how to write a sum function with infinite number of arguments, using currying in javascript?

Question

I have tried writing the below code to find sum of 'n' numbers using sum function. I am getting the correct response in output . But i am unable to return that using sum function, as i always have to return a function, which is required for curried effect.

Please help. Thanks in advance.

 var output = 0, chain; function sum() { var args = Array.prototype.slice.call(arguments); output += args.reduce(function(a, b) { return a + b; }); sumCurried = sum.bind(output); sumCurried.val = function() { return output; } return sumCurried; } debugger; document.getElementById('demo').innerHTML = sum(1, 2)(3)(4); // document.getElementById('demo').innerHTML = sum(1)(3)(4); 

 <p id='demo'></p> 

enter code here

Answer 1

You can add a stop condition to the curried function, for example - if the function is called without an argument return the output:

 var output = 0, chain; function sum() { var args = Array.prototype.slice.call(arguments); if(args.length === 0) { return output; } output += args.reduce(function(a, b) { return a + b; }); sumCurried = sum.bind(output); return sumCurried; } console.log(sum(1, 2)(3)(4)()); 

 <p id='demo'></p> 

The returned curry function has a val property, which is a function that returns the current value:

 var output = 0, chain; function sum() { var args = Array.prototype.slice.call(arguments); output += args.reduce(function(a, b) { return a + b; }); sumCurried = sum.bind(output); sumCurried.val = function() { return output; } return sumCurried; } console.log(sum(1, 2)(3)(4).val()); 

 <p id='demo'></p> 

Answer 2

Why would you use currying at all? However, here is a shorter version:

 const sum = (...args) => {
   const func = (...s)=> sum(...args,...s);
   func.value = args.reduce((a,b)=>a+b,0);
   return func;
 };

 //usable as
sum(1,2).value,
sum(1,1)(1).value,
sum(1,1)(1,1)(1,1).value

And you always need to end the currying chain. However, it can be shortified:

  func.valueOf = ()=> args.reduce((a,b)=>a+b,0);
  //( instead of func.value = ... )

So when called you can do:

  +sum(1,2,3)
 +sum(1)(1)(1)