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 $ch variable?
Answer
PHP does not have a type hint for resources because
No type hint for resources is added, as this would prevent moving from resources to objects for existing extensions, which some have already done (e.g. GMP).
However, you can use is_resource() within the function/method body to verify the passed argument and handle it as needed. A reusable version would be an assertion like this:
function assert_resource($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
}