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Function arg type inference (incorrectly) unions with null but excludes others #49850
Description
Bug Report
🔎 Search Terms
function arg generic inconsistent type inference
infer func arguments breaks on null
🕗 Version & Regression Information
This is the behavior in every version I tried, and I reviewed the FAQ for entries about functions and generics
⏯ Playground Link
Playground link with relevant code
💻 Code
const matchStrict = <T>(test: (x: unknown) => x is T) => {
return (...options: Array<T | undefined>): T | undefined => {
return options.find((o) => test(o));
};
};
const matchLoose = <T>(
test: (x: unknown) => x is T,
...options: Array<T | undefined>
): T | undefined => {
return options.find((o) => test(o));
};
const isStr = (x: unknown): x is string => {
return typeof x === 'string';
};
// Works as expected
matchStrict(isStr)(123, 'a'); // ✅ 'number' is not assignable
matchStrict(isStr)(null, 'a'); // ✅ 'null' is not assignable
// Inconsistent behavior
matchLoose(isStr, 123, 'a'); // ✅ 'number' is not assignable
matchLoose(isStr, null, 'a'); // ⛔️ WTF! matchLoose<string | null>
matchLoose<string>(isStr, null, 'a'); // ✅ 'null' is not assignable🙁 Actual behavior
matchLoose(isStr, 123) correctly excludes number, but matchLoose(isStr, null) incorrectly becomes matchLoose<string | null>.
🙂 Expected behavior
I expect matchLoose(isStr, null) and matchLoose(isStr, 123) to both throw errors, because isStr narrows <T> strictly. (I expect the intersection to work regardless of argument order too.)
Activity
whzx5byb commentedon Jul 10, 2022
A workaround could be using the type of type guard function as generic:
jcalz commentedon Jul 11, 2022
Seems like a duplicate of #14829. You don't want the
optionsparameter to be used as an inference site forT.TypeScript has some heuristics for accepting/rejecting/combining inference candidates;
stringfrom one andnumberfrom another tends to be rejected, butstringfrom one andnullfrom another tends to be accepted and combined via union. (See SO question, #19596, #44312, and probably a zillion other places I can't find right now).Workarounds here usually involve lowering the priority of the
optionsinference site, or otherwise blocking the inference. Your curried function is one way to do it, sinceTis already specified by the time the returned function is called. You can also define aNoInfer<T>utility and write the type ofoptionsasArray<NoInfer<T> | undefined>. Possibilities are listed in #14829. You can trytype NoInfer<T> = T & {}ortype NoInfer<T> = [T][T extends unknown ? 0 : 1]and see if that works for your use cases.mikestopcontinues commentedon Jul 11, 2022
Thanks Joe Calzaretta (@jcalz), good tips!
mikestopcontinues commentedon Jul 11, 2022
For other who arrive here, I found this to be an excellent solution:
typescript-bot commentedon Jul 13, 2022
This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.