The necessity for one type argument is apparent in a generic type, particularly when it is divided into a distinct type

I have a simple scenario that resembles the following and is functioning perfectly:

export interface CustomState {
  someBool: boolean;
  status: string;
}

function checkStateDifference<K extends keyof CustomState>(props: { stateKey: K, value: CustomState[K] }) {
  return this.state[props.stateKey] !== props.value;
}

I wanted to abstract this into its own type so I could reuse it in another function, but I encountered an error in my IDE 

Generic type 'CustomState' requires 1 type argument(s)
. Why does this not occur when defined inline and how can I resolve it?

type StateComparisonProps<K extends keyof CustomState> = { stateKey: K, value: CustomState[K] };

function checkStateDifference(props: StateComparisonProps) {
  return this.state[props.stateKey] !== props.value;
}

EDIT: corrected copy/paste error

typescripttypescript-typingstypescript-generics

Answer №1

When utilizing StateCompareProps without an optional generic type, it becomes necessary to specify the generic type like 

StateCompareProps<'status'>
. This means turning the statePropsDifferent function into a generic function once again, requiring you to provide the type when calling, such as 
statePropDifferent<'status'>({stateKey: status, value: 'pending'})

The only solution I am aware of for this issue is shown below:

export interface IProps {
  someBool: boolean;
  status: string;
}

export interface IState extends IProps {
...
// add extra state properties if necessary
}

function statePropDifferent(props: Partial<IProps>) {
  const state: IState = this;
  (Object.keys(props) as (keyof IProps)[]).some(
    (key) => state[key] !== props[key]
  );
}

This approach may not be ideal for your specific scenario, but it was how I handled a similar situation.

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