Following this answer, I have a function that removes null elements from an array (but not undefined):
// JavaScript
const dropNull = arr => {
return arr.flatMap((f) => (f === null ? [] : f));
};
And now I want to use TypeScript to strictly define the types of the function's input and output:
nullThis answer shows a very similar situation, but opposite to mine: whereas they want to allow null but restrict undefined, I want to restrict null and allow undefined. So inspired by the answer I did the reverse tweak:
type NonNull<T> = T extends null ? never : T[];
const dropNull = <T>(arr: T[]): NonNull<T>[] => {
return arr.flatMap((f) => (f === null ? [] : f));
};
But as this REPL shows, I get an error:
Type 'T[]' is not assignable to type 'NonNull[]'.
Type 'T' is not assignable to type 'NonNull'.(2322)
Researching this error I've seen answers such as this one, but I couldn't figure out how to adopt it to my case.
It's hard for typescript to enforce returning a conditional type, because it's hard to know how your logic corresponds to that type. But I think you're over complicating it, and you can sidestep the issue entirely.
const dropNull = <T>(arr: (T | null)[]): T[] => {
return arr.flatMap((f) => (f === null ? [] : f));
};
Here the generic parameter T is inferred as a type that does not contain null. And the argument is an array that can be filled with T or null.
Then you can simply return non null values from the array and typescript is happy.
const numsOrNull = [1,2,null] // (number | null)[]
const nums = dropNull(numsOrNull) // number[]