ecmascript 6 - Does a Javascript Object Destructure assignment use const, let, or var? - Stack Overflow

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In ES6/ES2015 I can use Object destructuring to pull fields from objects passed as function parameters:

function userId({id}) {
  return id;
}

var user = { 
  id: 42, 
};

console.log("userId: " + userId(user)); // "userId: 42"

In the above example, is the id variable set in the userId function equivalent to setting it via var, const, or let ?

So which of the below is the equivalent function:

Var

function userId(user) {
    var id = user.id
    return id;
}

Let

function userId(user) {
    let id = user.id
    return id;
}

Const

function userId(user) {
    const id = user.id
    return id;
}

(Example function from MDN page linked above, jump to "Pulling fields from objects passed as function parameter)

In ES6/ES2015 I can use Object destructuring to pull fields from objects passed as function parameters:

function userId({id}) {
  return id;
}

var user = { 
  id: 42, 
};

console.log("userId: " + userId(user)); // "userId: 42"

In the above example, is the id variable set in the userId function equivalent to setting it via var, const, or let ?

So which of the below is the equivalent function:

Var

function userId(user) {
    var id = user.id
    return id;
}

Let

function userId(user) {
    let id = user.id
    return id;
}

Const

function userId(user) {
    const id = user.id
    return id;
}

(Example function from MDN page linked above, jump to "Pulling fields from objects passed as function parameter)

• javascript
• ecmascript-6

Share Improve this question Follow asked Dec 22, 2016 at 1:31 Caleb JayCaleb Jay 2,19933 gold badges4141 silver badges7676 bronze badges 2

• Well you know it's not const, because within the function you can assign another value to id. – nnnnnn Commented Dec 22, 2016 at 1:36
• 1 There's no meaningful difference between let and var in this case, since the scope of function arguments are always the entire function. – PMV Commented Dec 22, 2016 at 1:37

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2 Answers 2

Sorted by: Reset to default 7

Declaring variables with destructuring uses the same scope as if you'd declared ordinary variables in the same position. So if you do:

let {id} = {id: 42};

then it's a let binding, if you do:

var {id} = {id: 42};

then it's a var binding.

Function parameters are like var bindings, since they're scoped to the entire function and they're not const. They're also like let bindings because they're scoped to the current block, which is the entire function body.

It will be a local variable scoped over the whole function (as all function parameters are), as in your first and second example (which are equivalent: there is no difference between let and var when at the top of the function).

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