Why Self-Executing Anonymous works?

Question

I know about Self-Executing Anonymous. And usually we create them as

(function(){ return 1;})()

the reason - parser feature which didn't run if we use

function(){ return 1}()

But today I found that next code works too ( check brackets order )

(function(){ return 1;}())

function(){ return 1; }() still give me SyntaxError, as it should

Please explain why? Thx for reference to get more details

P.S. the question is about (function(){ return 1;}()) variant!

you didn't close the paraenthesis in ( function(){ return 1; }() — Lunny, May 16 '16 at 14:44
Because that is not correct syntax. This: `function(){ return 1; }()` simply is not defined as valid syntax as it is. — XCS, May 16 '16 at 14:44
@LunZhang You mean in the last line? The first parenthesis isn’t part of the code. — Sebastian Simon, May 16 '16 at 14:45
@Cristy The questions is why it WORKS, not 'why in NOT works' — Vasiliy vvscode Vanchuk, May 16 '16 at 14:45
@VasiliyVanchuk Read this: https://en.wikipedia.org/wiki/Immediately-invoked_function_expression — XCS, May 16 '16 at 14:46
It’s for the same reason `{a: 1, b: 2}[a]` doesn’t work. Here, `{`…`}` is a block, not an object. It’s because the `{` is interpreted as a statement, not an expression. `({a: 1, b: 2})[a]` or `({a: 1, b: 2}[a])` do work, because `(`…`)` forces the `{`…`}` to be interpreted as an expression (object). It’s exactly the same for functions, because there are function statements and function expressions. Why? Because that’s how JavaScript was designed. — Sebastian Simon, May 16 '16 at 14:49
If you understand why the first one works, then you'll understand why the second works. The second should be no more mysterious than the first. — , May 16 '16 at 15:00
...and [this answer](http://stackoverflow.com/a/1634321/1106925) to a different question gives a pretty detailed explanation of why either work and the last one doesn't. — , May 16 '16 at 15:04

score 2 · Accepted Answer · answered May 16 '16 at 14:47

2

(function() {})()

and

(function() {}())

are equivalent.

To call second example you can include + operator before function

+function(){ return 1 }()

See Immediately-Invoked Function Expression (IIFE)

answered May 16 '16 at 14:47

guest271314

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There are better operators to use than `+`. The reason is that the `+` is overloaded as a binary operator, and so if you accidentally forget a semicolon after the previous expression, it will try to add that previous expression to the function. It's basically the same problem that people have with using `()`. If you're going to use an alternate operator to force an expression, use a unary operator, like `!` for example. – May 16 '16 at 15:06
@squint _"There are better operators to use than `+`"_ , _"If you're going to use an alternate operator to force an expression, use a unary operator, like `!` for example."_ Yes, though not for this particular case; `!function(){ return 1 }()` returns `false` instead of `1` – guest271314 May 17 '16 at 00:33
@squint Alternatively, `;null, function(){ return 1 }();` – guest271314 May 17 '16 at 00:52
If the return value is being captured, there's no need for any operator to make the function a valid expression. But yeah, anything to avoid unexpected evaluation of the operator is better. I've used `void` at times in the past. – May 17 '16 at 00:56

score 2 · Answer 2 · answered May 16 '16 at 14:55

The phrase IIFE is a better term for these functions .. Immediately Invoked Function Expressions.

As for why they are the same: The outer parens () simply make an expression and the () together do the invocation.

(function(){ return 1;})()
is the same as:
(function(){ return 1;}())


(function(){ return 1;})()
becomes
(functionexpression)()
becomes
functionexpression()

and

(function(){ return 1;}())
becomes
(functionExpression())
becomes   
functionExpression()

for the same reason that

(3)+2 is the same as ((3)+2).

EDIT

function(){ return 1; }()

Does NOT work because a function statement is different from a function expression. Function statements cannot be immediately invoked.

Why Self-Executing Anonymous works?

2 Answers2