Cannot pass null argument when using type hinting

Try:

function foo(Type $t = null) {

}

Check out PHP function arguments.

PHP 7.1 or newer (released 2nd December 2016)

You can explicitly declare a variable to be null with this syntax

function foo(?Type $t) {
}

this will result in

$this->foo(new Type()); // ok
$this->foo(null); // ok
$this->foo(); // error

So, if you want an optional argument you can follow the convention Type $t = null whereas if you need to make an argument accept both null and its type, you can follow above example.

You can read more here.

---

PHP 7.0 or older

You have to add a default value like

function foo(Type $t = null) {

}

That way, you can pass it a null value.

This is documented in the section in the manual about Type Declarations:

The declaration can be made to accept NULL values if the default value of the parameter is set to NULL.

As other answers already mentioned, this is only possible if you specify null as the default value.

But the cleanest type-safe object oriented solution would be a NullObject:

interface FooInterface
{
    function bar();
}
class Foo implements FooInterface
{
    public function bar()
    {
        return 'i am an object';
    }
}
class NullFoo implements FooInterface
{
    public function bar()
    {
        return 'i am null (but you still can use my interface)';
    }
}

Usage:

function bar_my_foo(FooInterface $foo)
{
    if ($foo instanceof NullFoo) {
        // special handling of null values may go here
    }
    echo $foo->bar();
}

bar_my_foo(new NullFoo);

Starting from PHP 7.1, nullable types are available, as both function return types and parameters. The type ?T can have values of the specified Type T, or null.

So, your function could look like this:

function foo(?Type $t)
{

}

As soon as you can work with PHP 7.1, this notation should be preferred over function foo(Type $t = null), because it still forces the caller to explicitly specify an argument for the parameter $t.