stackoverflow.com › questions › 34179897 › typescript-and-spread-operator

So there is a little clause you may have missed:

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

Without Rest Parameter

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];

(<any>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);

Answer from Fenton on Stack Overflow

Stack Overflow

stackoverflow.com › questions › 34179897 › typescript-and-spread-operator

Typescript and spread operator? - Stack Overflow

Without Rest Parameter

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];

(<any>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);

2 of 4

I think @Fenton explains it very well but I would like to add some more documentation and possible solutions.

Solutions:

Function overload. I prefer this solution in this case because it keeps some kind of type safety and avoids ignore and any. The original method and function call does not need to be rewritten at all.

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

foo(...args);

Use @ts-ignore to ignore specific line, TypeScript 2.3

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

Use as any.

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

(foo as any)(...args);

Link with documentation regarding the spread operator:

https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-1.html

Discussions regarding this:

https://github.com/Microsoft/TypeScript/issues/5296 https://github.com/Microsoft/TypeScript/issues/11780 https://github.com/Microsoft/TypeScript/issues/14981 https://github.com/Microsoft/TypeScript/issues/15375

GitBook

basarat.gitbook.io › typescript › future-javascript › spread-operator

Spread Operator | TypeScript Deep Dive

The spread operator allows you to easily place an expanded version of an array into another array. This is demonstrated in the example below:

Videos