Break down and extract the fully organized variable using a single expression

Is it possible to destructure a variable while still having access to the structured variable within the same function call?

For instance, what can be used in place of ??? below to achieve the desired result (and what other changes might need to be made in the code)?

const foo = ({ a, b }) => {
  console.log(a) // 1
  console.log(b) // 2
  console.log(???) // { a: 1, b: 2 }
}
const x = { a: 1, b: 2 }
foo(x)

I aim for both understanding and concise code - I am looking to avoid using const { a, b } = params as the first line within foo() when the entire params object may need to be passed on.

Answer №1

If you're looking for a way to reference an object after destructuring its argument, it's not possible. Once the argument is destructured, you can only access some of its properties, not the full object itself.

However, if the object and properties you need are part of a larger object, you can still achieve what you want by referencing the property name alone to get the object into a variable first, then destructure it later. Here's an example:

const foo = ({ obj: { a, b }, obj }) => {
  console.log(a) // 1
  console.log(b) // 2
  console.log(obj) // { a: 1, b: 2 }
}
const obj = { a: 1, b: 2 }
foo({ obj })

Your original code could potentially work if the object had a property that referenced itself, although this is quite unconventional:

const foo = ({ a, b, obj }) => {
  console.log(a) // 1
  console.log(b) // 2
  console.log(obj) // { a: 1, b: 2 }
}
const x = { a: 1, b: 2 }
x.obj = x;
foo(x)

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