Typescript and spread operator?

Question

function foo(x:number, y:number, z:number) { 
   console.log(x,y,z);
}
var args:number[] = [0, 1, 2];

foo(...args);

Why am i getting getting this

Answer 1

So there is a little clause you may have missed:

Type checking requires spread elements to match up with a rest parameter.

Without Rest Paramater

But you can use a type assertion to go dynamic... and it will convert back to ES5 / ES3 for you:

function foo(x:number, y:number, z:number) { 
 console.log(x,y,z);
}
var args:number[] = [0, 1, 2];

(foo)(...args);

This results in the same apply function call that you'd expect:

function foo(x, y, z) {
    console.log(x, y, z);
}
var args = [0, 1, 2];
foo.apply(void 0, args);

With Rest Parameter

The alternative is that it all works just as you expect if the function accepts a rest parameter.

function foo(...x: number[]) { 
 console.log(JSON.stringify(x));
}
var args:number[] = [0, 1, 2];

foo(...args);