CODEWITHSUNDEEP
javascript
user4841711
user4841711

Reputation:

Why does console.log paste the function instead of the value?

So I'm following Douglas Crockfords lectures on JS. One of his exercises is "write a function that takes one argument, and returns a function that returns the argument". The next question is "write a function that adds from two invocations".

I got the answer to both, what I don't get is why one doesn't post to the console and the other does. Here's what I mean. This is my answer to the first question:

function retFunc(x){
  return function(){
    return x;
  };
}

console.log(retFunc(4));

This posts the following to the console:

function(){
   return x;
}

However if I change it to this, it'll post 4 to the console (which is what I wanted):

var idf = retFunc(4);
console.log(idf());

Now this is my answer to the second question:

function addf(x){
  return function(y){
    return x+y;
  };
}

console.log(addf(3)(4));

And this works, it spits out 7 to the console. My question is, why does the first one not work? And this one does? Both are functions that return another function, that finally returns a value.

Upvotes: 0

Views: 1299

Answers (3)

behzad besharati
behzad besharati

Reputation: 6920

Instead of console.log(retFunc(4)); try console.log(retFunc(4)()); You have two nested functions. so you should call inner fucntion too.

Upvotes: 0

Jean-Bernard Pellerin
Jean-Bernard Pellerin

Reputation: 12670

The difference is that in your first case you are not calling the returned function, but in the second one you are

console.log(retFunc(4));

var idf = retFunc(4);  
console.log(idf()); // <-- note the parentheses.

Upvotes: 1

Sergio Tulentsev
Sergio Tulentsev

Reputation: 230521

why does the first one not work?

First one does work, exactly as it should. You call a function which returns you a function, but you don't call that second function. You print it instead.

Upvotes: 3

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