Spread syntax and Typescript — supplied parameters do not match?

Question

Looking at this simple code from MDN :

function myFunction(x, y, z):void { }
var args = [0, 1, 2];
myFunction(...args);

— I get an error :

Even If I'm being super clear :

function myFunction(x, y, z):void { }
var args:any[3] = [0, 1, 2];
myFunction(...args:any[3]);

It still doesn't work.

Question:

Why doesn't it work and did I miss something ?

I've already seen this answer which muted the error via :

function myFunction(x, y, z):void { }
var args  = [0, 1, 2];
(<any>myFunction)(...args);

Why did <any> mute the error ? It would've been clear if it was :

(<any>)(myFunction(...args)); but it's not.

I have run into the same issue before. myFunction.apply(null, args) works. — mhodges
– mhodges, Commented Jun 21, 2017 at 17:00
Is that the exact code for myFunction? Are you sure that x, y, z don't have types? — Bergi
– Bergi, Commented Jun 21, 2017 at 17:48

Heretic Monkey · Accepted Answer · 2017-06-21 17:31:38Z

2

As far as TypeScript is concerned, you're passing an array to a function that takes three arguments. Thus the signature mismatch error.

Let me be clear here: What you have is absolutely valid ES2015 JavaScript. It's just not valid TypeScript.

(<any>myFunction) casts myFunction as, well, "anything", so TypeScript doesn't look at the function definition. (<any>)(myFunction(...args)); would tell the compiler the result of calling myFunction is any.

edited Jun 21, 2017 at 17:31

answered Jun 21, 2017 at 17:15

Heretic Monkey

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6 Comments

Royi Namir Over a year ago

But I'm passing an array which is explicitly any[3]

Heretic Monkey Over a year ago

... and? It's still an array (okay, an array of three elements, but still an array) being passed to a function that takes three separate, required, arguments.

Royi Namir Over a year ago

I might be missing something but that's the job of ...args. to "break" the array of [0,1,2] into 0,1,2. So things should be fine , no? myfunction will still get what it expectes : 3 values not as an array - but as simple arguments. just like with .apply

Heretic Monkey Over a year ago

This looks pertinent: github.com/Microsoft/TypeScript/issues/238

Jared Smith Over a year ago

@Royi your understanding is correct: what's unclear here is how to get the TypeScript compiler to understand that and underscores the difficulty of layering types on top of extremely dynamic language features.

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