Why does re-declaring an argument inside of a try/catch throw a ReferenceError?

Question

I mistakenly wrote a re-declaration of an argument as a const in a function and instead of throwing SyntaxError: Identifier 'bar' has already been declared I ended up with ReferenceError: bar is not defined..

What causes this behaviour? It wasn't the expected error, and left me confused for a few minutes.

Example code:

function foo(bar) {   try {       console.log(bar);       const bar = 123;   } catch(err) { console.log(err) } } foo(456); 

If I don't wrap the declaration in a try/catch, I get (what I believe to be) the expected error.

Answer 1

Constants are block-scoped, much like variables defined using the let statement.

From this MDN article.

Since you wrapped bar inside a block of braces, its definition is relative to that block. And because you have another bar declaration inside of that block, despite being after the call to it, the compiler will attempt to use this newly defined bar instead of the passed-in parameter. Rename them as separate parameters to mitigate confusion, since one can assume they are holding different data because of your declaration.

Answer 2

It's a myth that const and let aren't hoisted at all. They're half-hoisted. :-) That is: Within a block, if const bar or let bar (or class bar { } for that matter) appears anywhere in that block, then bar cannot be used prior to that declaration in the block — even if it exists in a containing scope. This area between the beginning of the block and the declaration is called the Temporal Dead Zone:

function foo(bar) {   // `bar` is fine here   try {       // Temporal Dead Zone, `bar` cannot be used here       console.log(bar);       // End of TDZ       const bar = 123;       // `bar` would be fine here   } catch(err) { console.log(err) } } foo(456);