Node.js vs Javascript Closure

Question

Node.js vs Javascript Closure

I'm working my way through the Eloquent JavaScript Book and in it there is the following code:

function createFunction(){
  var local = 100;
  return function(){return local;};
}

When I run this via the node console (run node from command prompt) by calling createFunction(), I get [Function] as a returned value. However, according to the book I should get 100.

So my two questions: Why is this? and Second, is running these little examples in the node console a bad idea for testing JS code?

javascript
node.js
closures

Answer 1

You need to call the response of createFunction().

createFunction()();

The first invocation (()) calls createFunction() and returns the inner function, which the second invocation executes and returns the local variable which was closed over.

Running small examples in a node console (or any other) is fine, so long as you know the environment, e.g. a browser's console is generally eval()'d, which can create side effects, such as how delete can apparently delete variables, not just object properties.

Answer 2

You get 100 by invoking the return value of createFunction, which is itself a function.

createFunction()();

...or perhaps more clearly...

var new_func = createFunction();

new_func();

function createFunction(){
  var local = 100;

 //  v---v-----------------------v return a function from createFunction
  return function(){return local;};
}

  //  v------- the returned function is assigned to the new_func variable
var new_func = createFunction();

 //   v------- the returned function is invoked
new_func();

Answer 3

For those that have a similar problem, I completely missed the double () so the call looks like createFunction()().