TypeScript and optional destructured arguments

Question

I can't seem to get optional arguments works with destructured arguments in TypeScript.

The right code is generated for the argument, but Typescript doesn't seem to allow me to use the generated variables in my code, defeating the purpose.

Am I doing something wrong? Here is a reduction:

declare var lastDirectionWasDownish : boolean;

function goToNext(
    { 
        root: Node = document.activeElement, 
        actLikeLastDirectionWasDownish:boolean = lastDirectionWasDownish
    } = {}
) {
    return root && actLikeLastDirectionWasDownish;
}

which compiles into

function goToNext(_a) {
    var _b = _a === void 0 ? {} : _a, _c = _b.root, Node = _c === void 0 ? document.activeElement : _c, _d = _b.actLikeLastDirectionWasDownish, boolean = _d === void 0 ? lastDirectionWasDownish : _d;
    return root && actLikeLastDirectionWasDownish;
}

basarat · Accepted Answer · 2016-07-22 23:45:15Z

7

TypeScript is actually preventing you from making a mistake that you would miss in pure JS. The following pure JS:

function foo({root: Node}) {
   // the parameter `root` has been copied to `Node`
}

TypeScript understands this and doesn't let you use Node. To add a type annotation you would actually:

function foo({root}: {root: Node}) {
   // now you can use `root` and it is of type Node
}

Fix

You want

function foo({root = document.activeElement } : {root: Node}) {
    root;// Okay
}

answered Jul 22, 2016 at 23:45

basarat

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2 Comments

FremyCompany Over a year ago

Thanks, that is likely what I was looking for. That's a bit more repetitive and verbose, but it does work!

basarat Over a year ago

JavaScript chose : instead of = for assignment in object literals (structuring). That in turn was reused for destructuring. So we have : located externally for definition purposes as there isn't a clean way to plug it in.

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