I'm currently investigating some PHP-code and I'm trying to make it more up to date. I found a function, which retrieves a filehandle (via fopen).
The returntype is stream. I checked it with
$handle = fopen("myfile.csv", "r");
var_dump(get_resource_type($handle));
As mentioned, the type seems to be stream.
When I try to pass this object to a function, I can't force the type with stream.
public function parseHandle(stream $handle){}
ends up with this error-message:
Argument 1 passed to parseHandle must be an instance of stream, resource given.
When I try to change the signature to function parseHandle(resource $resource) it yields the following error:
Argument 1 passed to parseHandle must be an instance of resource, resource given
What is the correct datatype and how should I use it in the arguments?
I know I can use is_resource in the very first lines of my function to verify it, but I still wonder.
Unfortunately, you cannot use the primitive type resource as a type hint. The latest type hint system change to PHP, scalar type hints, only added int, float, string, and bool.
The error you quote has come up before as confusing, especially in the context of the scalar type hints. Let me template it for you:
Argument [N] passed to [Function] must be an instance of [Class], [Type] given
The confusion arises because PHP allows classes to have the same name as documented primitives (bool, float, etc.) for all primitives before PHP 7, and for some in PHP 7 and later †. So when you say stream $handle PHP is expecting $handle to be of class stream. Likewise resource $handle expects $handle to be of class resource.
If you want to type hint resources, I suggest using an \SplFileObject:
$handle = new \SplFileObject('myfile.csv', 'r');
function parseHandle(\SplFileObject $handle) { ... }
This is not the best thing in the world, as \SplFileObject has a few quirks, but at the end of the day, if you want to type hint it in PHP, you must either have an array, a scalar, or a class.