Variadic curried sum function

I need a js sum function to work l开发者_运维问答ike 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 that if adding + in front of sum can be done. Like +sum(1)(2)(3)(4).

Any ideas of how to do this?

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

This is an example of using empty brackets in the last call as a close key (from my last interview):

sum(1)(4)(66)(35)(0)()

function sum(firstNumber) {
    let accumulator = firstNumber;
    return function adder(nextNumber) {
        if (nextNumber === undefined) {
            return accumulator;
        } 
    
        accumulator += nextNumber;
        return adder;
    }
}

console.log(sum(1)(4)(66)(35)(0)());

I'm posting this revision as its own post since I apparently don't have enough reputation yet to just leave it as a comment. This is a revision of @Rafael 's excellent solution.

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

console.log( +sum(1)(2)(3)(4) ); //10

I didn't see a reason to keep the v.toString bit, as it didn't seem necessary. If I erred in doing so, please let me know in the comments why v.toString is required (it passed my tests fine without it). Converted the rest of the anonymous functions to arrow functions for ease of reading.

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)();

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))

Another slightly shorter approach:

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

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

Here's a solution with a generic variadic curry function in ES6 Javascript, with the caveat that a final () is needed to invoke the arguments:

const curry = (f) =>
   (...args) => args.length? curry(f.bind(0, ...args)): f();

const sum = (...values) => values.reduce((total, current) => total + current, 0)
curry(sum)(2)(2)(1)() == 5 // true

Here's another one that doesn't need (), using valueOf as in @rafael's answer. I feel like using valueOf in this way (or perhaps at all) is very confusing to people reading your code, but each to their own.

The toString in that answer is unnecessary. Internally, when javascript performs a type coersion it always calls valueOf() before calling toString().

// invokes a function if it is used as a value
const autoInvoke = (f) => Object.assign(f, { valueOf: f } );

const curry = autoInvoke((f) =>
   (...args) => args.length? autoInvoke(curry(f.bind(0, ...args))): f());

const sum = (...values) => values.reduce((total, current) => total + current, 0)
curry(sum)(2)(2)(1) + 0 == 5 // true

Try this

function sum (...args) {
  return Object.assign(
    sum.bind(null, ...args),
    { valueOf: () => args.reduce((a, c) => a + c, 0) }
  )
}

console.log(+sum(1)(2)(3,2,1)(16))

Here you can see a medium post about carried functions with unlimited arguments

https://medium.com/@seenarowhani95/infinite-currying-in-javascript-38400827e581

Try this, this is more flexible to handle any type of input. You can pass any number of params and any number of paranthesis.

function add(...args) {
    function b(...arg) {
        if (arg.length > 0) {
            return add(...[...arg, ...args]);
        }

        return [...args, ...arg].reduce((prev,next)=>prev + next);
    }

    b.toString = function() {
        return [...args].reduce((prev,next)=>prev + next);
    }

    return b;
}

// Examples
console.log(add(1)(2)(3, 3)());
console.log(+add(1)(2)(3)); // 6
console.log(+add(1)(2, 3)(4)(5, 6, 7)); // 28
console.log(+add(2, 3, 4, 5)(1)());    // 15

Here's a more generic solution that would work for non-unary params as well:

const sum = function (...args) {
  let total = args.reduce((acc, arg) => acc+arg, 0)
  function add (...args2) {
    if (args2.length) {
      total = args2.reduce((acc, arg) => acc+arg, total)
      return add
    }
    return total
  }

  return add
}

document.write( sum(1)(2)() , '<br/>') // with unary params
document.write( sum(1,2)() , '<br/>') // with binary params
document.write( sum(1)(2)(3)() , '<br/>') // with unary params
document.write( sum(1)(2,3)() , '<br/>') // with binary params
document.write( sum(1)(2)(3)(4)() , '<br/>') // with unary params
document.write( sum(1)(2,3,4)() , '<br/>') // with ternary params

ES6 way to solve the infinite currying. Here the function sum will return the sum of all the numbers passed in the params:

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

sum(1)(2)(3)(4)(5)() // 15

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

Here is another functional way using an iterative process

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

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

and one-line

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

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

You can make use of the below function

function add(num){
   add.sum || (add.sum = 0) // make sure add.sum exists if not assign it to 0
   add.sum += num; // increment it
   return add.toString = add.valueOf = function(){ 
      var rtn = add.sum; // we save the value
      return add.sum = 0, rtn // return it before we reset add.sum to 0
   }, add; // return the function
}

Since functions are objects, we can add properties to it, which we are resetting when it's been accessed.

we can also use this easy way.

function sum(a) {
    return function(b){
      if(b) return sum(a+b);
      return a;
    }
}

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

To make sum(1) callable as sum(1)(2), it must return a function.

The function can be either called or converted to a number with valueOf.

function sum(a) {

   var sum = a;
   function f(b) {
       sum += b;
       return f;
    }
   f.toString = function() { return sum }
   return f
}

   function sum(a){
    let res = 0;
    function getarrSum(arr){
            return arr.reduce( (e, sum=0) =>  { sum += e ; return sum ;} )
     }

    function calculateSumPerArgument(arguments){
            let res = 0;
            if(arguments.length >0){

            for ( let i = 0 ; i < arguments.length ; i++){
                if(Array.isArray(arguments[i])){
                    res += getarrSum( arguments[i]);
                }
                else{
                  res += arguments[i];
                }
             }
          }
            return res;
     }
    res += calculateSumPerArgument(arguments);



    return function f(b){
        if(b == undefined){
            return res;
        }
        else{
            res += calculateSumPerArgument(arguments);
            return f;
        }
    }

}

let add = (a) => {
  let sum = a;
  funct = function(b) {
    sum += b;
    return funct;
  };

  Object.defineProperty(funct, 'valueOf', {
    value: function() {
      return sum;
    }
  });
  return funct;
};


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

After looking over some of the other solutions on here, I would like to provide my two solutions to this problem.

Currying two items using ES6:

const sum = x => y => (y !== undefined ) ? +x + +y : +x
sum(2)(2) // 4

Here we are specifying two parameters, if the second one doesnt exist we just return the first parameter.

For three or more items, it gets a bit trickier; here is my solution. For any additional parameters you can add them in as a third

const sum = x => (y=0) => (...z) => +x + +y + +z.reduce((prev,curr)=>prev+curr,0)
sum(2)()()//2
sum(2)(2)()//4
sum(2)(2)(2)//6
sum(2)(2)(2,2)//8

I hope this helped someone