How can I make var a = add(2)(3); //5 work?

Question

I want to make this syntax possible:

var a = add(2)(3); //5

based on what I read at http://dmitry.baranovskiy.com/post/31797647

I\

Answer 1

This will handle both

add(2,3) // 5

or

add(2)(3) // 5

This is an ES6 curry example...

const add = (a, b) => (b || b === 0) ? a + b : (b) => a + b;

Answer 2

    let add = (a, b) => b === undefined ? add.bind(null, a) : a + b;
    console.log(add(10, 5)); //15
    console.log(add(10)(5)); //15

/* In the arrow function which returns a ternary expression, we explicitly check if the 2nd argument(b) is passed in or not b==="undefined" in both cases of add(a,b) and add(a)(b).

1. if "undefined" in the case of add(a)(b) it uses "add.bind(null,a)", the "bind" method is associated with all functions(functions are objects) which creates a new function that is returned by the arrow function and passes in:
   • "null" for the first argument "the context its binding to which allows it to default to the window object"
   • "a" the second argument to be returned with the new function.
   • so when the new function is called at this point add(a)>>>(b) "b" is no longer undefined so the ternary switches to a+b.
2. If not b==="undefined" in the case of add(a, b) the ternary switches to a+b */

Answer 3

You need add to be a function that takes an argument and returns a function that takes an argument that adds the argument to add and itself.

var add = function(x) {
    return function(y) { return x + y; };
}

Answer 4

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

Ah, the beauty of JavaScript

This syntax is pretty neat as well

function add(x) {
    return function(y) {
        if (typeof y !== 'undefined') {
            x = x + y;
            return arguments.callee;
        } else {
            return x;
        }
    };
}
add(1)(2)(3)(); //6
add(1)(1)(1)(1)(1)(1)(); //6

Answer 5

function add(a, b){
 return a && b ? a+b : function(c){return a+c;}
}

console.log(add(2, 3));
console.log(add(2)(3));

Answer 6

This is concept of currying in JS.
Solution for your question is:

function add(a) {
  return function(b) {
    return a + b;
  };
}

This can be also achieved using arrow function:

let add = a => b => a + b;

solution for add(1)(2)(5)(4)........(n)(); Using Recursion

function add(a) {
  return function(b){
    return b ? add(a + b) : a;
  }
}

Using ES6 Arrow function Syntax:

let add = a => b => b ? add(a + b) : a;