add words on why generics don't help.

[Danack]

https://en.reddit.com/r/PHP/comments/rv11fc/operator_overloading_rfc_is_in_voting_what_are/hr316wy/

–]john7 points 2 days ago*
I can't wait for it. The reason is that I use lots of lazy evaluations; old version of example is here.

So I could have a function like this:

​

function doSomething(int|LazyValue $param): int {
return $param + 42;
}
​

but without operator overload:

​

function doSomething(int|LazyValue $param): int {
if ($param instanceof LazyValue) {
$param = $param->getValue();
}

return $param + 42;

}
​

Updated:

I would prefer for RFC to require interfaces, if possible:

​

interface Addable
{
public function add(int|float|Addable $param): int|float|Addable
}
​

End result:

​

function doSomething(int|Addable $param): int {
return $param + 42;
}
​

Interfaces would make easier job for static analysis, allow ctrl+click to find all Addable implementations and avoid some WTF situations in older apps with __add() method.

[–]Danack 2 points 2 days ago
I would prefer for RFC to require interfaces, if possible:

Did read the reasons why that isn't appropriate in the RFC ?

https://wiki.php.net/rfc/user_defined_operator_overloads#why_not_interfaces

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[–]john 1 point 1 day ago*
I did, but I think it is mostly because of lack of generics.

For example:

interface Addable
{
public function add(T $param): T
}

class Money implements Addable<int|Money>{}
​

But I could be wrong, that's why I said if possible; maybe other contributors can come up with some idea.

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[–]Danack 3 points 1 day ago
That doesn't help anything. If you're using an interface as a type e.g.

function foo (Addable $x, int $y)
{
return $x + $y;
}
You can't tell if the code is valid or not from the interface.