Develop Reference

What this function (probably callback) is doing in this code?

So after 3 days I've started to understand what callback function actually is and just come across this code:
const timeFuncRuntime = funcParameter => {
let t1 = 120;
funcParameter();
let t2 = 200;
return t2 - t1;
}
const addOneToOne = () => 1 + 1;
timeFuncRuntime(addOneToOne);
console.log(timeFuncRuntime(addOneToOne))
With just my 3 months experience in JS (which I'm aware is nothing), I can tell that in this line of code "timeFuncRuntime(addOneToOne);" another function is being passed as an argument which is 'addOneToOne' however, I don't understand what it's actually doing here? What's the point of having it there?
In the last line when the main function is called with another function being passed as an argument, that function (probably that's what a callback function is called) is returning 1+1 = 2 however, it's just returning and I don't think there's any need for that, or is there?
In the first arrow function, two local variables are declared with different values with a callback function (I think it's a callback) so what is that callback function doing there?

This code all appears to be useless.
The function passed is a pure function, but its result is never used, so it isn't doing anything meaningful.
The wrapper timeFuncRuntime will always return 80, which also isn't very useful.
That means this code is essentially equivalent to:
console.log(80)
Just with extra processing.
This code is likely attempting to show use of Higher Order Functions, but isn't doing so very effectively. The t2 - t1 looks like it's attempting to show how you'd write a benchmarking function, but they hardcoded in numbers which defeats the purpose.
Study functions like map and reduce to get better examples of why you'd want to pass functions around.
What they were likely going for is something like this:
const benchmark = (f) => {
let t1 = Date.now();
f();
let t2 = Date.now();
return t2 - t1;
}
Then you can do:
let t = benchmark(() => {
// Something expensive
for(let i = 0; i < 100000; i++) {
let d = new Date()
}
});
// Prints how long it took the function to execute.
// A number around 50 for me.
console.log(t);

Not understanding higher order functions (problem example using javascript)

I'm studying Javascript basics, particularly higher order functions, at the moment. I have read many articles and watched as many videos where people explain the basic definition and demonstrate the most basic construction of a higher order function. However, when I encounter an actual problem, I am lost. Here is an example (this is just for my personal study, not for a grade or for work):
Write a maybe function that, given a predicate (a function that returns a boolean value) and any other function, only calls the latter if the former returns true: maybe(x => x > 100, myFn). If the predicate returns true, the value of x should be passed to myFn. If the predicate returns false, x should be returned unchanged.
I don't understand how to pass the value of x from one function to another...
I solved this by adding a number argument to maybe, in addition to the predicate function and callback function. However, only two parameters are specified in the prompt, so I guess I'm not doing it correctly. Here is what I did:
//predicate (evaluates to a boolean)
const notZero = function(x) {
//console.log('predicate runs!');
return x !== 0;
}
//callback (does something within the main function)
const plusOne = function(x) {
//console.log('callback runs!');
return x + 1;
}
//checking if my predicate works
//test(notZero(1), true); // => test passed!
//another callback
const myFn = function(x) {
return x - 100;
}
//main function
function maybe(number, predicate, callback) {
if (predicate(number) === true) {
//console.log('predicate === true');
//console.log(callback(number));
return callback(number);
} else {
return number;
}
}
test(maybe(1, notZero, plusOne), 2);
test(maybe(0, notZero, plusOne), 0);
test(maybe(101, x => x > 100, myFn), 1);
test(maybe(99, x => x > 100, myFn), 99);
EDIT: As shared below, maybe can now take only 2 parameters (the predicate and the callback) because it now returns a function whose parameter is number. That's the idea I was looking for.
function maybe(predicate, callback) {
return function (number) {
if (predicate(number) === true) {
return callback(number);
} else {
return number;
}
}
}

