constt return same closure when called without arguments

[thomasvargiu]

Hi,

I don't know if it's correct, but let's see this example:

$result = f\constt('foo')(); // returns the closure
$result = f\constt('foo')('bar); // returns "foo"

Is it correct? I think that it should return "foo" in both cases, evaluating it even when there are no parameters

[widmogrod]

When implementing curried functions I assume existence of explicit arguments.

$result = f\constt('foo')(null);

Under this assumption implementation is correct.

For me this was never an issue. But I'm open for suggestions.

[ilario-pierbattista]

Perhaps it could be introduced the concept of thunk.

I guess there is not a haskell equivalent function since it uses lazy evaluation by default.

It could be something like this:

function lazy($a): \Closure {
    return function() use ($a) {
        return $a;
    }
}

I would not change the current behaviour of constt since it would mean introducing a big BC.