How can I write a function for mod(3)(9)? [duplicate]

Question

I was in an interview and I got all the questions right except for this one.

The first question leading up to it was how do you write a function for mod(3,9) so that it returns 0.

Ok, easy:

function mod(a,b){
    return b%a;
}

After that was how do you write the function mod(3)(9) so that it returns 0?

I was stumped...

Answer 1

You write a function that returns a closure.

function mod(a) {
  return function(b) {
    return b % a;
  }
}

alert(mod(3)(9));

The alert expression is short for:

var tempfun = mod(3);
alert(tempfun(9));

When you call mod(3), it returns a function that takes an argument b, and performs the modulus with the saved binding of a, which contains 3. We can then use that in the same way we'd use any other function: we can assign it to a variable and then call that as a function, or we can just call it directly by putting another pair of parentheses after it.

Answer 2

First of all, look closely at the usage of this construction:

mod(3)(9);

You can split it into two steps:

var fn = mod(3);
fn(9);

From here is it obvious that mod(3) alone must return a new function so that later it could be invoked again. This new function needs to preserve the value passed in with the first invocation. This is the key part: you should store that value in the closure (well it's stored automatically due to nature of closures):

function mod(x) {
    return function(y) {
        return y % x;
    };
}

Here comes good illustration of the term "closure". If someone asks you (on interview, for example) you may say: closure is the function with the scope it was originally created in. So in the function above, new inner function always has internal access to the outer function parameter x.

Answer 3

You can use a function that returns another function

function mod(a) {
    return function(b) {
        return b % a;
    }
}