It's technically impossible. x is locked out of the world in predicate's scope. There's no way you may extract it out of this function.
Apart from that, as you correctly assume in your code, we logically need to communicate x to both predicate & callback. Otherwise, what's the point of maybe at all?
Since then, your solution is one of very few possible ones.
You may "decorate" it nicer with currying. The idea is precisely identical as in your code but if you'll do so, you will be able to call the final function exactly with 2 arguments.
const setMaybeBase => x => (predicate, callback) => predicate(x) ? callback(x) : x;
// use it later this way
const maybe42 = setMaybeBase(42);
maybe42(yourFn, yourFnTwo);
This is a huge overkill unless the argument you pass to setMaybeBase is for example a complex object you are going to work with.
Alternatively, you might go wild & use a function to get the x as well.
Nevertheless, always remember that the easiest solution is the best one.
Here is a real-world example of maybe function taken straight from node.js repo:
function maybeCallback(cb) {
if (typeof cb === 'function')
return cb;
throw new ERR_INVALID_CALLBACK(cb);
}

Memoize function passes function and returns function JavaScript

I am having multiple problems with this function. It's part of a bonus question for a Data Structures and Algorithms class and I've invested so much time in this one problem, that I'd really like to get it to work and understand what's going on.
There is one main problem, which has caused several little ones...this problem's name is JavaScript. We've never programmed in JavaScript before, but for some reason we have to use it.
The function has to pass tests (this one and fibonacci), which are structured like this:
var fn = (n) => 2 * n
var m_fn = memoize(fn)
expect(m_fn(18)).to.equal(fn(18))
So I have to pass the function I want to memoize as a parameter of the memoize function and the memoize function has to return a function. I am not allowed to do it any other way.
I did all of the reading and researched the memoize function, but all of the implementations take a different approach.
Basically, I understand what I have to do, but I don't quite understand HOW. I know what the memoize function should do, but I don't understand how to adjust the original function using my memoize function. This is what I have so far/what I don't have:
I know it's wrong. But I think I'm missing something major. I am supposed to return a function, but I am returning values...
In the test, it's writen var m_fn = memoize(fn), so the memoize function passes fn, then returns a new function, but in my memoize, I am returning values for fn(n), so I AM doing something wrong...
/**
* Creates a memoized version of the function fn. It is assumed that fn is a referentially transparent
* function.
* #param {function} fn Some referentially transparent function that takes a basic datatype (i.e. number / string)
* #returns {function} A new function that is the memoized version of fn. It never calculates the result of
* a function twice.
*/
memoize: (fn) => { //here we enter the function that we want to memoize
var memory = []; //we need to create an array to hold the previously calculated values, length n (parameter of fn)
if(this.n > memory.length){ //Check to see if this particular value is in the array already.
return memory[this.n]; //How do I access the integer parameter that was passed through fn though? Is this correct?
} else{ // if not, we want to save it and return it
var result = fn(this.n);
memory.push(result);
return result;
}
}

Indeed, you need to return a function.
Secondly, an array is not the ideal structure for memory, because it takes linear time to find an argument value in it. I would suggest to use a Map for this, which is ideal for such purposes. It has has(), get() and set() methods which run in near-constant time:
function memoize(fn) {
var memory = new Map();
return function(arg) {
if (memory.has(arg)) {
console.log('using memory');
return memory.get(arg);
} else {
var result = fn(arg);
memory.set(arg, result);
return result;
}
};
}
var fn = (n) => 2 * n
var m_fn = memoize(fn)
console.log(fn(18));
console.log(m_fn(18));
console.log(m_fn(18)); // outputs also "using memory"

You could use a Map as memory.
var memoize = f =>
(map => v => (!map.has(v) && map.set(v, f(v)), map.get(v)))(new Map),
fn = (n) => 2 * n,
m_fn = memoize(fn);
console.log(m_fn(18), fn(18));

