Why does redeclaring a function identifier within a try block throw a SyntaxError?

Question

The following lines of JavaScript

try {
    function _free() {}
    var _free = 1;
} finally { }

result in the following error:

 Uncaught SyntaxError: Identifier '_free' has already been declared

However, the following two blocks of JavaScript code don't:

Without the try block scope:
```
function _free() {}
var _free = 1;
```

Within a function scope:

function a() {
    function _free() {}
    var _free = 1;
}

But why?

(Testing environment: Chromium 61.0.3126.0)

@Jai Nothing to do with strict mode, also it's called `use strict` :) — Mr. Alien, Jul 04 '17 at 04:54

Bergi · Accepted Answer · 2017-07-04T04:58:40.730

15

Because block-scoped function declarations are a new ES6 feature and were made safe (i.e. throw an error on name collisions, similar to let and const), but the other cases (which are programmer mistakes regardless) needed to stay backwards compatible and silently overwrite the function.

edited Jul 04 '17 at 04:58

answered Jul 04 '17 at 04:49

Bergi

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I wonder does this happen in other browsers and older browsers that don't ES6 – user93 Jul 04 '17 at 04:50
@Bergi would this be due to `strict mode`. May be somewhere OP used it. – Jai Jul 04 '17 at 04:52
I tested it in IE nothing happens. But firefox and other new browsers have it – user93 Jul 04 '17 at 04:54
@user93 Treatment of block-level function declarations [was very inconsistent](http://kangax.github.io/nfe/#function-declarations-in-blocks) before ES6 – Bergi Jul 04 '17 at 04:54
@user93 IE implements the least amount of ES6 features out of all the major browsers. – Andrew Li Jul 04 '17 at 04:55
@user93—there is history to "block scoped function declarations", best to just not use them. Use function expressions if you want to conditionally create functions. – RobG Jul 04 '17 at 04:56
@Jai No, [strict mode doesn't really affect this](https://stackoverflow.com/a/31461615/1048572) – Bergi Jul 04 '17 at 04:56
Thanks! Though the code including these redeclaring statements are generated by emscripten and it seems a little outdated...:/ – zhouhy Jul 04 '17 at 05:29

score 5 · Answer 2 · answered Jul 04 '17 at 08:56

To expand on Bergis answer, there is a difference in how the code is interpreted in ES5 and ES6 since block-scoped function declarations were added.

Input:

function test() {
    try {
        function _free() { }
        var _free = 1;
    } finally { }
}

Since ES5 does not support block-level functions, _free is hoisted to the parent function:

function test() {
    var _free = function _free() { }
    try {
        var _free = 1;
    } finally { }
}

In ES6, the function is declared at block-level, and semantically equal to a let/const declaration:

function test() {
    try {
        let _free = function _free() { }
        var _free = 1;
    } finally { }
}

This throws an error because var _free tries to declare a variable which is already declared. For example, this throws in ES6 as well:

let _free;
var _free = 1;    // SyntaxError: Indentifier '_free' has already been declared

While this is fine:

var _free;
var _free = 1;    // No SyntaxError

Setting the value of the already declared identifier is fine:

let _free;
_free = 1;

Therefore, to set the declared identifier _free to 1, you need to skip the second declaration:

try {
    function _free() { }
    _free = 1;
} finally { }

"*Since ES5 does not support block-level functions, _free is hoisted to the parent function*" doesn't really make sense. It did **not support them**, so the code simply was a syntax error. Nothing got hoisted. — Bergi, Jul 04 '17 at 11:52
An ES6 function declaration isn't completely semantically equivalent to `let`/`const`, as it is still hoisted. — Bergi, Jul 04 '17 at 11:53

Why does redeclaring a function identifier within a try block throw a SyntaxError?

2 Answers2

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