update
solution works in foreach loop but not in for loop
function x(number){
return number - 10;
}
var i = 0
var runtimefunctions = {};
var allLevels = {"1":"State","2":"Educational_Services","3":"Principal_Networks","4":"Schools"}
for (var key in allLevels) {
runtimefunctions[i] = function() { return x(i); };
i++;
};
console.log(runtimefunctions[1]()); // -6
console.log(runtimefunctions[2]()); // -6
console.log(runtimefunctions[3]()); // -6
tried hard to make functions but it's first time to create such thing so cant understand the proper way...
I have a function..
function x(number){
return number - 10;
}
runtimefunctions = {};
now I have a loop to run
[1,2,3].forEach(function(y){
//here I want to create a function.. which will make a function x(y) -- like this
runtimefunctions[x] = new Function("return function x_" + levelIterator + "(levelIterator){ console.log(levelIterator); x(" + y + ") }")();
});
so basically..want to make functions like this.
runtimefunctions= {
"1": x(1),
"2": x(2),
and so on
}
Is this what you need?
function x(number){
return number - 10;
}
var runtimefunctions = {};
[1,2,3].forEach(function(y){
runtimefunctions[y] = function() { return x(y); };
});
console.log(runtimefunctions[1]()); // -9
console.log(runtimefunctions[2]()); // -8
console.log(runtimefunctions[3]()); // -7
To satisfy your next (for-in) requirement, you need to closure the index variable with additional function call:
var runtimefunctions = {}, i = 0;
var allLevels = {"1":"State","2":"Educational_Services","3":"Principal_Networks","4":"Schools"}
for (var key in allLevels) {
runtimefunctions[i] = function(index){ return function() { return x(index); } }(i++);
};
It is much easier.
For example:
const createFunctionWith = (x) => {
return (param) => console.log(x, param)
}
let a = [1,2,3].map(x => createFunctionWith(x));
console.log(a[1]("bebe")); // 2, "bebe"
https://jsfiddle.net/muLxoxLd/
You could do something like this
// Found in your code
var x = (a) => {
console.log(a)
};
var runtimefunctions = {};
[1, 2, 3].forEach(function(y) {
//Create a function with a parameter named "levelIterator"
runtimefunctions[y] = Function("levelIterator", "{ console.log(levelIterator); x(" + y + ") }");
});
runtimefunctions[1]('test')
I have tried writing the below code to find sum of 'n' numbers using sum function. I am getting the correct response in output. But i am unable to return that using sum function, as i always have to return a function, which is required for curried effect.
Please help. Thanks in advance.
var output = 0,
chain;
function sum() {
var args = Array.prototype.slice.call(arguments);
output += args.reduce(function(a, b) {
return a + b;
});
sumCurried = sum.bind(output);
sumCurried.val = function() {
return output;
}
return sumCurried;
}
debugger;
document.getElementById('demo').innerHTML = sum(1, 2)(3)(4);
// document.getElementById('demo').innerHTML = sum(1)(3)(4);
<p id='demo'></p>
enter code here
You can add a stop condition to the curried function, for example - if the function is called without an argument return the output:
var output = 0,
chain;
function sum() {
var args = Array.prototype.slice.call(arguments);
if(args.length === 0) {
return output;
}
output += args.reduce(function(a, b) {
return a + b;
});
sumCurried = sum.bind(output);
return sumCurried;
}
console.log(sum(1, 2)(3)(4)());
<p id='demo'></p>
The returned curry function has a val property, which is a function that returns the current value:
var output = 0,
chain;
function sum() {
var args = Array.prototype.slice.call(arguments);
output += args.reduce(function(a, b) {
return a + b;
});
sumCurried = sum.bind(output);
sumCurried.val = function() {
return output;
}
return sumCurried;
}
console.log(sum(1, 2)(3)(4).val());
<p id='demo'></p>
Why would you use currying at all? However, here is a shorter version:
const sum = (...args) => {
const func = (...s)=> sum(...args,...s);
func.value = args.reduce((a,b)=>a+b,0);
return func;
};
//usable as
sum(1,2).value,
sum(1,1)(1).value,
sum(1,1)(1,1)(1,1).value
And you always need to end the currying chain. However, it can be shortified:
func.valueOf = ()=> args.reduce((a,b)=>a+b,0);
//( instead of func.value = ... )
So when called you can do:
+sum(1,2,3)
+sum(1)(1)(1)
I have stacked in the sum of an array. The code is bellow
function User(name,email) {
this.name = name;
this.email = email;
this.cartAmount = [];
this.total = 0;
}
User.prototype = {
constructor: User,
addCart: function(mrp){
this.cartAmount.push(mrp);
},
changeEmail: function(newmail){
this.email = newmail;
},
showCart: function() {
var cart = this.cartAmount.length >0 ? this.cartAmount.join("tk,") : "No product in the cart";
return this.name+" has "+cart+" in his cart.";
},
intotal: function(){
for(var n in this.cartAmount){
this.total += this.cartAmount[n];
return this.total;
}
}
};
athar= new User("Athar Jamil", "atharjamil#gmail.com");
console.log(athar.name);
athar.changeEmail("atharjamil#yahoo.com");
console.log(athar.email);
athar.addCart(20);
athar.addCart(50);
athar.addCart(80);
console.log(athar.intotal());
It shows me only 20 as the result of the sum. What is the problem?
