CODEWITHSUNDEEP
typescript
yuppies-networking
yuppies-networking

Reputation: 1

Calling function with spread operator

Why this simple code

const arr: [string, number] = ['str', 1];

function fn(arg1?: string, arg2?: number): void {
    alert(arg1);
}

fn(...arr);

produces this error:

Argument of type 'string | number' is not assignable to parameter of type 'string'. Type 'number' is not assignable to type 'string'.

When I change my call to 

fn(arr[0], arr[1])

everything works fine.

Why spread does not work as expected? Why type is also converted to 'string | number'? Shouldn't it detect that the first one is a string and the latter is a number?

Upvotes: 0

Views: 389

Answers (1)

jpetitte
jpetitte

Reputation: 620

Good find. This has already been picked up as a bug by the Typescript community:

https://github.com/Microsoft/TypeScript/issues/4130

It looks like they are still working on it.

Upvotes: 1

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