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php 7 – PHP 7 and strict "resource" types

Posted by: admin April 23, 2020 Leave a comment


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

For example:

    declare (strict_types=1);

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

    function test (resource $ch)


The above will give the error:

Fatal error: Uncaught TypeError: Argument 1 passed to test() must be an instance of resource, resource given

A var_dump on $ch reveals it to be resource(4, curl), and the manual says curl_init () returns a resource.

Is it at all possible to strictly type the test() function to support the $ch variable?

How to&Answers:

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(
                'Argument must be a valid resource type. %s given.',

which you could then use within your code like that:

function test($ch)
    // do something with resource


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().