type SelectData = {
name?: boolean;
address?: boolean;
}
const selectData: SelectData = {
name: true
}
const untypedSelectData = {
name: true
}
I am looking to enforce TypeScript to throw an error if there is an attempt to assign a property that does not exist in the SelectData type:
const selectData: SelectData = {
age: true
}
How can I ensure that the type of selectData remains consistent with untypedSelectData ({name: boolean})?
To ensure that an object is compatible with a weaker type at compile time while preserving a stronger inferred type for the variable or constant it's assigned to, you can employ a generic identity function.
function validateData<T extends SelectData>(data: T): T {
return data;
}
// type is { name: true }
const selectData = validateData({
name: true
})
// error: object literal may only specify known properties
const errorData = validateData({
age: 23
})
As @jcalz points out, there is a feature request on the official GitHub tracker for achieving this functionality without the need for additional function declarations and calls, but as of now, such a feature has not been implemented.
Check out the example you can copy and test in TypeScript Playground: class GreetMessage { message: {}; constructor(msg: string) { let newTyping: { textMsg: string }; // reassign necessary this.message = newTyping; this.message.textMsg = msg; ...
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