How can I make var a = add(2)(3); //5 work?

Question

I want to make this syntax possible:

var a = add(2)(3); //5

based on what I read at http://dmitry.baranovskiy.com/post/31797647

I've got no clue how to make it possible.

Answer 1

You need add to be a function that takes an argument and returns a function that takes an argument that adds the argument to add and itself.

var add = function(x) {
    return function(y) { return x + y; };
}

Answer 2

function add(x) {
    return function(y) {
        return x + y;
    };
}

Ah, the beauty of JavaScript

This syntax is pretty neat as well

function add(x) {
    return function(y) {
        if (typeof y !== 'undefined') {
            x = x + y;
            return arguments.callee;
        } else {
            return x;
        }
    };
}
add(1)(2)(3)(); //6
add(1)(1)(1)(1)(1)(1)(); //6

Answer 3

It's about JS curring and a little strict with valueOf:

function add(n){
  var addNext = function(x) {
    return add(n + x);
  };

  addNext.valueOf = function() {
    return n;
  };

  return addNext;
}

console.log(add(1)(2)(3)==6);//true
console.log(add(1)(2)(3)(4)==10);//true

It works like a charm with an unlimited adding chain!!

Answer 4

function add(x){
  return function(y){
    return x+y
  }
}

First-class functions and closures do the job.

Answer 5

function add(n) {
  sum = n;
  const proxy = new Proxy(function a () {}, {
    get (obj, key) {
      return () => sum;
    },
    apply (receiver, ...args) {
      sum += args[1][0];
      return proxy;
    },
  });
  return proxy
}

Works for everything and doesn't need the final () at the end of the function like some other solutions.

console.log(add(1)(2)(3)(10));    // 16
console.log(add(10)(10));         // 20

Answer 6

try this will help you in two ways add(2)(3) and add(2,3)

1.)

 function add(a){ return function (b){return a+b;} }

    add(2)(3) // 5

2.)

function add(a,b){
        var ddd = function (b){return a+b;};
        if(typeof b =='undefined'){
            return ddd;
        }else{
            return ddd(b);
        }
    }

add(2)(3) // 5
add(2,3) // 5

Answer 7

ES6 syntax makes this nice and simple:

const add = (a, b) => a + b;

console.log(add(2, 5)); 
// output: 7

const add2 = a => b => a + b;

console.log(add2(2)(5));
// output: 7

Answer 8

Arrow functions undoubtedly make it pretty simple to get the required result:

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

console.log(Sum(3)(4)(2)(5)()); //14

console.log(Sum(3)(4)(1)()); //8

Answer 9

This is a generalized solution which will solve add(2,3)(), add(2)(3)() or any combination like add(2,1,3)(1)(1)(2,3)(4)(4,1,1)(). Please note that few security checks are not done and it can be optimized further.

function add() {
 var total = 0;

 function sum(){
  if( arguments.length ){
   var arr = Array.prototype.slice.call(arguments).sort();
   total = total + arrayAdder(arr);
   return sum;
  }
  else{
   return total;
  }
 }

 if(arguments.length) {
  var arr1 = Array.prototype.slice.call(arguments).sort();
  var mytotal = arrayAdder(arr1);
  return sum(mytotal);
 }else{
  return sum();
 }

 function arrayAdder(arr){
  var x = 0;
  for (var i = 0; i < arr.length; i++) {
   x = x + arr[i];
  };
  return x;
 }
}
add(2,3)(1)(1)(1,2,3)();

Answer 10

This will handle both

add(2,3) // 5

or

add(2)(3) // 5

This is an ES6 curry example...

const add = (a, b) => (b || b === 0) ? a + b : (b) => a + b;

Answer 11

This is concept of currying in JS.
Solution for your question is:

function add(a) {
  return function(b) {
    return a + b;
  };
}

This can be also achieved using arrow function:

let add = a => b => a + b;

solution for add(1)(2)(5)(4)........(n)(); Using Recursion

function add(a) {
  return function(b){
    return b ? add(a + b) : a;
  }
}

Using ES6 Arrow function Syntax:

let add = a => b => b ? add(a + b) : a;

Answer 12

in addition to what's already said, here's a solution with generic currying (based on http://github.com/sstephenson/prototype/blob/master/src/lang/function.js#L180)

