TypeScript not detecting type in filter() - Stack Overflow

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 const values: (string | null)[] = [];

 const filtered = values.filter((v) => v !== null && v.trim() !== "");
 filtered // still (string | null)[]

Why does TypeScript not detect the filtering of null values?

 const values: (string | null)[] = [];

 const filtered = values.filter((v) => v !== null && v.trim() !== "");
 filtered // still (string | null)[]

Why does TypeScript not detect the filtering of null values?

• typescript

Share Improve this question Follow edited Mar 21 at 22:55 jonrsharpe 122k3030 gold badges268268 silver badges476476 bronze badges asked Mar 21 at 22:21 AlexAlex 66.2k185185 gold badges459459 silver badges651651 bronze badges 1

• This question is similar to: Why can't my filter method narrow the type and eliminate the undefined and false from the array?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – jonrsharpe Commented Mar 21 at 22:55

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2 Answers 2

Sorted by: Reset to default 1

Typescript can't infer what your function is actually doing.

However, you can give it a predicate :

const filtered = values.filter((v): v is string => v !== null && v.trim() !== "");

const filtered = values.filter((v) => v !== null && v.trim() !== "");

Please observe below the return type of the callback passed into filter. This will get by hovering over the literal filter in the above statement. The callback returns the type (string | null). Therefore the resultant array is also typed accordingly. This is the reason for the undesired type (string | null)[].

(method) Array<string | null>.filter(predicate: (value: string | null, index: number, array: (string | null)[]) => unknown, thisArg?: any): (string | null)[] (+1 overload)

Solution: using a type predicate

The expression v is string included in the below statement asserts the return type of the callback as string. This can be checked again by hovering the literal filter. However, this kind of assertion is safe since the condition guards v !== null && v.trim() !== "" this assertion. This assertion effectively narrows the type from (string | null)[] to string[].

const filtered = values.filter((v) : *v is string* => v !== null && v.trim() !== "");

By hovering the literal filter

(method) Array<string | null>.filter<string>(predicate: (value: string | null, index: number, array: (string | null)[]) => value is string, thisArg?: any): string[] (+1 overload)

For more, please read Using type predicates.

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