Writing a javascript function in curry, which can be called an arbitrary number of times, which returns a value on the most recent function call

Question

I am currently working on a programming problem when I ask that I create a javascript function that can be called this way.

add(1) // 1 add(1)(2) // 3 add(1)(2)(3); // 6 add(1)(2)(3)(4); // 10 add(1)(2)(3)(4)(5); // 15 

I am having trouble figuring out how to return a value on the last call.

For example, in order for add(1)(2) work, then add(1) must return a function, but in accordance with add(1) instructions, when the caller itself returns 1 .

I suggest that you can overcome this to find out how many times the add function is called, but I cannot think of a way to achieve it. Does anyone have any hints that can point me in the right direction?

I read these two articles ( 1 , 2 ) on the currying function, and I understand them, but I'm not sure how to do currying when dealing with a variable number of arguments.

Answer 1

This is not possible, use the value of ().

 function add(initNum) { var sum = initNum; var callback = function (num) { sum += num; return callback; }; callback.valueOf = function () { return sum; }; return callback; }; console.log(add(1)(2)==3); //true console.log(add(1)(1)+1); //3 console.log(add(1)(2)(3).valueOf()); //6 

Answer 2

It is not possible to execute a variational function with an unknown number of arguments.

Where add is a variational function, you can do something like

 var add5 = curryN(add, 5); add5(1)(2)(3)(4)(5); //=> 15 var add3 = curryN(add, 3); add3(1)(2)(3); //=> 6 

There simply cannot be avoided, because the curries function will continue to return the function until the last argument is received, after which the calculation will be started.

---

The only other option is to create a way to "short circuit" the arguments and notify the function that the arguments pass. This will require something like

 var xadd = curryUntilUndefined(add); xadd(1)(2)(3)(4)(5)(undefined); //=> 15 

Here, undefined signals the end of the variational arguments. I really do not recommend this because of other problems that it may create for you. Not to mention that it is not particularly pleasant to watch.