Adding type in a function in Typescript

Question

Suppose I define a function:

const nonEmpty = (str: string) => str.length > 0;

Now I'm adding a type to nonEmpty like that:

const nonEmpty : (str: string) => boolean = (str: string) => str.length > 0;

It compiles but I don't like that the argument name (str) appears twice. Moreover, the argument name does not make sense to add an argument name to the function type.

I would prefer const nonEmpty : string => boolean = ... or similar. Is it possible in Typescript ?

P.S. I found another way to declare a function type:

 const nonEmpty = (str: string): boolean => str.length > 0;

It looks more concise but a bit strange for an inexperienced Typescript developer. Would you prefer this type declaration ?

captain-yossarian from Ukraine · Accepted Answer · 2020-12-15 10:03:23Z

1

Feel free to omit second string type:

const nonEmpty : (str: string) => boolean = (str) => str.length > 0;

nonEmpty('hello') // ok
nonEmpty(1) // error

Second option:

type NonEmpty = (a: string) => boolean
const nonEmpty: NonEmpty = (str) => str.length > 0;

answered Dec 15, 2020 at 10:03

captain-yossarian from Ukraine

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Hoff · Accepted Answer · 2020-12-15 10:08:05Z

1

Also note that typescript can automatically infer the type definition, so you might not need to explicitly add it at all.

answered Dec 15, 2020 at 10:08

Hoff

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1 Comment

Michael Over a year ago

Sure but I do want to add the type (and use a "linter" to check it)

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