What is the difference between ?int $number and int $number = null? - php

Question

Given this explanation
Nullable types: 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.
https://www.php.net/manual/en/migration71.new-features.php
The following code :
public function test(?int $var) {
}
Means that test() can be called with $var either as int or as null.
And the following code :
public function test(int $var = null) {
}
Means that test() can be called with $var either as int or as null as well.
What are the differences between those two methods ? Is any of these more performant than the other?

Answer 1

It is important to distinguish between the two language features being discussed here, that is, type declarations and default argument values.
The first function is only using type declarations, this means that the input argument has to be of the type int or NULL.
The second function is using both type declarations and default argument values, this means that the argument has to be of the type int or NULL but if omitted it will default to NULL.
Take your first function, if you simply called test() without passing anything to it, you'd get:
PHP Fatal error: Uncaught ArgumentCountError: Too few arguments to function test() [...]
which is correct as the function expects either int or NULL but got neither whereas for the second, as you have defined the argument with a default value, it would run without errors.
Code
function test(?int $var) {
var_dump($var);
}
function test2(int $var = null) {
var_dump($var);
}
test(1); // fine
test(); // error
test2(1); // fine
test2(); // fine
As far as performance goes, the difference is probably negligible, nothing significant enough that would be a cause for concern.
Live Example
Repl

Answer 2

If the language were designed today, int $var = null would probably be an error, and should really be written ?int $var = null. The two parts mean different things:
The ? indicates that null is a valid value for that parameter.
The = null indicates that null is the default if the parameter is not passed.
However, before the ?type syntax was introduced, there was a special case in the language: if null is given as the default for a parameter, then it is legal to pass null to that parameter, even if the type declaration would otherwise prevent it.

Answer 3

The difference is how you can call the function:
// public function test(?int $var)
$foo->test("x"); // does not work (Argument 1 passed to Foo::test() must be of the type int or null, string given)
$foo->test(123); // works
$foo->test(null); // works
$foo->test(); // does not work (Too few arguments to function Foo::test(), 0 passed)
// public function test(int $var = null)
$foo->test("x"); // does not work (Argument 1 passed to Foo::test() must be of the type int or null, string given)
$foo->test(123); // works
$foo->test(null); // works
$foo->test(); // works
The difference is that you cannot call the function as ->test() using the first syntax.