You're returning too early, hence your for loop runs only once and returns the first item in the cart.
Try this instead:
intotal: function(){
for(var n in this.cartAmount){
this.total += this.cartAmount[n];
}
return this.total;
}
Do not use this.total for that. If you call this method several times, your total will increase each time you call it. You should at least put a this.total = 0 at the top of your method.
I personally would have written it like this instead :
intotal: function(){
var out = 0;
for(var n in this.cartAmount){
out += this.cartAmount[n];
}
return out;
}
Using Array.prototype.reduce() you can simplify that function a lot:
intotal: function() {
return this.cartAmount.reduce((a, b)=> a + b)
}
From MDN:
The reduce() method applies a function against an accumulator and each value of the array (from left-to-right) to reduce it to a single value.
Here you pass an arrow function to the reduce method, which takes two arguments: a and b, and returns sum of them.
intotal function returning the first element of cartAmount array.
Place the return statement of intotal function outside of for loop.
function sum(){
var args = Array.prototype.slice.call(arguments);
return args.reduce(function(pre,curr){
if(!isNaN(curr)){
return pre+curr;
}
else
{
throw Error("Non-Numeric arguments"+curr);
}
},0)
}
var result = sum(12,13,14); //pass any number of parameter to sum
alert("Sum Is:"+result);
https://www.npmjs.com/package/mnjs
First import sum function from mnjs package
<script src="https://cdn.jsdelivr.net/npm/mnjs/browser/index.js"></script>
or
const { sum } = require('mnjs')
or
import { sum } from 'mnjs'
Then use sum function as follow:
const cartAmount = [1, 4, 2]
sum(cartAmount) // returns 7
https://runkit.com/dr-montasir/mnjs-sum-function
I am trying to write a function that takes functions as arguments (as many as it gets) and returns them. The function funcArg should return 'Called me'. I used Array.prototype.slice.call(arguments); to create an array but I don't know how to call die functions in that array. Any ideas? Thanks!!
var caller = function() {
return "Called ";
};
var adder = function() {
return " me";
};
var funcArgs = function() {
var myArray = Array.prototype.slice.call(arguments);
}
funcArgs(caller);
funcArgs(calleradder);
You can do this using reduce.
var funcArgs = function() {
var functions = Array.prototype.slice.call(arguments);
return functions.reduce(function(total, f) {
return total + f();
}, '');
};
The way this works if you start off with an array of functions. We then go through each function one at a time. We then call that function and append it to the result of the previous function. Breaking this down into simpler code would look like this:
var funcArgs = function() {
var functions = [caller, adder];
var result = '';
result += functions[0](); // caller();
result += functions[1](); // adder();
return result;
};
If you have an array of functions you can loop over them with forEach.
var caller = function() {
return "Called "
}
var adder = function() {
return " me"
}
var funcArgs = function() {
var myArray = Array.prototype.slice.call(arguments);
myArray.forEach(function (fn) {
console.log(fn())
})
}
funcArgs(caller, adder); // "Called me"
If you want to actually return the values, rather than just console.log them, you can use reduce to return the strings concatenated (or whatever else)
var funcArgs = function() {
var myArray = Array.prototype.slice.call(arguments);
return myArray.reduce(function (acc, fn) {
return acc + fn()
}, '')
}
Using ES5, how do you curry a function that takes infinite arguments.
function add(a, b, c) {
return a + b + c;
}
The function above takes only three arguments but we want our curried version to be able to take infinite arguments.
