Question

when I code:

var a = function() {     alert('44')     return function(){alert(33)} }()(); 

is this expression evaluated in the following order?

1. define the function;
2. pass its reference pointer to a
3. a() is invoked
4. return in a a new function pointer
5. a() is invoked again

and if so why do I have a syntax error if I do:

function() {     alert('44')     return function(){alert(33)} }(); 

the interpreter wants a left operand first…

but this syntax works:

(   function()   {     alert('44')     return function(){alert(33)}   };  )() 

the outer parenthesis what does meaning???

Thanks

Answer 1

It’s the syntax of the language. If you want to in-place execute an anonymous function, you must enclose it in parens.

JS has these edge cases where the syntax is weirder than you expect. Take for example, evaling a string that has a JSON doesn’t work unless it’s wrapped with parens.

// Wrong eval('{ ... }'); // Right eval('({ ... })'); 

It’s the syntax of the language.

That said, I think (and this is strictly IMHO), the steps you’ve outlined are not accurate.

1. Function is defined and invoked. alert(’44’); happens as a result.
2. The function returns another function which is also invoked. alert(’33’); happens.
3. The innermost function doesn’t return anything, so a is effectively undefined. typeof a returns ‘undefined’.