Type guards in array find

[jer-sen]

TypeScript Version: 3.9.2

Search Terms:

Expected behavior: No error

Actual behavior: Error Type 'string | null | undefined' is not assignable to type 'string | undefined'. Typescript doesn't take into account the type guard in find arrow function.

Related Issues:

Code

const res: string | undefined = ["a", null].find((v) => v !== null);

Output

"use strict";
const res = ["a", null].find((v) => v !== null);

Compiler Options

{
  "compilerOptions": {
    "noImplicitAny": true,
    "strictNullChecks": true,
    "strictFunctionTypes": true,
    "strictPropertyInitialization": true,
    "strictBindCallApply": true,
    "noImplicitThis": true,
    "noImplicitReturns": true,
    "useDefineForClassFields": false,
    "alwaysStrict": true,
    "allowUnreachableCode": false,
    "allowUnusedLabels": false,
    "downlevelIteration": false,
    "noEmitHelpers": false,
    "noLib": false,
    "noStrictGenericChecks": false,
    "noUnusedLocals": false,
    "noUnusedParameters": false,
    "esModuleInterop": true,
    "preserveConstEnums": false,
    "removeComments": false,
    "skipLibCheck": false,
    "checkJs": false,
    "allowJs": false,
    "declaration": true,
    "experimentalDecorators": false,
    "emitDecoratorMetadata": false,
    "target": "ES2017",
    "module": "ESNext"
  }
}

Playground Link: Provided

[MartinJohns]

(v) => v !== null is not a type guard. It's a function that returns boolean. A type could would return v is string.

You can fix it by adding a type annotation to make it a type guard: find((v): v is string => v !== null)

Alternative just add a comfort function:

function isNotNull<T>(value: T | null): value is T { return value !== null; }

find(isNotNull)

---

Also your playground link does not work.

[jer-sen]

v !== null is a type guard, no ?
Typescript should do the same assumption on find result than inside a if (res !== null) { ... }
Ideally a developer should not have to do anything special for this code to work. Don't you agree ?

But your code works fine thanks !

As for the playground link I used the "Export->Report GitHub issue on TypeScript" menu of typescript Playground...

[Barbiero]

v !== null is a type guard, no ?
Typescript should do the same assumption on find result than inside a if (res !== null) { ... }
Ideally a developer should not have to do anything special for this code to work. Don't you agree ?

But your code works fine thanks !

As for the playground link I used the "Export->Report GitHub issue on TypeScript" menu of typescript Playground...

no, v !== null is an expression which typescript can use to infer a more restricted type, but it's not an actual type guard.

The problem is that typescript can't really return guard-like expressions from a function without explicitly stating that the function is a guard. I've wrote the #37868 issue in an attempt to address that situation; hopefully someone more knowledgable than me can work that into the language.

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.