Function.prototype.curry = function() {
    if (!arguments.length) return this;
    var __method = this, args = [].slice.call(arguments, 0);
    return function() {
      return __method.apply(this, [].concat(
        [].slice.call(args, 0),
        [].slice.call(arguments, 0)));
   }
}


add = function(x) {
    return (function (x, y) { return x + y }).curry(x)
}

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

Answer 13

Concept of CLOSURES can be used in this case.
The function "add" returns another function. The function being returned can access the variable in the parent scope (in this case variable a).

function add(a){

    return function(b){
        console.log(a + b);
    }

}


add(2)(3);

Here is a link to understand closures http://www.w3schools.com/js/js_function_closures.asp

Answer 14

With ES6 spread ... operator and .reduce function. With that variant you will get chaining syntax but last call () is required here because function is always returned:

function add(...args) {
    if (!args.length) return 0;
    const result = args.reduce((accumulator, value) => accumulator + value, 0);
    const sum = (...innerArgs) => {
        if (innerArgs.length === 0) return result;
        return add(...args, ...innerArgs);    
    };
    return sum;
}




// it's just for fiddle output
document.getElementById('output').innerHTML = `
<br><br>add() === 0: ${add() === 0 ? 'true' : 'false, res=' + add()}
<br><br>add(1)(2)() === 3: ${add(1)(2)() === 3 ? 'true' : 'false, res=' + add(1)(2)()}
<br><br>add(1,2)() === 3: ${add(1,2)() === 3 ? 'true' : 'false, res=' + add(1,2)()}
<br><br>add(1)(1,1)() === 3: ${add(1)(1,1)() === 3 ? 'true' : 'false, res=' + add(1)(1,1)()}
<br><br>add(2,3)(1)(1)(1,2,3)() === 13: ${add(2,3)(1)(1)(1,2,3)() === 13 ? 'true' : 'false, res=' + add(2,3)(1)(1)(1,2,3)()}
`;

<div id='output'></div>

Answer 15

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

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

Or (`${a} ${b}`) for strings.

Answer 16

can try this also:

let sum = a => b => b ? sum(a + b) :a
console.log(sum(10)(20)(1)(32)())   //63

Answer 17

const sum  = function (...a) {
    const getSum = d => {
        return d.reduce((i,j)=> i+j, 0);
    };

    a = getSum(a);
    return function (...b) {
        if (b.length) {
            return sum(a + getSum(b)); 
        }
        return a;
    }
};
console.log(sum(1)(2)(3)(4,5)(6)(8)())

Answer 18

function add(a, b){
 return a && b ? a+b : function(c){return a+c;}
}

console.log(add(2, 3));
console.log(add(2)(3));

Answer 19

This question has motivated so many answers already that my "two pennies worth" will surely not spoil things.

I was amazed by the multitude of approaches and variations that I tried to put "my favourite" features, i. e. the ones that I would like to find in such a currying function together, using some ES6 notation:

const add=(...n)=>{
 const vsum=(a,c)=>a+c;
 n=n.reduce(vsum,0);
 const fn=(...x)=>add(n+x.reduce(vsum,0));
 fn.toString=()=>n; 
 return fn;
}

let w=add(2,1); // = 3
console.log(w()) // 3
console.log(w); // 3
console.log(w(6)(2,3)(4)); // 18
console.log(w(5,3)); // 11
console.log(add(2)-1); // 1
console.log(add()); // 0
console.log(add(5,7,9)(w)); // 24

.as-console-wrapper {max-height:100% !important; top:0%}

Basically, nothing in this recursively programmed function is new. But it does work with all possible combinations of arguments mentioned in any of the answers above and won't need an "empty arguments list" at the end.

You can use as many arguments in as many currying levels you want and the result will be another function that can be reused for the same purpose. I used a little "trick" to also get a numeric value "at the same time": I redefined the .toString() function of the inner function fn! This method will be called by Javascript whenever the function is used without an arguments list and "some value is expected". Technically it is a "hack" as it will not return a string but a number, but it will work in a way that is in most cases the "desired" way. Give it a spin!

Answer 20

Simple Recursion Solution for following use cases

add(); // 0

add(1)(2)(); //3

add(1)(2)(3)(); //6

function add(v1, sum = 0) {
  if (!v1) return sum;
  sum += v1
  return (v2) => add(v2, sum);
}

Answer 21

function add() { var sum = 0;

    function add() {
        for (var i=0; i<arguments.length; i++) {
            sum += Number(arguments[i]);
        }
        return add;
    }
    add.valueOf = function valueOf(){
        return parseInt(sum);
    };
    return add.apply(null,arguments);
}

// ...

console.log(add() + 0);               // 0
console.log(add(1) + 0);/*                 // 1
console.log(add(1,2) + 0);               // 3

Answer 22

function A(a){
  return function B(b){
      return a+b;
  }
}

I found a nice explanation for this type of method. It is known as Syntax of Closures

please refer this link Syntax of Closures

Answer 23

Simply we can write a function like this

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

    sum(2)(3)(4)//Output->9

Answer 24

Don't be complicated.

var add = (a)=>(b)=> b ? add(a+b) : a;
console.log(add(2)(3)()); // Output:5

it will work in the latest javascript (ES6), this is a recursion function.

