PHP 7 and strict “resource” types

Question

Does PHP 7 support strict typing for resources? If so, how?

For example:

    declare (strict_types=1);

    $ch = curl_init ();
    test ($ch);

             


        

Answer 1

resource is not a valid type so it's assumed to be a class name as per good old PHP/5 type hints. But curl_init() does not return an object instance.

As far as I know there's not way to specify a resource. It probably wouldn't be so useful since not all resources are identical: a resource generated by fopen() would be useless for oci_parse().

P.S. If you want to check the resource in the function body, you can use get_resource_type() (with is_resource() to prevent errors), as in:

is_resource($ch) && get_resource_type($ch) === 'curl'