for initiating variables. Make sure you are not using PHP4. PHP4 does not support the visibility keywords (private, protected and public), use var in replacement:

 var $arr=array();
var $mystr;

 

What results do you get?

 

EDIT: Didn't see the replies above

but i want to be array $arr but php treats it like string

Even if the you pass the produceStr function an array for the $arr argument $arr will still be an array, without 'array' before $arr.

[Daniel0]

wildteen: You can use array for type hinting as well.

 

Type Hints can only be of the object and array (since PHP 5.1) type. Traditional type hinting with int and string isn't supported.

[trq]

wildteen: You can use array for type hinting as well.

 

Type Hints can only be of the object and array (since PHP 5.1) type. Traditional type hinting with int and string isn't supported.

 

That was me. And thanks, something I was not aware of.