javascript - understanding crockford's function that returns a function with a variable value -

i watching douglas crockford videos , gave following exercise :

write function, when passed variable, returns function if called , returns value of variable.

so wrote following function :

function funcky(o) {     return function send(o){ // notice o in send       return o;     }   }    var x = funcky(3);    console.log(x()); // undefined why ??  

notice o in send. have been programming javascript while , still don't understand why undefined ??

crockfords solution follows :

  function funcky(o) {     return function send(){       return o;     }   }    var x = funcky(3);    console.log(x()); // 3 . 

now how come solution works , mine does't ? don't see of difference in solution , nothing wrong see. can explain please ?

this has scope of o. when write:

return function send(o){ // notice o in send   return o; } 

the scope of o local function send. but, if write:

return function send(){   return o; } 

the scope of o not local function send, local scope of funcky.

so, when write function send(o){/*...*/} happening o becomes argument, , need called this: funcky()(10), want able funcky(10)().

edit:

for more information variable scope in javascript, please see this detailed answer on so.