Sum with an arbitrary amount of brackets

Write function sum that would work like this:

          sum(1)(2) == 3; // 1 + 2
sum(1)(2)(3) == 6; // 1 + 2 + 3
sum(5)(-1)(2) == 6
sum(6)(-1)(-2)(-3) == 0
sum(0)(1)(2)(3)(4)(5) == 15
        

P.S. Hint: you may need to setup custom object to primitive conversion for your function.

Open a sandbox with tests.

For the whole thing to work anyhow, the result of sum must be function.
That function must keep in memory the current value between calls.
According to the task, the function must become the number when used in ==. Functions are objects, so the conversion happens as described in the chapter Object to primitive conversion, and we can provide our own method that returns the number.

Now the code:

function sum(a) {

  let currentSum = a;

  function f(b) {
    currentSum += b;
    return f;
  }

  f.toString = function() {
    return currentSum;
  };

  return f;
}

alert( sum(1)(2) ); // 3
alert( sum(5)(-1)(2) ); // 6
alert( sum(6)(-1)(-2)(-3) ); // 0
alert( sum(0)(1)(2)(3)(4)(5) ); // 15

Please note that the sum function actually works only once. It returns function f.

Then, on each subsequent call, f adds its parameter to the sum currentSum, and returns itself.

There is no recursion in the last line of f.

Here is what recursion looks like:

          function f(b) {
  currentSum += b;
  return f(); // <-- recursive call
}
        

And in our case, we just return the function, without calling it:

          function f(b) {
  currentSum += b;
  return f; // <-- does not call itself, returns itself
}
        

This f will be used in the next call, again return itself, as many times as needed. Then, when used as a number or a string – the toString returns the currentSum. We could also use Symbol.toPrimitive or valueOf here for the conversion.

Open the solution with tests in a sandbox.