How come
function(){ alert("test123");}()
produces SyntaxError: Unexpected token (
while
!function(){ alert("test123");}()
alerts "test123"
?
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Alon
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4http://benalman.com/news/2010/11/immediately-invoked-function-expression/ – zerkms Aug 27 '13 at 08:13
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@zerkms You should probably post that as an answer, as it actually explains the heart of the matter quite well. – Kippie Aug 27 '13 at 08:15
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1c'mon, there _has_ to be a duplicate... – John Dvorak Aug 27 '13 at 08:15
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1@Kippie: link-only answers aren't welcomed here :-) – zerkms Aug 27 '13 at 08:15
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@zerkms what about heavy quoting? – John Dvorak Aug 27 '13 at 08:16
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For even more fun. Anything that must be evaluated in an expression also does this: `-function(){}()`, `+function(){}()`,`0|function(){}()` etc – slebetman Aug 27 '13 at 08:22
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The most evil has got to be `0?0:function(){}()` – slebetman Aug 27 '13 at 08:25
1 Answers
3
It's because by adding ! sign you convert the declaration into an expression and invoke it immediately.
By enclosing your function into brackets you will make first example working without errors:
(function(){ alert("test123");})()
To make it clearer you can think about first expression as something like:
if (false || !function(){ return false; }())
And as @zerkms noticed there is a complete explanation of Immediately-invoking functions.
Anatolii Gabuza
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too bad I can't write a proper answer, but... "identifier" is not the correct term. Rather the distinction is between a definition and an expression. The former cannot be immediately invoked. – John Dvorak Aug 27 '13 at 08:18
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1Right track but wrong explanation. The ! converts the declaration into an expression. – slebetman Aug 27 '13 at 08:18
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I can define a function and _not_ invoke it: `!function(){alert("will not be called")}` – John Dvorak Aug 27 '13 at 08:20