I'm trying to create a function that accepts two callbacks and a value that will return a boolean indicating if the passing the value into the first function, and then passing the resulting output into the second function, yields the same output as the same operation in reverse (passing the value into the second function and then passing the output into the first function).
I get the following error and I believe it is probably due to scoping but not sure how to resolve it: Reference Error on line 206: ouput2 is not defined
function commutative(func1, func2, value) {
//check to see if invoking cb1 on value then passing output to cb2 as cb2 => cb1
function func1() {
let output1 = func1(value);
return output1;
}
function func2() {
let output2 = func2(output1);
return output2;
}
function reverseOrder() {
let output3 = func2(value);
let output4 = func1(output3);
}
//return boolean
if (ouput2 === output4) {
return true;
} else {
return false;
}
}
// Test cases:
const multBy3 = n => n * 3;
const divBy4 = n => n / 4;
const subtract5 = n => n - 5;
console.log(commutative(multBy3, divBy4, 11)); // should log: true
console.log(commutative(multBy3, subtract5, 10)); // should log: false
console.log(commutative(divBy4, subtract5, 48)); // should log: false
You are never calling your inner functions, and output1, output2, etc. are defined locally to those inner functions, not accessible in the scope of commutative. Also, you are overwriting the func1 and func2 parameters. Try this:
function commutative(func1, func2, value) {
return func2(func1(value)) === func1(func2(value));
}
// Test cases:
const multBy3 = n => n * 3;
const divBy4 = n => n / 4;
const subtract5 = n => n - 5;
console.log(commutative(multBy3, divBy4, 11)); // should log: true
console.log(commutative(multBy3, subtract5, 10)); // should log: false
console.log(commutative(divBy4, subtract5, 48)); // should log: false
Related
Let's consider I have the following function call,
function add(){
x = 0 ;
for(i = 0 i < ##; i++){ // need to run a loop four times
x+=1
}
}
Let's consider I am trying to Implement the function that will add one on each subsequent call, like below
console.log(add()()().getValue()); // 3
console.log(add().getValue()); // 1
console.log(add()().getValue()); // 2
A call to add must return a function which also has a getValue method, and each call to that function must return the same thing. So:
function add() {
var x = 1;
function inner() {
x += 1;
return inner;
}
inner.getValue = function () {
return x;
}
return inner;
}
console.log(add()()().getValue()); // 3
console.log(add().getValue()); // 1
console.log(add()().getValue()); // 2
My guess is they were expecting you to use toString() which is not the greatest way of doing this.
function add(x = 0) {
function next() {
return add(x+1);
}
next.toString = function () {
return x;
};
return next;
}
console.log("example 1", add()()()());
console.log("example 2", add()()()()()()()()());
I think you are trying to emulate the behavior of generator functions. Here is a snippet that illustrates one way you could do it with a generator.
function* adder() {
let x = 0;
while (true) {
yield x + 1;
x++;
}
}
const add = adder();
const firstValue = add.next();
const secondValue = add.next();
const thirdValue = add.next().value;
const _debounce = (n, func) => {
//code here
}
const originFun = () => {
console.log('hit')
}
const _call = () => _debounce(2, originFun)
_call() //The originFun not executes
_call() //hit
_call() //The originFun not executes
_call() //hit
I do not know how to implement it, even after a test.
_debounce should accept the sort argument, and the originFun function, and return its own function (or closure) that you can assign to _call. You can then call _call with it's own argument - the number.
function originFun(sort) {
console.log('hit,', sort);
}
function _debounce(sort, func) {
// Set a counter
let count = 1;
// Because closures retain the variables from the
// preceding lexical environment we can
// still use and update `count` when the function is returned
// This function (which accepts a number argument) is what's
// assigned to _call.
return function(n) {
// Call originFun with the number if the remainder of dividing
// sort by count is zero
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Remainder
if (count % sort === 0) func(n);
count++;
}
}
// _debounce returns a function that accepts a number
// argument. We assign that function to _call.
const _call = _debounce(2, originFun);
_call(1) //not execute
_call(2) //hit,2
_call(3) //not execute
_call(4) //hit,4
This implementation of debounce should help your use case. No global variable, and is implemented in pure javascript.
function _debounce(func, n){
let count = 0;
return (...args) => {
count++;
if (count % n === 0) {
count = 0;
func.apply(this, args);
}
};
}
const originFun = (sort) => {
console.log('hit,', sort)
}
const _call = _debounce((sort) => originFun(sort), 2);
_call(1) //not execute
_call(2) //hit,2
_call(3) //not execute
_call(4) //hit,4 */
_call(1) //not execute
_call(2) //hit,2
_call(3) //not execute
_call(4) //hit,4 */
To keep track of the number of calls you could use a global counter.
