How come words2 and words are giving the same result? I know some currying but I can't understand this example below.

function curry(fn) {
  const arity = fn.length;
  return function $curry(...args) {
    if (args.length < arity) {
      return $curry.bind(null, ...args);
    }
    return fn.call(null, ...args);
  };
}

const split = curry((sep, str) => str.split(sep));
const words = split(' ');
const words2 = str => split(' ', str)

console.log(words('something cool'))
console.log(words2('something cool'))

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words equals to split(' ') function but what happens when I call words again with a string and how can it be equal to words2 function?

CodePudding user response：

When you apply curry to the split function it's like getting back two versions of the split function:

function split(sep, str){
  str.split(sep);
} 

and

function split(sep) {
  return function(str) {
   return str.split(sep);
  }
}

and which one is called is based on the number of parameters provided. So for the case of words you call the 2nd version with ' ' and then you call the returned function with 'something cool' which means that the actual code running is:

'something cool'.split(' ')

In the other case of words2 you call the 1st version and the actual code running:

'something cool'.split(' ')

That's why you get the same result.

CodePudding user response：

Because split(' ', str) and split(' ')(str) gives the same result
And calling words2 calls the former while calling words calls the latter