Variadic curried sum function

Question

I need a js sum function to work like this:

sum(1)(2) = 3
sum(1)(2)(3) = 6
sum(1)(2)(3)(4) = 10 
etc.

I heard it can\'t be done. But heard

Answer 1

Not sure if I understood what you want, but

function sum(n) {
  var v = function(x) {
    return sum(n + x);
  };

  v.valueOf = v.toString = function() {
    return n;
  };

  return v;
}

console.log(+sum(1)(2)(3)(4));

JsFiddle

Answer 2

Here is another functional way using an iterative process

const sum = (num, acc = 0) => {
    if (!num) return acc;
    return x => sum(x, acc + num)
}

sum(1)(2)(3)()

and one-line

const sum = (num, acc = 0) => !num ? acc : x => sum(x, acc + num)

sum(1)(2)(3)()

Answer 3

Another slightly shorter approach:

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

Usable as:

console.log(
  sum(1)(2)(),
  sum(3)(4)(5)()
);

Answer 4

Here is a solution that uses ES6 and toString, similar to @Vemba

function add(a) {
  let curry = (b) => {
    a += b
    return curry
  }
  curry.toString = () => a
  return curry
}

console.log(add(1))
console.log(add(1)(2))
console.log(add(1)(2)(3))
console.log(add(1)(2)(3)(4))

Answer 5

New ES6 way and is concise.

You have to pass empty () at the end when you want to terminate the call and get the final value.

const sum= x => y => (y !== undefined) ? sum(x + y) : x;

call it like this -

sum(10)(30)(45)();

Answer 6

function add(a) {
    let curry = (b) => {
        a += b
        return curry;
    }
    curry[Symbol.toPrimitive] = (hint) => {
        return a;
    }
    return curry
}

console.log(+add(1)(2)(3)(4)(5));        // 15
console.log(+add(6)(6)(6));              // 18
console.log(+add(7)(0));                 // 7
console.log(+add(0));                    // 0