Reseting a const variable, even though it's const (ES6)

Question

Understanding variable scopes

const { response } = somecall();
if (response) {
   const { response } = someOtherCallThatReturnsUndefined(response);
   // the value of response being passed in to **someOtherCallThatReturnsUndefined** is actually undefined
}

no, you are declaring two differents objects with the a variable of the same name inside — Mister Jojo, Nov 18 '19 at 14:34
They're two different variables. ([Example](https://jsfiddle.net/erd0znc9/1/)) — Tyler Roper, Nov 18 '19 at 14:41
okay this all makes sense, I'm going to go ahead and delete this question — Free Lancer, Nov 19 '19 at 12:52

score 2 · Accepted Answer · answered Nov 18 '19 at 15:12

What you're doing is called variable shadowing.

When you declare a variable with const (or let) it's block-scoped: the second time you're declaring a new response constant you're in a different scope, so they're different variables. But the fact that they shares the same name, means you're shadowing the outer one (that could potentially lead to bugs); and preventing the inner scope to access to the variable of the outer scope.

To be clear:

  const foo = "1";
  {
     const foo = "bar";
     console.log(foo); // "bar"
  }
  console.log(foo); // "1";

The variable foo in the inner block is a totally different variable than the foo in the outer block, in fact when you exit the block you still have the right value. They just "shares" the same name, that's it.

This really cleared things up, thanks! – Free Lancer Nov 19 '19 at 12:52 — Free Lancer, Nov 19 '19 at 12:52

score 1 · Answer 2 · answered Nov 18 '19 at 14:33

1

By saying const { response } you are declaring the variable in a different context than the original variable hence why you can do so.

answered Nov 18 '19 at 14:33

Dylan

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Reseting a const variable, even though it's const (ES6)

2 Answers2