Develop Reference

Understanding pipe function in JavaScript (functional programming)

I'm working through question 11 on bfe.dev's JavaScript coding exercises. The question asks to chain multiple basic math functions together via composition.
Here's some sample code that passes the provided test cases.
function pipe(funcs) {
let result;
return function (arg) {
result = arg;
for (let fn of funcs) {
result = fn(result);
}
return result;
}
}
And some driver code with the intended result, according to bfe.dev
pipe([
times(2),
plus(3),
times(4)
])
// (x * 2 + 3) * 4
I'm struggling to understand the implementation on a code-level. pipe receives an array of functions, and I understand the approach is to walk through the array and compute the result of each function, store that in the result variable, similar to an array reduce.
But I cannot wrap my head around how the return function works, and what arg would actually be. I would appreciate a line-by-line walkthrough if possible.
Thanks in advance!

The example probably needs a bit more to make sense:
const doMath = pipe([
times(2),
plus(3),
times(4)
])
const finalResult = doMath(5)
console.log(finalResult);
pipe's job is to create a new function. That new function has the text:
function (arg) {
result = arg;
for (let fn of funcs) {
result = fn(result);
}
return result;
}
And that is what gets assigned to doMath in my example. Then when i call doMath(5), it's going to execute the code, with arg equal to 5. It assigns 5 to result, and then starts looping through the functions (which it still has access to because this is a closure). Each function calculates a new value, which gets assigned to result, and then eventually the final result is returned. So in this case the calculation goes 5 => 10 => 13 => 52. 52 gets returned and then logged out.

Arguments Optional FreeCodeCamp Challenge

in need of help with a codecamp challenge:
Arguments Optional - The challenge
https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/arguments-optional
My Question
I know this can be done with the arguments object (see figure 1), where I can call the function again when the second value is undefined so I've gone for a different approach; However, my code of using currying works but becomes an issue with 'addTogether(5)(7);'
Issue ->
I get the returned value of 12 but doesn't get approved in the code challenge.
I did originally return the value inside the sum function but the challenge required the sum value to be returned in addTogether function, which I did so now it resulting in the following
addTogether(2,3); // now working
addTogether(5)(7); // NOT working
addTogether(23, 30); // now working
Is there something I'm doing wrong that's resulting in the test case failing even though the correct value is returned?
let val = 0;
function sum(a, b) {
val = a + b;
}
function sumTwoAnd(sumFunc) {
return function addTogether(...params) {
let numsArr = [...params];
const res = numsArr.every(el => typeof el === 'number');
if (res === false) return;
if (numsArr.length >= sumFunc.length) {
sumFunc(...numsArr);
} else {
return function(...args2) {
let newArr = numsArr.concat(args2);
addTogether(...newArr);
}
}
console.log(val);
return val;
}
}
let addTogether = sumTwoAnd(sum);
addTogether(2,3);
addTogether(5)(7);
addTogether(23, 30);
Figure 1
Shows how I can get 'test(5)(7)' the second parameter from the function
function test() {
const [f, s] = arguments;
console.log(f, s)
if (s === undefined) {
return s => test(f, s)
}
}
test(23, 30);
test(5)(7);

You declared addTogether using let, so the declaration won't be hoisted above the point where it was defined. This is making your recursive call in the else statement fail, since addTogether() doesn't exist that far up.
You might want to extract the function you're returning in sumTwoAnd() as a separate function definition, so it can freely call itself, similar to your Figure 1 example.
Or you can call sumTwoAnd() instead to regain the function, then pass newArr to said function.

Instead of calling the function I have now returned it...
return addTogether(...newArr);
This now works :)

Can I mock functions with specific arguments using Jest?

I want to mock a function with Jest, but only if it is called with specific arguments, for example:
function sum(x, y) {
return x + y;
}
// mock sum(1, 1) to return 4
sum(1, 1) // returns 4 (mocked)
sum(1, 2) // returns 3 (not mocked)
There is a similar feature implemented in Ruby's RSpec library:
class Math
def self.sum(x, y)
return x + y
end
end
allow(Math).to receive(:sum).with(1, 1).and_return(4)
Math.sum(1, 1) # returns 4 (mocked)
Math.sum(1, 2) # returns 3 (not mocked)
What I'm trying to achieve in my tests is a better decoupling, let's say I want to test a function that relies on sum:
function sum2(x) {
return sum(x, 2);
}
// I don't want to depend on the sum implementation in my tests,
// so I would like to mock sum(1, 2) to be "anything I want",
// and so be able to test:
expect(sum2(1)).toBe("anything I want");
// If this test passes, I've the guarantee that sum2(x) is returning
// sum(x, 2), but I don't have to know what sum(x, 2) should return
I know that there is a way to implement this by doing something like:
sum = jest.fn(function (x, y) {
if (x === 1 && y === 2) {
return "anything I want";
} else {
return sum(x, y);
}
});
expect(sum2(1)).toBe("anything I want");
But it would be nice if we had some sugar function to simplify it.
Does it sounds reasonable? Do we already have this feature in Jest?
Thanks for your feedback.

I found this library that a colleague of mine wrote recently: jest-when
import { when } from 'jest-when';
const fn = jest.fn();
when(fn).calledWith(1).mockReturnValue('yay!');
const result = fn(1);
expect(result).toEqual('yay!');
Here's the library: https://github.com/timkindberg/jest-when

jest-when mentioned by STeve Shary is probably the best option.
If you don't want to install a new library, when you don't care about the original function, here's a one-line solution:
sum.mockImplementation((x, y) => x === 1 && y === 2 && "my-return-value")

I thought I needed a way to mock with arguments too, but for me, I could solve my problem by simply knowing the order of calls.
Taken from the jest documentation
const filterTestFn = jest.fn();
// Make the mock return `true` for the first call,
// and `false` for the second call
filterTestFn.mockReturnValueOnce(true).mockReturnValueOnce(false);
So in the example above, provided you know that on first pass the function should return true and on second pass it should return false, you're good to go!

No there is no way to do this in Jest yet. You could use sinons stubs for this. from the docs:
stub.withArgs(arg1[, arg2, ...]);
Stubs the method only for the
provided arguments. This is useful to be more expressive in your
assertions, where you can access the spy with the same call. It is
also useful to create a stub that can act differently in response to
different arguments.
"test should stub method differently based on arguments": function () {
var callback = sinon.stub();
callback.withArgs(42).returns(1);
callback.withArgs(1).throws("TypeError");
callback(); // No return value, no exception
callback(42); // Returns 1
callback(1); // Throws TypeError
}

import * as helper from "../helper"; //file where all functions are
jest.spyOn(helper, "function_name").mockImplementation((argument) => {
// This argument is the one passed to the function you are mocking
if (argument === "something") {
return "something"
} else {
return "something else"
}
});

This may help...
I had something similar whereby I had the same method called with different parameters requiring different returned result from the stubbed/mocked call. I've used a variable with a list of functions when I've made a call to the mocked service i take function off the top of the queue and execute function. It requires knowledge of the order of execution you are testing and doesn't really handle varying the response by argument but has allowed me to work around the restriction in jest.
var mockedQueue = [];
mockedQueue.push(() => {return 'A';})
mockedQueue.push(() => {return 'B';})
service.invoke = jest.fn(()=>{
serviceFunctionToCall = mockedQueue.shift();
return serviceFunctionToCall();
})

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.

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);