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'

Answer 2

PHP does not have a type hint for resources because

No type hint for resources is added, as this would prevent moving from resources to objects for existing extensions, which some have already done (e.g. GMP).

However, you can use is_resource() within the function/method body to verify the passed argument and handle it as needed. A reusable version would be an assertion like this:

function assert_resource($resource)
{
    if (false === is_resource($resource)) {
        throw new InvalidArgumentException(
            sprintf(
                'Argument must be a valid resource type. %s given.',
                gettype($resource)
            )
        );
    }
}

which you could then use within your code like that:

function test($ch)
{
    assert_resource($ch);
    // do something with resource
}