Looking at your code and in-code comments and assuming I'm interpreting correctly, you're really close to the solution. As you've said in the question, you need to return a function that returns the values rather than returning the values.
See comments for explanation:
function memoize(f) {
// An array in which to remember objects with the input arg and result
var memory = [];
// This is the function that will use that array; this is the
// return value of memoize
return function(arg) {
// This code runs when the function returned by memoize is called
// It's *here* that we want to process the argument, check the `memory`
// array, call `f` if necessary, etc.
var entry;
// See if we have a previously-saved result for `arg`
var entry = memory.find(entry => entry.arg === arg);
if (!entry) {
// No -- call `fn`, remember the `arg` and result in an object
// we store in memory``
entry = {arg, result: f(arg)};
memory.push(entry);
}
// We have it (now), return the result
return entry.result;
};
}
function fn(arg) {
console.log("fn called with " + arg);
return 2 * arg;
}
var m_fn = memoize(fn);
console.log(m_fn(18));
console.log(m_fn(18));
console.log(m_fn(20));
console.log(m_fn(20));
Note: There was an arrow function in your code, so I've assumed it's okay to use ES2015 features above. There's not actually very much of it, though, just the arrow function passed to memory.find, the assumption that Array#find will be available, and the syntax used to create the entry object (in ES5 we' need entry = {arg: arg, result: f(arg)} instead).
Note that if we can assume that arg will be a string or number or other value that can reliably be converted to string, we can use an object to store the data rather than an array.
And actually, given this is ES2015, we can use a Map:
function memoize(f) {
// An Map in which to remember objects with the input arg and result
const memory = new Map();
// This is the function that will use that array; this is the
// return value of memoize
return function(arg) {
// This code runs when the function returned by memoize is called
// It's *here* that we want to process the argument, check the `memory`
// array, call `f` if necessary, etc.
let result;
// See if we have a previously-saved result for `arg`
if (!memory.has(arg)) {
// No -- call `fn`, remember the `arg` and result in an object
// we store in memory``
result = f(arg);
memory.set(arg, result);
} else {
// Yes, get it
result = memory.get(arg);
}
// We have it (now), return the result
return result;
};
}
function fn(arg) {
console.log("fn called with " + arg);
return 2 * arg;
}
var m_fn = memoize(fn);
console.log(m_fn(18));
console.log(m_fn(18));
console.log(m_fn(20));
console.log(m_fn(20));
Note that in both cases, I've written the code verbosely to allow for comments and easy comprehension. The ES2015 version with Map, in particular, can be quite a lot shorter.

javascript es6 double arrow functions

I want to better understand es6 arrow functions.
Given the following example:
export default function applyMiddleware(...middlewares) {
return (createStore) => (reducer, preloadedState, enhancer) => {
// snip actual enhancer logic
return {
...store,
dispatch
}
}
}
Describing the above in words:
Our exported function (applyMiddleware) takes an array parameter with spread.
Then applyMiddleware returns a nameless function with a createStore parameter, which returns another nameless function this time with three parameters.
So without the arrows it would look like this:
export default function applyMiddleware(...middlewares) {
return function(createStore){
return function(reducer,preloadedState,enhancer){
//some logic
return{...store,dispatch}
}
}
}
My questions:
Am I correct?
What is the common use case/code paradigm we see here?

The answer to your first question is more or less (see my comment). The answer to your second question is that the pattern you are seeing is a combination of using a closure and currying. The original parameters to the exported function get gathered into an array called 'middlewares' that the returned functions close over (i.e. have access to). The function then can be called again with yet another parameter 'createStore' then another function is returned that can accept even more parameters. This allows one to partially apply the parameters. For a more trivial (and perhaps more easily comprehended) example, lets take a function called 'add' that adds two numbers:
let add = (x, y) => x + y;
Not very interesting. But lets break it up so it can take the first number and return a function that takes the second:
let add = x => y => x + y;
let add3 = add(3);
let seven = add3(4); // 7
This may not seem like a big win for our add function, but you started with a much more reasonable real-world example. Additionally, rather than currying manually it is possible (and desirable) to use a curry function that does it for you, many popular libraries (lodash, underscore, ramda) implement curry for you. An example using Ramda:
let add = R.curry((x, y) => x + y);
let add3 = add(3);
let five = add3(2);
let also5 = add(3, 2);
let still5 = add(3)(2); // all are equivalent.

This answer is for those who still have some doubts in double arrow functions. Lets dig deep into it.
const doubleArrowFunc = param1 => param2 => {
console.log('param1', param1);
console.log('param2', param2);
}
If you call this function
const executeFunc = doubleArrowFunc('Hello');
If you print executeFunc in console you'll get an output like this
ƒ (param2) {
console.log('param1', param1);
console.log('param2', param2);
}
Now this is like a half executed code. If you wanna execute it fully, you've to do it like this
executeFunc('World');
//And the output will be
param1 Hello
param2 World
If you want even more understanding. I can execute the same without arrow function
function doubleArrowFunc(param1) {
return function innerFunction(param2) {
console.log('param1', param1);
console.log('param2', param2);
}
}