Additionally, you can pass arguments to your function using the spread syntax. That way you can have multiple arguments and all of them will be passed to the function in order.
let counter = 1;
const _debounce = (n, func, ...args) => {
if (counter === n) {
counter = 1;
func(...args);
} else {
counter += 1;
}
}
const originFun = (sort) => {
console.log('hit,', sort)
}
const _call = (sort) => _debounce(2, originFun, sort)
_call(1) //not execute
_call(2) //hit,2
_call(3) //not execute
_call(4) //hit,4
I don't understand how all those f() function work, can someone explain why it prints two '1', I know it prints '1' for every '()' after f(f), but I don't know why.
function f(y) {
let x = y;
var i = 0;
return () => {
console.log(++i);
return x(y);
};
}
f(f)()();
And why does the 'i' doesn't increase?
Thank you.
function f(y) {
let x = y;
var i = 0;
return () => {
console.log(++i);
return x(y);
};
}
f(f)()();
is equivalent to
function f() {
var i = 0;
return () => {
console.log(++i);
return f();
};
}
const t1 = f();
const t2 = t1();
t2();
is equivalent to
function f() {
var i = 0;
return () => {
console.log(++i);
};
}
const t1 = f();
t1();
const t2 = f();
t2();
If you did call each of t1 or t2 multiple times instead of just once, you'd increment the i from the respective closure some more. But if you instead just chain them, they call f again and initialise a new var i = 0 for a different closure.
First, the f(y) function essentially calling y onto itself. Executing f(y) would return a new function, which when executed, would execute x(y) and return the results.
To reduce the confusion, it's just calling f(f) for each f() you executed in this example, as x === y and y === f. The important part of why i never seems to increase, is that every execution creates a new i.
What happens in behind are:
f(f)()();
// is same as
const f1 = f(f);
const f2 = f1();
const f3 = f2();
// f(f) would execute first, and returns () => {
// console.log(++i);
// return x(y); // which is same as returning f(f) in this example
// }
Notice that executing f(f) returns x(y) which x(y) seems to be equal to f(f). Seems as it is similar in code, but different instance. Another point is that i was never carried to this new function, nor are shared to the other instances. Each f(f) creates a new i, never passed to the next function.
Let's name the function at line 4 , function A.
return result of function f() is function A.(at line 4 you define a function, you don't call it.)
the body of A is gonna be this and since you use those variables in an inner function, they are not gonna be exposed(i = 0, x = y = f ):
function A(){
console.log(++i);
return x(y);
so what you have now is: A()().
First parenthesis: A() prints a '1' and returns result of f(f) which is function A.(the first i is exposed and a new i is created)
Second parenthesis : A is executed like I said and return another A, since there are no any parenthesis left, there is no more call.
You declare in f always a new i.
Instead, you could store the count into a property of the function itself.
function f(y) {
let x = y;
f.i = f.i || 0;
return () => {
console.log(++f.i);
return x(y);
};
}
f(f)()()();
I am interested in the scenario where we have some function f which is recursive and which we are not provided the source code to.
I would like a function memoizer: Function -> Function which takes in say f and returns a function g such that g = f (in the sense they return the same value given the same arguments) which when called first checks if the called arguments are in its 'cache' (memory of results it has calculated before) and if so returns the result from this, otherwise it should compute f, should f call itself with some arguments, this is tantamount to calling g with those arguments and I would like that f first check if the cache of g contains those arguments and if so return the result from this, otherwise ...
This is easy (in Javascript) to do given the source code of f, I simply define memoize in the obvious way and do something like
let f = memoize((...args) => {/* source code of f */});
But this doesn't appeal to me at all (mainly because I might want a memoized and non memoized version of the same function and then I'd have to write the same function twice) and won't work if I don't know how to implement f.
