JavaScript immediate function invocation - why are parenthesis needed around the function?

Question

Why does a function declaration need to be wrapped in parentheses to be immediately invoked? I'm curious as to how the interpreter reads the immediately invoked function when wrapped in parentheses.

I.e.

Why must I do this...

(function() { 
   // Logic 
 })();

and not this...

function() { 
   // Logic 
 }();

The second is OK where an expression is allowed, e.g. `var x = function(){return 'foo'}();` — RobG, Feb 18 '13 at 03:36

David G · Accepted Answer · 2013-02-18T00:47:25.397

6

When a function is wrapped in parenthesis it's parsed as an expression - a function expression. Otherwise without them it's parsed as a function declaration. A function declaration requires a name which it sees you have not given it, which in turn causes a syntax error. Moreover, you can't apply () inline to a function declaration in order to call it. The empty parenthesis is a syntax error, but a non-empty parenthesis is an expression which will be evaluated separately from the function.

edited Feb 18 '13 at 00:47

answered Feb 18 '13 at 00:39

David G

94,763
41
167
253

2

`function foo(){}(bar);` is valid though, but it does not do what one might intend. – Felix Kling Feb 18 '13 at 00:43
Wonderful answer that makes total sense, thank you. – contactmatt Feb 18 '13 at 17:39
@FelixKling, what would your example do? – contactmatt Feb 18 '13 at 17:41
@contactmatt: It would define a function `foo` and then *evaluate* the expression inside the parenthesis. In `(bar)`, the parenthesis are the grouping operator (as in `(5 + 3) * 2`), they are not calling parenthesis. – Felix Kling Feb 18 '13 at 18:04

JavaScript immediate function invocation - why are parenthesis needed around the function?

1 Answers1