## 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.

- 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.