Cannot assert Partial<T>[keyof T] not to be undefined using ! (regression from 2.5)

[mprobst]

TypeScript Version: 2.6.2

Code

See also in the playground example here.

function createComponent<T>(
    comp: T, inputs: Partial<T> = {}) {
    for (const key in inputs) {
        if (inputs.hasOwnProperty(key)) {
            const partialVal = inputs[key];
            const value = partialVal!;
            comp[key] = value;
        }
    }
}

Expected behavior:

parialVal! should strip the |undefined from value, and the assignment should succeed.

Actual behavior:

  Type 'T[keyof T] | undefined' is not assignable to type 'T[keyof T]'.
    Type 'undefined' is not assignable to type 'T[keyof T]'.
      Type 'undefined' is not assignable to type 'T[string]'.
        Type 'Partial<T>[keyof T]' is not assignable to type 'T[string]'.
          Type 'T[keyof T] | undefined' is not assignable to type 'T[string]'.
            Type 'undefined' is not assignable to type 'T[string]'.

[ghost]

I think this would essentially require us to have #14366 working so we could have a NonNull<Partial<T>[keyof T]> type.

[mprobst]

For the record, this used to succeed in 2.5, so this is a regression.

[ghost]

OK, I must be wrong about a NonNull type being needed.

[ahejlsberg]

This is caused by #18042. Previously we would immediately transform an indexed access applied to a mapped type { [P in keyof T]?: T[P] | undefined }[X] into a simpler form T[X] | undefined, but we now defer that operation. For that reason the type appears to not include undefined and the ! operator has no effect.

[mhegazy]

duplicate of #14366

[typescript-bot]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.