CODEWITHSUNDEEP
javascriptfunctionarguments
anotherOne
anotherOne

Reputation: 1573

A function creating variable names for parameters that I never gave

I am learning JavaScript from an eloquent book and in the chapter about higher order functions I found this code:

function noisy(f) {
  return function(...args) => {
    console.log("calling with", args);
    let result = f(...args);
    console.log("called with", args, "returned", result);
    return result;
  };
}

noisy(Math.min)(3,2,1);
// calling with [3,2,1]
// called with [3,2,1] returned 1

I know that the rest parameter ...args takes a number of arguments and groups them into an array, but when did I gave any parameter to the arrow function?
Does args automatically contain all the extra parameters passed to noisy() (which is expecting only f)?
If yes what are the rules of this way of using parameters?
Could I have used even only the first two extra parameters?
Shouldn't the original code be like the following?

function noisy(f, ...args) {
    return function(args) => { // with the rest of the program

Upvotes: 1

Views: 40

Answers (1)

Alex
Alex

Reputation: 3345

when did I gave any parameter to the arrow function

You passed them in the (3,2,1) part of noisy(Math.min)(3,2,1). noisy() is returning a function which you are then calling immediately with parameters (3,2,1)

it might be clearer to break out that call

var myArrowFunction = noisy(Math.min)
myArrowFunction(3,2,1)

Upvotes: 3

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