CODEWITHSUNDEEP
typescripttype-declaration
Michael
Michael

Reputation: 42100

Adding type in a function in Typescript

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 ?

Upvotes: 0

Views: 624

Answers (2)

Hoff
Hoff

Reputation: 39896

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

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Upvotes: 1

captain-yossarian from Ukraine
captain-yossarian from Ukraine

Reputation: 33101

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;

Upvotes: 1

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