Why don't you just let the inference do it job ?

input(true, { transform: v => v === 'append' ? 'append' : booleanAttribute(v) }); is actually perfectly valid.

Why don't you just let the inference do it job ?

input(true, { transform: v => v === 'append' ? 'append' : booleanAttribute(v) }); is actually perfectly valid.

Oh wow, I actually wasn't aware of this! Thank you! But where is this magic? How does the signal know that it's value is not of the type of the first argument only? Never the less, would a default value hurt?

When the generic isn't specified, typescript will try to infer it.
But when there are multiple generics, and you specified only one of them, it won't do that and fail to find matching overload (with a single generic).

I wouldn't expect T to equal TransformT as the goal of the transform is to coerce a type, often unknown to another.

When the generic isn't specified, typescript will try to infer it. But when there are multiple generics, and you specified only one of them, it won't do that and fail to find matching overload (with a single generic).

I wouldn't expect T to equal TransformT as the goal of the transform is to coerce a type, often unknown to another.

I know about this infering but the values type is T - not TransformT. Never the less is the value of type TransformT - where / when is this happening?

But: right now, specifying only one type T, TransformT will fallback to undefined - that behavior would never be usefull, is it?

Edit: by the way, here the reason I created this PR: https://stackoverflow.com/questions/79082672/unable-to-add-types-to-input-with-transform-functionality

But: right now, specifying only one type T, TransformT will fallback to undefined - that behavior would never be usefull, is it?

This isn't the case; TypeScript requires that if some type arguments are present, then all non-optional type parameters must also have a corresponding type argument. A default type as introduced in this PR makes it valid, but only for atypical usages of transform: the output type is expected to be different from the input type.

Concretely, in your example you mention to have

multiple= input<boolean | 'append', boolean | 'append'>(true, { transform: v => v === 'append' ? 'append' : coerceBooleanProperty(v) });

this is unlikely what you intent to represent; e.g. the coerceBooleanProperty supports converting '' (the empty string) to a true value in support of attributes without value binding (which are assigned using the empty string), yet the '' is not part of your input type, so it's invalid according to this type. Hence, the entire transform becomes a no-op, as all values that inhabit the boolean | 'append' type are returned as-is.

This change also hinders potential advances in TypeScript that would be true solutions for this case, namely microsoft/TypeScript#54047 and microsoft/TypeScript#26349. It's unclear if such change will ever make it into TypeScript, but introducing a default value in Angular now will make such future feature inapplicable. Hence we do not want to go forward with this change.

This issue has been automatically locked due to inactivity.
Please file a new issue if you are encountering a similar or related problem.

Read more about our automatic conversation locking policy.

_{This action has been performed automatically by a bot.}