Why do you need to invoke an anonymous function on the same line?

Question

I was reading some posts about closures and saw this everywhere, but there is no clear explanation how it works - everytime I was just told to use it...:

//          


        

Answer 1

examples without brackets:

void function (msg) { alert(msg); }
('SO');

(this is the only real use of void, afaik)

or

var a = function (msg) { alert(msg); }
('SO');

or

!function (msg) { alert(msg); }
('SO');

work as well. the void is causing the expression to evaluate, as well as the assignment and the bang. the last one works with ~, +, -, delete, typeof, some of the unary operators (void is one as well). not working are of couse ++, -- because of the requirement of a variable.

the line break is not necessary.