Understanding the syntax of a deferred execution chain

I'm getting around to learning JavaScript - really learning JavaScript. I come from a PHP background so some JavaScript concepts are still new to me, especially asynchronous programming. This question might have already been answered many times before but I have not been able to find an answer. It might be because I don't really even know how to ask the question other than by showing an example. So here it is:
When using the deferred package from npm, I see the following example:
delayedAdd(2, 3)(function (result) {
return result * result
})(function (result) {
console.log(result); // 25
});
They refer to this as chaining and it actually works as I'm currently using this code to check when a promise is resolved or is rejected. Even though they call it chaining, it reminds me of trailing closures like in Swift.
I don't really understand what type of chaining this is since we have a function invocation and then immediately after, an anonymous function enclosed in parentheses.
So I guess I have two questions.
What pattern is this?
How does it work? This may be a loaded question but I like to know how something works so when someone asks me about this I can give them a detailed explanation.
Here is the delayedAdd function:
var delayedAdd = delay(function (a, b) {
return a + b;
}, 100);
which uses the following function:
var delay = function (fn, timeout) {
return function () {
var def = deferred(), self = this, args = arguments;
setTimeout(function () {
var value;
try {
value = fn.apply(self, args));
} catch (e) {
def.reject(e);
return;
}
def.resolve(value);
}, timeout);
return def.promise;
};
};

It's actually really easy to understand. Let's look at what's going on here when the expression is evaluated:
First the delayedAdd(2, 3) function will be called. It does some stuff and then returns. The "magic" is all about its return value which is a function. To be more precise it's a function that expects at least one argument (I'll get back to that).
Now that we evaluated delayedAdd(2, 3) to a function we get to the next part of the code, which is the opening parenthesis. Opening and closing parenthesis are of course function calls. So we're going to call the function that delayedAdd(2, 3) just returned and we're going to pass it an argument, which is what gets defined next:
That argument is yet another function (as you can see in your example). This function also takes one argument (the result of the computation) and returns it multiplied by itself.
This function that was returned by the first call to delayedAdd(2, 3) returns yet another function, which we'll call again with an argument that is another function (the next part of the chain).
So to summarize we build up a chain of functions by passing our code to whatever function delayedAdd(2, 3) returned. These functions will return other functions that we can pass our functions again.
I hope this makes the way it works somewhat clear, if not feel free to ask more.

mhlz's answer is very clear. As a supplementary, here I compose a delayedAdd for your to better understand the process
function delayedAdd(a, b) {
var sum = a + b
return function(f1) {
var result1 = f1(sum)
return function(f2) {
f2(result1)
}
}
}
Where in your example code, the function you passed as f1 is:
function (result) {
return result * result
}
and f2 is:
function (result) {
console.log(result)
}

Functions are first-class citizens in JS - that means (among others), they can take the role of actual parameters and function return values. Your code fragment maps functions to functions.
The signatures of the functions in your chained call might look like this.
delayedAdd: number -> fn // returns function type a
a: fn ( number -> number) -> fn // returns function type b
b: fn ( number -> void ) -> void // returns nothing ( guessing, cannot know from your code portion )
General setting
Of course, JS is a weakly typed language, so the listed signatures are derived from the code fragment by guessing. There is no way to know whether the code actually does what is suggested above apart from inspecting the sources.
Given that this showed up in the context of 'chaining', the signatures probably rather look like this:
delayedAdd: number x number -> fn (( fn T -> void ) -> ( fn T -> void ))
Which means that delayedAdd maps two numbers to a function x, which maps functions of arbitrary signatures to functions of the same signature as itself.
So who would do anything like this ? And why ?
Imagine the following implementation of x:
//
// x
// Collects functions of unspecified (possibly implicit) signatures for later execution.
// Illustrative purpose only, do not use in production code.
//
// Assumes
function x ( fn ) {
var fn_current;
if (this.deferred === undefined) {
this.deferred = [];
}
if (fn === undefined) {
// apply functions
while ( this.deferred.length > 0 ) {
fn_current = this.deferred.shift();
this.accumulator = fn_current(this.accumulator);
}
return this.accumulator;
}
else {
this.deferred.push ( fn );
}
return this;
}
Together with a function delayedAdd that actually returns an object of the following kind ...:
function delayedAdd ( a1, a2) {
return x ( function () { a1 + a2; } );
}
... you'll effectively register a chain of functions to be executed at some later point of time (e.g. in a callback to some event).
Notes and reminders
JS functions are JS objects
The signatures of the registered functions may actually be arbitrary. Considering them to be unified just serves to keep this exposition simpler (well ...).
Caveat
I do not know whether the outlined codeis what node.js does (but it could be ... ;-))

