总和函数currying

Question

I'm trying to write a sum function that does the following:

sum(1)(2)(3) => returns 6

However, I am having hard time with my solution. I know i'm making a silly mistake, can someone point me in the right direction?

My Implementation:

 function add(args) { let sum = args[0]; let func = function(...args2) { if (!args2[0]) return sum; sum += args2[0]; return func; } return func; } add(1)(2)(3); 

Additionally, can I write a generic function that does the following? add(1)(2)(3) or add (1)(2)(3) () => 6

Answer 1

To have an arbitrary amount of calls, all which take a number, you'd need the return value to behave both as a function and a number depending. So that:

const five = add(2)(3);
console.log(five(10)); // behaves like a function
console.log(five + 10); // behaves like a number

I can't think of anyway to do that (or if there's a good reason one should do that) other than what I'd call a hack.

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With that said, for fun I was able to do the following by abusing valueOf() :

 const add = (num1) => { const func = (num2) => add(num1 + num2); func.valueOf = () => num1; return func; } console.log(add(1)(2)(3)); // *logs function* [output varies by environment] console.log(add(1)(2)(3) + 10); // 16 console.log(add(1)(2) + 10); // 13 console.log(add(1) + 10); // 11 console.log(add(1)(2)(3) == 6); // true console.log(add(1)(2)(3) === 6); // false console.log(typeof add(1)(2)(3)); // function console.log(typeof (add(1)(2) + 3)); // number 

But that's obviously not very kosher.

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Edit: Switched from using toString() to valueOf() per @PatrickRoberts's comment .

Answer 2

Try this, you omitted the spread operator

 function add(...args) { let sum = args[0]; let func = function(...args2) { if(args2[0] === undefined) return sum; sum += args2[0]; return func; } return func; } console.log(add(1)(2)(3)()); 

cheers!!

Answer 3

Recursion anyone? 😂

function add(n) {
  return function(m) {
    if (isNaN(m)) return n;
    return add(n + m);
  };
}

// add(1)(2)(3)() => 6
// add(1)(2)(3)("")() => "6"
// add(1)("2")(3)() => "123"

// functional one liner
const add = n => m => isNaN(m) ? n : add(n + m);

Answer 4

function sum(x){
   return function (y){
      return function(z){
        return x+y+z;
     }
   }
}

You can call like : sum(1)(2)(3)