In case it's not clear what I'm asking,
I would like a function memoize which takes a function such as
fact = n => n === 0 ? 1 : n * fact(n - 1);
And returns some new function g such that fact(n) = g(n) for all n and which for example when g(10) is computed stores the values of fact(0), ..., fact(10) which are computed while computing g(10) and then if I ask for say g(7) it finds the result in the cache and returns it to me.
I've thought that conceptually it's possible to detect when f is called since I have it's address and maybe I could replace all calls to f with a new function where I compute f and store the result and then pass the value on to where it would normally go. But I don't know how to do this (and it sounds unpleasant).
maybe I could replace all calls to f with a new function where I compute f and store the result and then pass the value on to where it would normally go.
This is actually very easy to do, as Bergi referred to in a comment.
// https://stackoverflow.com/questions/24488862/implementing-automatic-memoization-returns-a-closured-function-in-javascript/
function memoize(func) {
var memo = {};
var slice = Array.prototype.slice;
return function() {
var args = slice.call(arguments);
if (args in memo)
return memo[args];
else
return (memo[args] = func.apply(this, args));
}
}
function fib(n) {
if (n <= 1) return 1;
return fib(n - 1) + fib(n - 2);
}
fib = memoize(fib);
console.log(fib(100));
I might want a memoized and non memoized version of the same function and then I'd have to write the same function twice
Yes, you need to. The recursive call to fact(n - 1) inside the function can only refer to one fact function - either a memoized or an unmemoized one.
So what you need to do to avoid code duplication is define fact with the Y combinator:
const makeFact = rec => n => n === 0 ? 1 : n * rec(n - 1);
// ^^^ ^^^
const factA = Y(makeFact);
const factB = memoizingY(makeFact);
function Y(make) {
const f = make((...args) => f(...args)); // const f = make(f) is easier to understand
return f; // but doesn't work with eager evaluation
}
I'll leave the definition of memoizingY as an exercise to the reader :-)
Possibly simpler approach:
const makeFact = annotate => {
const f = annotate(n => n === 0 ? 1 : n * f(n - 1));
return f;
}
const factA = makeFact(identity);
const factB = makeFact(memoize);
In my limited experience, we do have access to JavaScript source code. We could thus attempt to generate new source code for the memoized function.
// Redefine Function.prototype.bind
// to provide access to bound objects.
// https://stackoverflow.com/questions/7616461/generate-a-hash-from-string-in-javascript
var _bind = Function.prototype.apply.bind(Function.prototype.bind);
Object.defineProperty(Function.prototype, 'bind', {
value: function(obj) {
var boundFunction = _bind(this, arguments);
boundFunction.boundObject = obj;
return boundFunction;
}
});
// Assumes the parameters for the function,
// f, can be consistently mapped.
function memo(f){
if (!(f instanceof Function))
throw TypeError('Argument is not an instance of Function.');
// Generate random variable names
// to avoid conflicts with unknown
// source code
function randomKey(numBytes=8){
let ranges = [[48, 10], [65, 26], [97, 26]];
let key = '_';
for (let i=0; i<numBytes; i++){
let idx = Math.floor(Math.random() * ranges.length);
key += String.fromCharCode(ranges[idx][0] + Math.random() * ranges[idx][1]);
}
return key;
}
let fName = f.name;
let boundObject;
let fCode;
const nativeCodeStr = '(){[nativecode]}';
// Possible Proxy
try {
fCode = f.toString();
} catch(error){
if (error.constructor == TypeError){
if (Function(`return ${ fName }.toString()`)() != nativeCodeStr){
throw TypeError(`Possible Proxy detected: function has a name but no accessible source code. Consider memoizing the target function, ${ fName }.`);
} else {
throw TypeError(`Function has a name but no accessible source code. Applying toString() to its name, ${ fName }, returns '[native code]'.`);
}
} else {
throw Error('Unexpected error calling toString on the argument.');
}
}
if (!fName){
throw Error('Function name is falsy.');
// Bound functions
// Assumes we've monkey-patched
// Function.prototype.bind previously
} else if (fCode.replace(/^[^(]+|\s+/g, '') == nativeCodeStr){
if (/^bound /.test(fName)){
fName = fName.substr(6);
boundObject = f.boundObject;
// Bound functions return '[native code]' for
// their toString method call so get the code
// from the original function.