To be fair this pattern can be either chaining or currying(or partial application). Depending how it's implemented. Note this is a theoretical answer to provide more information about the pattern and not your specific use case.
Chaining
There is nothing special here because we can just return a function that will be called again. Functions in javascript are first class citizens
function delayedAdd(x, y) {
// In here work with x and y
return function(fn) {
// In here work with x, y and fn
return function(fn2) {
//Continue returning functions so long as you want the chain to work
}
}
}
This make it unreadable in my opinion. There is a better alternative.
function delayedAdd(x, y) {
// In here work with x and y
return {
then: function(fn) {
// In here work with x, y and fn
return {
then: function(fn2) {
//Continue returning functions so long as you want the chain to work
}
}
}
}
}
This changes the way your functions are called from
delayedAdd(..)(..)(..); // 25
is transformed to
delayedAdd().then().then()
Not only is more readable when you are passing several callback functions but allows a distinction from the next pattern called currying.
Currying
The term cames after the mathematician Haskell Curry. The definition is this
In mathematics and computer science, currying is the technique of translating the evaluation of a function that takes multiple arguments (or a tuple of arguments) into evaluating a sequence of functions, each with a single argument (partial application). It was introduced by Moses Schönfinkel and later developed by Haskell Curry.
Basically what it does is take several arguments and merge with the subsecuents and apply them to the original function passed in the first argument.
This is a generic implementation of this function taken from Stefanv's Javascript Patterns.
{Edit}
I changed my previous version of the function to one which has partial application included to make a better example. In this version you must call the function with no argument to get the value returned or you will get another partially applied function as result. This is a very basic example, a more complete one can be found on this post.
function schonfinkelize(fn) {
var slice = Array.prototype.slice,
stored_args = [],
partial = function () {
if (arguments.length === 0){
return fn.apply(null, stored_args);
} else {
stored_args = stored_args.concat(slice.call(arguments));
return partial;
}
};
return partial;
}
This are the results of the application of this function
function add(a, b, c, d, e) {
return a + b + c + d + e;
}
schonfinkelize(add)(1, 2, 3)(5, 5)(); ==> 16
Note that add (or in your case delayedAdd) can be implemented as the curying function resulting in the pattern of your example giving you this
delayedAdd(..)(..)(..); // 16
Summary
You can not reach a conclusion about the pattern just by looking at the way the functions are called. Just because you can invoke one after the other it doens't mean is chaining. It could be another pattern. That depends on the implementation of the function.

All excellent answers here, especially #mhlz and #Leo, I'd like to touch on the chaining part you've mentioned. Leo's example shows the idea of calling functions like foo()()() but only works for fixed number of callbacks. Here's an attempt to imlpement unlimited chaining:
delayedAdd = function da(a, b){
// a function was passed: call it on the result
if( typeof a == "function" ){
this.result = a( this.result )
}
else {
// the initial call with two numbers, no additional checks for clarity.
this.result = a + b;
}
// return this very function
return da;
};
Now you can chain any number of functions in () after the first call:
// define some functions:
var square = function( number ){ return number * number; }
var add10 = function( number ){ return number + 10; }
var times2 = function( number ){ return number * 2; }
var whatIs = function( number ){ console.log( number ); return number; }
// chain them all!
delayedAdd(2, 3)(square)(whatIs)(add10)(whatIs)(times2)(whatIs);
// logs 23, 35 and 70 in the console.
http://jsfiddle.net/rm9nkjt8/3/

If we expand this syntax logically we would reach something like this:
var func1 = delayedAdd(2, 3);
var func2 = function (result) {
return result * result
};
var func3 = function (result) {
console.log(result);
};
var res = func1(func2); // variable 'res' is of type 'function'
res(func3);