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

function add() { var sum = 0;

    function add() {
        for (var i=0; i<arguments.length; i++) {
            sum += Number(arguments[i]);
        }
        return add;
    }
    add.valueOf = function valueOf(){
        return parseInt(sum);
    };
    return add.apply(null,arguments);
}

// ...

console.log(add() + 0);               // 0
console.log(add(1) + 0);/*                 // 1
console.log(add(1,2) + 0);               // 3

Answer 2

let total = 0;
const add = (n) => {
if (n) {
    total += n;
    return add;
 }
}

add(1)(2)(3);
console.log(total);

Is this wrong?

Answer 3

function add(n) {
  sum = n;
  const proxy = new Proxy(function a () {}, {
    get (obj, key) {
      return () => sum;
    },
    apply (receiver, ...args) {
      sum += args[1][0];
      return proxy;
    },
  });
  return proxy
}

Works for everything and doesn't need the final () at the end of the function like some other solutions.

console.log(add(1)(2)(3)(10));    // 16
console.log(add(10)(10));         // 20

Answer 4

Arrow functions undoubtedly make it pretty simple to get the required result:

const Sum = a => b => b ? Sum( a + b ) : a;

console.log(Sum(3)(4)(2)(5)()); //14

console.log(Sum(3)(4)(1)()); //8

Answer 5

Here we use concept of closure where all the functions called inside main function iter refer and udpate x as they have closure over it. no matter how long the loop goes , till last function , have access to x.

function iter(x){    
return function innfunc(y){
//if y is not undefined
if(y){
//closure over ancestor's x
x = y+x;
return innfunc;
}
else{
//closure over ancestor's x
return x;
    }
  }
}

iter(2)(3)(4)() //9 iter(1)(3)(4)(5)() //13

Answer 6

we can do this work using closure.

    function add(param1){
      return function add1(param2){
      return param2 = param1 + param2;
    }
  }
  console.log(add(2)(3));//5