Hence, of all the following test cases should pass:
var test = add(1);
test(2); //should return 3
test(2,3); //should return 6
test(4,5,6); //should return 16
Here is the solution that I came up with:
function add(a, b, c) {
var args = Array.prototype.slice.call(arguments);
return function () {
var secondArgs = Array.prototype.slice.call(arguments);
var totalArguments = secondArgs.concat(args);
var sum = 0;
for (i = 0; i < totalArguments.length; i++) {
sum += totalArguments[0];
}
return sum;
}
}
However, I have been told that it's not very “functional” in style.
Part of the reason your add function is not very "functional" is because it is attempting to do more than just add up numbers passed to it. It would be confusing for other developers to look at your code, see an add function, and when they call it, get a function returned to them instead of the sum.
For example:
//Using your add function, I'm expecting 6
add(1,2,3) //Returns another function = confusing!
The functional approach
The functional approach would be to create a function that allows you to curry any other functions, and simplify your add function:
function curry(fn) {
var args = Array.prototype.slice.call(arguments, 1);
return function () {
return fn.apply(this, args.concat(
Array.prototype.slice.call(arguments, 0)
));
}
}
function add() {
var args = Array.prototype.slice.call(arguments);
return args.reduce(function (previousValue, currentValue) {
return previousValue + currentValue;
});
}
Now, if you want to curry this function, you would just do:
var curry1 = curry(add, 1);
console.log(
curry1(2), // Logs 3
curry1(2, 3), // Logs 6
curry1(4, 5, 6) // Logs 16
);
//You can do this with as many arguments as you want
var curry15 = curry(add, 1,2,3,4,5);
console.log(curry15(6,7,8,9)); // Logs 45
If I still want to add 1, 2, 3 up I can just do:
add(1,2,3) //Returns 6, AWESOME!
Continuing the functional approach
This code is now becoming reusable from everywhere.
You can use that curry function to make other curried function references without any additional hassle.
Sticking with the math theme, lets say we had a multiply function that multiplied all numbers passed to it:
function multiply() {
var args = Array.prototype.slice.call(arguments);
return args.reduce(function (previousValue, currentValue) {
return previousValue * currentValue;
});
}
multiply(2,4,8) // Returns 64
var curryMultiply2 = curry(multiply, 2);
curryMultiply2(4,8) // Returns 64
This functional currying approach allows you take that approach to any function, not just mathematical ones. Although the supplied curry function does not support all edge cases, it offers a functional, simple solution to your problem that can easily be built upon.
Method 1: Using partial
A simple solution would be to use partial as follows:
Function.prototype.partial = function () {
var args = Array.prototype.concat.apply([null], arguments);
return Function.prototype.bind.apply(this, args);
};
var test = add.partial(1);
alert(test(2)); // 3
alert(test(2,3)); // 6
alert(test(4,5,6)); // 16
function add() {
var sum = 0;
var length = arguments.length;
for (var i = 0; i < length; i++)
sum += arguments[i];
return sum;
}
Method 2: Single Level Currying
If you only want one level of currying then this is what I would do:
var test = add(1);
alert(test(2)); // 3
alert(test(2,3)); // 6
alert(test(4,5,6)); // 16
function add() {
var runningTotal = 0;
var length = arguments.length;
for (var i = 0; i < length; i++)
runningTotal += arguments[i];
return function () {
var sum = runningTotal;
var length = arguments.length;
for (var i = 0; i < length; i++)
sum += arguments[i];
return sum;
};
}
Method 3: Infinite Level Currying
Now, here's a more general solution with infinite levels of currying:
var add = running(0);
var test = add(1);
alert(+test(2)); // 3
alert(+test(2,3)); // 6
alert(+test(4,5,6)); // 16
function running(total) {
var summation = function () {
var sum = total;
var length = arguments.length;
for (var i = 0; i < length; i++)
sum += arguments[i];
return running(sum);
}
summation.valueOf = function () {
return total;
};
return summation;
}
A running total is the intermediate result of a summation. The running function returns another function which can be treated as a number (e.g. you can do 2 * running(21)). However, because it's also a function you can apply it (e.g. you can do running(21)(21)). It works because JavaScript uses the valueOf method to automatically coerce objects into primitives.
