Question

Why does

var spo = function(){
   var qq = function(){};
}

throw the error undefined is not a function when

var spo = function(){
   function qq(){};
}

does not?

---

More elaborate example throws the exact error: TypeError: undefined is not a function

node version v0.6.10 compiled on Ubuntu

spo = function(car){

        var q = 10;
        var s = 'fraggle';

        var qq = function(){
                console.log(s);
        }

        (function(){

                while(q){
                        console.log(q);
                        q--;
                }
                        qq.call();
                        car.call();
        })();


};


spo(function(){console.log('as intended');});

Answer 1

In this qq is assigned the function dynamically.

var spo = function(){
   // qq is undefined here
   var qq = function(){};
   // qq is defined here
}

On the other hand, in this case, qq is defined as a function which is visible everywhere in spo

var spo = function(){
   // qq is defined here
   function qq(){};
   // qq is defined here also
}

EDIT:
In your updated code the actual problem is visible, no semi colon after function which results in a wrong statement like given below

var qq = function(){}()(); // This is causing TypeError

Put a semi colon after function.

var qq = function(){
    console.log(s);
};  // you missed this semi colon

(function(){

     while(q){
         console.log(q);
         q--;
     }
     qq.call();
     car.call();
})();