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

Drop the semicolon after the function definition.

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

Above should work.

DEMO Page: https://jsfiddle.net/e7ooeq6m/

I have discussed this kind of pattern in this post:

jQuery and $ questions

EDIT:

If you look at ECMA script specification, there are 3 ways you can define a function. (Page 98, Section 13 Function Definition)

1. Using Function constructor

var sum = new Function('a','b', 'return a + b;');
alert(sum(10, 20)); //alerts 30

2. Using Function declaration.

function sum(a, b)
{
    return a + b;
}

alert(sum(10, 10)); //Alerts 20;

3. Function Expression

var sum = function(a, b) { return a + b; }

alert(sum(5, 5)); // alerts 10

So you may ask, what's the difference between declaration and expression?

From ECMA Script specification:

FunctionDeclaration : function Identifier ( FormalParameterListopt ){ FunctionBody }

FunctionExpression : function Identifieropt ( FormalParameterListopt ){ FunctionBody }

If you notice, 'identifier' is optional for function expression. And when you don't give an identifier, you create an anonymous function. It doesn't mean that you can't specify an identifier.

This means following is valid.

var sum = function mySum(a, b) { return a + b; }

Important point to note is that you can use 'mySum' only inside the mySum function body, not outside. See following example:

var test1 = function test2() { alert(typeof test2); }

alert(typeof(test2)); //alerts 'undefined', surprise! 

test1(); //alerts 'function' because test2 is a function.

Live Demo

Compare this to

 function test1() { alert(typeof test1) };

 alert(typeof test1); //alerts 'function'

 test1(); //alerts 'function'

---

Armed with this knowledge, let's try to analyze your code.

When you have code like,

    function(msg) { alert(msg); }

You created a function expression. And you can execute this function expression by wrapping it inside parenthesis.

    (function(msg) { alert(msg); })('SO'); //alerts SO.

It's called a self-invoked function.

What you are doing when you call (function(){}) is returning a function object. When you append () to it, it is invoked and anything in the body is executed. The ; denotes the end of the statement, that's why the 2nd invocation fails.

One thing I found confusing is that the "()" are grouping operators.

Here is your basic declared function.

Ex. 1:

var message = 'SO';

function foo(msg) {
    alert(msg);
}

foo(message);

Functions are objects, and can be grouped. So let's throw parens around the function.

Ex. 2:

var message = 'SO';

function foo(msg) {  //declares foo
    alert(msg);
}

(foo)(message);     // calls foo

Now instead of declaring and right-away calling the same function, we can use basic substitution to declare it as we call it.

Ex. 3.

var message = 'SO';

(function foo(msg) {
    alert(msg);
})(message);          // declares & calls foo

Finally, we don't have a need for that extra foo because we're not using the name to call it! Functions can be anonymous.

Ex. 4.

var message = 'SO';

(function (msg) {   // remove unnecessary reference to foo
    alert(msg);
})(message);

To answer your question, refer back to Example 2. Your first line declares some nameless function and groups it, but does not call it. The second line groups a string. Both do nothing. (Vincent's first example.)

(function (msg){alert(msg)});  
('SO');                       // nothing.

(foo); 
(msg); //Still nothing.

But

(foo)
(msg); //works