为什么在try块中重新声明函数标识符会抛出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:

1. Without the try block scope:

    function _free() {} var _free = 1; 

2. Within a function scope:

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

But why?

(Testing environment: Chromium 61.0.3126.0)

Answer 1

因为块范围的函数声明是新的ES6特性并且是安全的（即在名称冲突上抛出错误，类似于let和const ），但是其他情况（无论是程序员错误）都需要保持向后兼容并静默覆盖功能。

Answer 2

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 { }