ES6 object destructuring mandatory parameters

Question

Here is the function, password is mandatory but other id and name have default values:

function name({id = null, name = 'user', password}) { }

Results:

name();     //throws error as expected
name({});   //passes 'password' as 'undefined' - should've thrown error

How can I make this function throw an error if password is not provided using ES6 functionality?

Why don't you just validate incoming parameters in the function itself and throw if anything is invalid? Heck an empty string or even a non-string could be passed for a password and that would pass any automatic ES6 stuff so you're going to have to validate the password field anyway. Even if you had a hard typed language, you're still going to want to validate. — jfriend00, Jan 08 '17 at 19:05
@jfriend00 I agree, but I just want to save code for scenarios where I only have to care about `undefined`. It is just `mcve`. — Shaharyar, Jan 08 '17 at 19:10

Nina Scholz · Answer 1 · 2017-01-08T18:27:10.363

4

You could assign the same variable, it throws an error Use before declaration, if not given.

function name({ id = null, name = 'user', password = password }) {
    console.log(password);
}
name({});

If set, then it works without error.

function name({ id = null, name = 'user', password = password }) {
    console.log(password);
}
name({ password: 'foo' });

edited Jan 08 '17 at 18:27

answered Jan 08 '17 at 18:24

Nina Scholz

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This article is also worth checking out: http://www.2ality.com/2014/04/required-parameters-es6.html – Bartosz Gościński Jan 08 '17 at 18:28
why is `,password = password}` different than `,password}` ? – dandavis Jan 08 '17 at 18:32
because its a default assignment, with a variable which does not exist. and if it exist, then no assignment occurs. – Nina Scholz Jan 08 '17 at 18:33
default assignment is also `undefined`, how is that `undefined` is not equal to `an undefined variable`? I hope I am clear, this `undefined` hell is confusing me. – Shaharyar Jan 08 '17 at 18:48
the variable `password` does not exists., even if RHS changed to `_`, it does not exists. it is not defined, and therefore it throws an error. – Nina Scholz Jan 08 '17 at 18:52
I know, but if I leave it as `,password` the default assignment is performed and that default assignment is also `undefined` *(I'm using vscode debugger to see this, in nodejs env)* and no there is no error in default. Error is only thrown when I `password = password` and this creates hell lot of confusion for me. – Shaharyar Jan 08 '17 at 19:12

Scimonster · Answer 2 · 2017-01-08T18:21:45.563

2

You didn't make it required, you simply didn't provide a default value.

function name({id = null, name = 'user', password}) {
     if (password == undefined) {
        throw new Error('password is required');
    }

The reason it threw an error the first time is because it tried to destructure undefined, which is impossible.

edited Jan 08 '17 at 18:21

answered Jan 08 '17 at 18:19

Scimonster

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I know it can be achieved using ES5 or traditional javascript code, but is there anyway I don't have to write this extra validation and use `ES6 functionality` just like `usecase 1`? – Shaharyar Jan 08 '17 at 18:21
No, the "validation" is simply a side effect of destructuring. – Scimonster Jan 08 '17 at 18:22

score 2 · Accepted Answer · edited May 23 '17 at 11:52

Scimonster's answer is spot-on about what the probelm is.

If you like, you can handle it a bit more declaratively by using a utility function to throw an Error in the function parameter list:

// Utility function:
function throwRequired(name) {
  throw new Error(`'${name}' is required`);
}

function name({id = null, name = 'user', password = throwRequired("password")}) {
  console.log("All is good, password is: " + password);
}

try {
  name();     //throws error as expected
} catch (e) {
  console.log("Got error from name():", e.message);
}
try {
  name({});     //throws error as expected
} catch (e) {
  console.log("Got error from name({}):", e.message);
}
name({password: "foo"}); // works

That works because the initializer on a default parameter is only evaluated if it's required to fill in the default.

*(Doh! If you don't see `throwRequired` above, hit refresh.)* — T.J. Crowder, Jan 08 '17 at 18:31

score 2 · Answer 4 · edited May 23 '17 at 12:13

Note: This is an ugly hack using the default parameters, and I prefer @Scimonster's answer.

You can actually run a function, and even an IIFE as a default param. The IIFE can throw an error:

function name({
  id = null, 
  name = 'user',
  password = (() => { throw new Error('password not defined'); })()
}) { 
  console.log(id, name, password);
}
  
name({ id: 1, name: 'the name', password: 'the password' }); // works

name({}); // throws an error

ES6 object destructuring mandatory parameters

4 Answers4

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