Eloquent Javascript: Higher Order Functions

Question

I'm going through Eloquent Javascript: Higher Order Functions example below and already read questions and answers here and here. But I'm still very confused.

function noisy(f) {
  return function(arg) {
    console.log("calling with", arg);
    var val = f(arg);
    console.log("called with", arg, "- got", val);
    return val;
  };
}
noisy(Boolean)(0);
// → calling with 0
// → called with 0 - got false

How can (0) be passed into noisy(f) since noisy() only takes one parameter and that is (Boolean)? I can see the inner function f(arg) is basically Boolean(0), but I don't understand how two parameters can get passed into a function that only allow one parameter. Would "noisy(Boolean)(0)(1)(2)(3);" be a valid function call? If so, how would you differentiate each value after Boolean within the noisy function? Which value will be referenced by "arg"?
The book noted the example function is modifying another function. Which function is being modified? I'm not understanding what the author meant by "modified".

noisy(f) returns a function that takes one argument, arg. Calling this function with the sole argument is perfectly valid. — Wiktor Zychla, Jun 28 '16 at 06:00
Possible duplicate of [Higher-order functions in Javascript](http://stackoverflow.com/questions/23535316/higher-order-functions-in-javascript) — Aditya Singh, Jun 28 '16 at 06:10
Maybe `var ModifiedBoolean = noisy(Boolean); console.log(ModifiedBoolean(0))` is easier to understand. Though instead of "modified" the term "wrapped" might be better. — Bergi, Jun 28 '16 at 06:52
Depending on what function you pass to `noisy` the behavior of the returned function `function(arg)` is **modified**. Pass an `inc` function for instance and you get a completely different behavior: `noisy(inc)(0)` yields 1 — , Jun 28 '16 at 09:31

score 1 · Answer 1 · edited Jul 22 '21 at 02:55

but I don't understand how two parameters can get passed into a function that only allow one parameter

noisy returns a function, Boolean is passed to noisy, 0 is passed to anonymous function returned from noisy, where f is Boolean, val becomes Boolean(0).

For example

function fn1(arg1) {
  return function fn2(arg2) {
    console.log(arg1, arg2)
  }
}

// Call fn1, inside fn1 fn2 is called with `"b"` as parameter.
fn1("a")("b") // `a b`, `fn2`

score 0 · Accepted Answer · answered Jun 28 '16 at 06:09

This is the concept of currying in JavaScript where you can curry functions to return partially applied functions or pass in other functions

How can (0) be passed into noisy(f) since noisy() only takes one parameter and that is (Boolean)?

The answer to this is the curried function noisy() which expects a function f as parameter and returns another function. The returned function has a closure over noisy and as a result it can identify that Boolean was passed as parameter to noisy even after it was returned. That's why calling noisy(Boolean)(0) basically substitutes f=Boolean, arg=0

Refer this for more on currying: http://javascript.crockford.com/www_svendtofte_com/code/curried_javascript/ and closures: https://developer.mozilla.org/en/docs/Web/JavaScript/Closures

Thanks for mentioning currying. Another explanation is here: https://www.sitepoint.com/currying-in-functional-javascript/ — j7an, Jun 30 '16 at 06:28

Eloquent Javascript: Higher Order Functions

2 Answers2