type SelectData = {
name?: boolean;
address?: boolean;
}
const selectData: SelectData = {
name: true
}
const untypedSelectData = {
name: true
}
I want typescript to throw an error if I try to do something like this:
const selectData: SelectData = {
age: true
}
But I also want the type of selectData to be the same as untypedSelectData (i.e. {name: boolean})?
If you want a compile-time check that an object is assignable to some weaker type, while also retaining a stronger inferred type for the variable or constant it's assigned to, you can use a generic identity function.
function checkData<T extends SelectData>(data: T): T {
return data;
}
// type is { name: true }
const selectData = checkData({
name: true
})
// error: object literal may only specify known properties
const errorData = checkData({
age: 23
})
As @jcalz notes, there is a feature request on the official GitHub tracker for a way to do this without adding the extra function declaration and function call, but at present such a feature does not exist.