fCode = Function(`return ${ fName }.toString()`)();
} else {
throw Error("Cannot access source code, '[native code]' provided.");
}
}
const fNameRegex = new RegExp('(\\W)' + fName + '(\\W)', 'g');
const cacheName = randomKey();
const recursionName = randomKey();
const keyName = randomKey();
fCode = fCode.replace(/[^\(]+/,'')
.replace(fNameRegex, '$1' + recursionName + '$2')
.replace(/return/g, `return ${ cacheName }[${ keyName }] =`)
.replace(/{/, `{\n const ${ keyName } = Array.from(arguments);\n\n if (${ cacheName }[${ keyName }])\n return ${ cacheName }[${ keyName }];\n`);
const code = `function(){\nconst ${ cacheName } = {};\n\nfunction ${ recursionName + fCode }\n\nreturn ${ recursionName }.apply(${ recursionName }, arguments);}`;
let g = Function('"use strict";return ' + code)();
if (boundObject){
let h = (g).bind(boundObject);
h.toString = () => code;
return h;
} else {
return g;
}
} // End memo function
function fib(n) {
if (n <= 1) return 1;
return fib(n - 1) + fib(n - 2);
}
const h = fib.bind({a: 37});
const g = memo(h);
console.log(`g(100): ${ g(100) }`);
console.log(`g.boundObject:`, g.boundObject);
console.log(`g.toString():`, g.toString());
try{
memo(function(){});
} catch(e){
console.log('Caught error memoizing anonymous function.', e)
}
const p = new Proxy(fib, {
apply: function(target, that, args){
console.log('Proxied fib called.');
return target.apply(target, args);
}
});
console.log('Calling proxied fib.');
console.log(`p(2):`, p(2));
let memoP;
try {
memoP = memo(p);
} catch (e){
console.log('Caught error memoizing proxied function.', e)
}
I have a 'twice' function that return 2 of the argument passed into it. I also have another function 'runTwice' that counts the number of times it called the 'twice' function (the idea being that I want the 'twice' function to only run 'twice' no matter how often it is called via the 'runTwice' function). Can you please help?
Functions are given below:
var count = 1;
function twice(num){
return num*2;
}
function runTwice(func){
if (count<3){
count++;
return func;
} else {
return 'Function cannot run!';
}
}
var myFunc = runTwice(twice)
var output = [];
for (var i = 0; i < 3; i++){
output.push(myFunc(i));
}
console.log(output);
I would like the output to be [0, 2, 'Function cannot run!'].
I can make this work if I count the 'twice' function directly but I am looking to understand why this doesn't work as presented above.
Just for fun I'll make a generic expireAfter(invocable[, times[, message]]) function:
function expireAfter(invocable, times = 2, message = 'Function cannot run!') {
return function expires() {
if (times > 0) {
times--;
return invocable.apply(this, arguments);
}
return message;
}
}
function twice(n) {
return n * 2;
}
var myFunc = expireAfter(twice);
console.log(Array(3)
.fill()
.map((_, index) => myFunc(index))
);
The function runTwice should return another function that will decide whether to call the function func (using Function.prototype.apply) or to return a string message instead:
function twice(num){
return num * 2;
}
function runTwice(func){
var count = 0; // this will be trapped in a closure along with func
return function() { // this is the function that gets called
count++; // it increments its version of the count variable
if(count <= 2) // if count is less than 2
return func.apply(this, arguments); // then it calls the function func with whatever arguments passed into it and return the returned value of that call
return "Not available anymore!"; // otherwise (count > 2), then it returns a string
}
}
var myFunc = runTwice(twice);
for (var i = 0; i < 3; i++){
console.log(myFunc(i));
}
Even better:
You can pass in the number of times allowed as well:
function double(num) {
return num * 2;
}
function triple(num) {
return num * 3;
}
function run(func, times){
var count = 0; // this will be trapped in a closure along with func and times
return function() { // this is the function that gets called
count++; // it increments its version of the count variable
if(count <= times) // if count is less than times
return func.apply(this, arguments); // then it calls the function func with whatever arguments passed into it and return the returned value of that call
return "Not available anymore!"; // otherwise (count > times), then it returns a string
}
}
var double2times = run(double, 2); // double2times can only be called 2 times
var triple5times = run(triple, 5); // triple5times can only be called 5 times
for (var i = 0; i < 10; i++){
console.log("Double:", double2times(i));
console.log("Triple:", triple5times(i));
}