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Adjust Type guard behavior in iterators #29916
Description
Activity
I don't know why, but the playground does not show the correct typing, please test in your code.
You need to enable strictNullChecks in the options on the playground to see your issue reproduced there.
Reacted by Daniel ESame as #29805 and duplicated/related to #9998.
Basically, typescript don't know if the predicate on
forEachis called synchronously or not.Workaround:
const myFunc = () => { let myVal: string | undefined; if (!myVal) return; const v = myVal; // Type = string new Array(2).forEach(x => v); // Type = string }
Playground (enable
strictNullChecks).Reacted by Vladimir ĐokićThank you Jordi Oliveras Rovira (@j-oliveras) for pointing the relations.
I read about the workaround with the
constdeclaration, thanks for that.My feature-request here is to add logic for basic Javascript iterators. So, if it is a standard callback-function (not overwritten), then you could teach Typescript that these functions run synchronously, right?
This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.
- locked as resolved and limited conversation to collaborators
on Oct 21, 2025
TypeScript Version: 3.2.2
Search Terms: Conditional Types, Type Guards, iterators, loop, callback functions
Suggestion:
When having values that are of union type and conditions result in telling the Typescript Type Guards to adjust the type, in basic Javascript iterator functions this adjusted type is reset to its initial type.
Code Example:
Expected behavior:
The type of
myValin the last line should only be of typestring.This would be nice for all iterators (e.g. for, while, map, …)
Actual behavior:
The type of
myValisstring | undefinedPlayground Link:
https://www.typescriptlang.org/play/#src=const%20myFunc%20%3D%20()%20%3D%3E%20%7B%0D%0A%20%20%20%20let%20myVal%3A%20string%20%7C%20undefined%3B%0D%0A%20%20%20%20if%20(!myVal)%20return%3B%0D%0A%20%20%20%20myVal%3B%20%2F%2F%20Typ%20%3D%20string%0D%0A%20%20%20%20new%20Array(2).forEach(x%20%3D%3E%20myVal)%3B%20%2F%2F%20Typ%20%3D%20string%20%7C%20undefined%0D%0A%7D
Info:
I don't know why, but the playground does not show the correct typing, please test in your code.
Checklist:
My suggestion meets these guidelines: