What is the purpose of the question marks before type declaration in PHP7 (?string or ?int)?
It's called a Nullable type, introduced in PHP 7.1.
You could pass a NULL
value if there is a Nullable type (with ?
) parameter, or a value of the same type.
Parameters :
function test(?string $parameter1, string $parameter2) {
var_dump($parameter1, $parameter2);
}
test("foo","bar");
test(null,"foo");
test("foo",null); // Uncaught TypeError: Argument 2 passed to test() must be of the type string, null given,
Return type :
The return type of a function can also be a nullable type, and allows to return null
or the specified type.
function error_func():int {
return null ; // Uncaught TypeError: Return value must be of the type integer
}
function valid_func():?int {
return null ; // OK
}
function valid_int_func():?int {
return 2 ; // OK
}
Property type (as of PHP 7.4) :
The type of a property can be a nullable type.
class Foo
{
private object $foo = null; // ERROR : cannot be null
private ?object $bar = null; // OK : can be null (nullable type)
private object $baz; // OK : uninitialized value
}
See also :
Nullable union types (as of PHP 8.0)
As of PHP 8, "?T
notation is considered a shorthand for the common case of T|null
"
class Foo
{
private ?object $bar = null; // as of PHP 7.1+
private object|null $baz = null; // as of PHP 8.0
}
References
As of PHP 7.1 :
Type declarations for parameters and return values can now be marked as nullable by prefixing the type name with a question mark. This signifies that as well as the specified type, NULL can be passed as an argument, or returned as a value, respectively.
As of PHP 7.4 : Class properties type declarations.
As of PHP 8.0 : Nullable Union Type
The question mark before the string
in your function parameter denotes a nullable type. In your above example, $parameter1
must is allowed to have a NULL
value, whereas $parameter2
is not; it must contain a valid string.
Parameters with a nullable type do not have a default value. If omitted the value does not default to null and will result in an error:
function f(?callable $p) { }
f(); // invalid; function f does not have a default