Furthermore, the function produced by running is recursively curried allowing you to apply it as many times to as many arguments as you wish.
var resultA = running(0);
var resultB = resultA(1,2);
var resultC = resultB(3,4,5);
var resultD = resultC(6,7,8,9);
alert(resultD + resultD(10)); // 100
function running(total) {
var summation = function () {
var sum = total;
var length = arguments.length;
for (var i = 0; i < length; i++)
sum += arguments[i];
return running(sum);
}
summation.valueOf = function () {
return total;
};
return summation;
}
The only thing you need to be aware of is that sometimes you need to manually coerce the result of running into a number by either applying the unary plus operator to it or calling its valueOf method directly.
Similar to the above problem. Sum of nth level curry by recursion
Trick: To stop the recursion I'm passing last () as blank**
function sum(num1) {
return (num2) => {
if(!num2) {
return num1;
}
return sum(num1 + num2);
}
}
console.log('Sum :', sum(1)(2)(3)(4)(5)(6)(7)(8)())
There is more generic approach by defining a curry function that takes minimum number of arguments when it evaluates the inner function. Let me use ES6 first (ES5 later), since it makes it more transparent:
var curry = (n, f, ...a) => a.length >= n
? f(...a)
: (...ia) => curry(n, f, ...[...a, ...ia]);
Then define a function that sums all arguments:
var sum = (...args) => args.reduce((a, b) => a + b);
then we can curry it, telling that it should wait until at least 2 arguments:
var add = curry(2, sum);
Then it all fits into place:
add(1, 2, 3) // returns 6
var add1 = add(1);
add1(2) // returns 3
add1(2,3) // returns 6
add1(4,5,6) // returns 16
You can even skip creating add by providing the first argument(s):
var add1 = curry(2, sum, 1);
ES5 version of curry is not as pretty for the lack of ... operator:
function curry(n, f) {
var a = [].slice.call(arguments, 2);
return a.length >= n
? f.apply(null, a)
: function () {
var ia = [].slice.call(arguments);
return curry.apply(null, [n, f].concat(a).concat(ia));
};
}
function sum() {
return [].slice.call(arguments).reduce(function (a, b) {
return a + b;
});
};
The rest is the same...
Note: If efficiency is a concern, you may not want to use slice on arguments, but copy it to a new array explicitly.
Bit late in this game, but here is my two cents. Basically this exploits the fact that functions are also objects in JavaScript.
function add(x) {
if (x === undefined) {
return add.numbers.reduce((acc, elem) => acc + elem, 0);
} else {
if (add.numbers) {
add.numbers.push(x);
} else {
add.numbers = [x];
}
}
return add;
}
Infinite sum with currying, you can pass a single parameter or multiple up-to infinite:
function adding(...arg) {
return function clousureReturn(...arg1) {
if (!arguments.length) {
let finalArr = [...arg, ...arg1];
let total = finalArr.reduce((sum, ele) => sum + ele);
return total;
}
return adding(...arg, ...arg1)
}
}
This is my solution for single level currying
function sum() {
let args = [...arguments];
let total = args.reduce((total,num) => total + num,0);
return total;
}
console.log(sum(1,2,3,4)) // 10
and the solution for infinite level currying
let sum= function (...args1) {
let total =args1.reduce((total,num) => total + num,0)
return function(...args2) {
if(args2.length!== 0) {
let total2 = args2.reduce((total,num)=>total + num,0);
return sum(total,total2);
}
return total;
};
};
console.log(sum(2,3,4)(2,3)(1)()); // 15
Simple solution
const add = (one) => { // one: Parameter passed in test
return (...args) => {
// args: Array with all the parameters passed in test
return one + args.reduce((sum, i) => sum + i, 0) // using reduce for doing sum
}
}
var test = add(1);
console.log(test(2)); //should return 3
console.log(test(2, 3)); //should return 6
console.log(test(4, 5, 6)); //should return 16