Spread arguments should appropriately spread into optional parameters

[DanielRosenwasser]

Adapted from #4755.

Currently the following is fine:

function f(...things: any[]) {
    let [a, ...bs] = things;

    // do things with 'a' and 'bs'
}

function g(xs: any[]) {
    f(...xs);
}

However, it might be the case that you want to be clear about how things will be used when your function gets called, so it should be fine to use an optional parameter for a:

function f(a?: any, ...bs: any[]) {
    // do things with 'a' and 'bs'
}

function g(xs: any[]) {
    f(...xs);
//  ~~~~~~~~
}

Currently, with the invocation of f above you'll get the error message Supplied parameters do not match any signature of call target. even though a is optional.

[bootstraponline]

👍

[mhegazy]

should be subsumed by a fix to #4130

[mhegazy]

on a closer look, this is already fixed.

[yordis]

@mhegazy @DanielRosenwasser using "typescript": "^2.7.0-dev.20171123" I am still having this issue everywhere I use spread operator I have to actually do functioName.call(null, value)

I am confused because most of the tickets said close but this still happening

[bootstraponline]

Yeah, I don''t think this works.

let str = "abc";
let arr = [...str];

Maybe it requires some configuration.

• https://stackoverflow.com/a/33233956