Answer 25

Here we use concept of closure where all the functions called inside main function iter refer and udpate x as they have closure over it. no matter how long the loop goes , till last function , have access to x.

function iter(x){    
return function innfunc(y){
//if y is not undefined
if(y){
//closure over ancestor's x
x = y+x;
return innfunc;
}
else{
//closure over ancestor's x
return x;
    }
  }
}

iter(2)(3)(4)() //9 iter(1)(3)(4)(5)() //13

Answer 26

let multi = (a)=>{
 return (b)=>{
  return (c)=>{
   return a*b*c
  }
 }
}
multi (2)(3)(4) //24

let multi = (a)=> (b)=> (c)=> a*b*c;
multi (2)(3)(4) //24

Answer 27

we can do this work using closure.

    function add(param1){
      return function add1(param2){
      return param2 = param1 + param2;
    }
  }
  console.log(add(2)(3));//5

Answer 28

I came up with nice solution with closure, inner function have access to parent function's parameter access and store in its lexical scope, when ever we execute it, will get answer

    const Sum = function (a) {
        return function (b) {
            return b ? Sum(a + b) : a;
        }
    };

    Sum(1)(2)(3)(4)(5)(6)(7)() // result is 28
    Sum(3)(4)(5)() // result is 12
    Sum(12)(10)(20) // result is 42

enter image description here

Answer 29

You should go in for currying to call the function in the above format.

Ideally, a function which adds two numbers will be like,

let sum = function(a, b) {
  return a + b;
}

The same function can be transformed as,

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

console.log(sum(2)(3));

Let us understand how this works.

When you invoke sum(2), it returns

function(b) {
    return 2 + b;
}

when the returned function is further invoked with 3, b takes the value 3. The result 5 is returned.

More Detailed Explanation:

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

let func1 = sum(2);
console.log(func1);

let func2 = func1(3)
console.log(func2);

//the same result can be obtained in a single line

let func3 = sum(2)(3);
console.log(func3);

//try comparing the three functions and you will get more clarity.

Answer 30

This is a short solution:

const add = a => b => {
  if(!b) return a;
  return add(a + b);
}

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

Answer 31

// executor function
const sum = (a,b,c) => a+b+c;

// curried function
function curry(fn){
  return function curried(...args) {
    if(args.length >= fn.length){
      return fn.apply(this,args);
    } else {
      return function (...arg) {
        return curried(...arg, ...args)
      }
    }
  }
}

const curriedSum = curry(sum);
console.log(curriedSum(1,2,3));
console.log(curriedSum(1)(2)(3));
console.log(curriedSum(1)(2,3));

Answer 32

function add () {
    var args = Array.prototype.slice.call(arguments);
 
    var fn = function () {
        var arg_fn = Array.prototype.slice.call(arguments);
        return add.apply(null, args.concat(arg_fn));
    }
 
    fn.valueOf = function () {
        return args.reduce(function(a, b) {
            return a + b;
        })
    }
 
    return fn;
}

console.log(add(1));
console.log(add(1)(2));
console.log(add(1)(2)(5));

from http://www.cnblogs.com/coco1s/p/6509141.html

Answer 33

    let add = (a, b) => b === undefined ? add.bind(null, a) : a + b;
    console.log(add(10, 5)); //15
    console.log(add(10)(5)); //15

/* In the arrow function which returns a ternary expression, we explicitly check if the 2nd argument(b) is passed in or not b==="undefined" in both cases of add(a,b) and add(a)(b).

1. if "undefined" in the case of add(a)(b) it uses "add.bind(null,a)", the "bind" method is associated with all functions(functions are objects) which creates a new function that is returned by the arrow function and passes in:
   • "null" for the first argument "the context its binding to which allows it to default to the window object"
   • "a" the second argument to be returned with the new function.
   • so when the new function is called at this point add(a)>>>(b) "b" is no longer undefined so the ternary switches to a+b.
2. If not b==="undefined" in the case of add(a, b) the ternary switches to a+b */