I need a js sum function to work like this:
sum(1)(2) = 3
sum(1)(2)(3) = 6
sum(1)(2)(3)(4) = 10
etc.
I heard it can't be done. But heard that if adding + in front of sum can be done.
Like +sum(1)(2)(3)(4). Any ideas of how to do this?
Not sure if I understood what you want, but
function sum(n) {
var v = function(x) {
return sum(n + x);
};
v.valueOf = v.toString = function() {
return n;
};
return v;
}
console.log(+sum(1)(2)(3)(4));
JsFiddle
This is an example of using empty brackets in the last call as a close key (from my last interview):
sum(1)(4)(66)(35)(0)()
function sum(firstNumber) {
let accumulator = firstNumber;
return function adder(nextNumber) {
if (nextNumber === undefined) {
return accumulator;
}
accumulator += nextNumber;
return adder;
}
}
console.log(sum(1)(4)(66)(35)(0)());
I'm posting this revision as its own post since I apparently don't have enough reputation yet to just leave it as a comment. This is a revision of #Rafael 's excellent solution.
function sum (n) {
var v = x => sum (n + x);
v.valueOf = () => n;
return v;
}
console.log( +sum(1)(2)(3)(4) ); //10
I didn't see a reason to keep the v.toString bit, as it didn't seem necessary. If I erred in doing so, please let me know in the comments why v.toString is required (it passed my tests fine without it). Converted the rest of the anonymous functions to arrow functions for ease of reading.
New ES6 way and is concise.
You have to pass empty () at the end when you want to terminate the call and get the final value.
const sum= x => y => (y !== undefined) ? sum(x + y) : x;
call it like this -
sum(10)(30)(45)();
Here is a solution that uses ES6 and toString, similar to #Vemba
function add(a) {
let curry = (b) => {
a += b
return curry
}
curry.toString = () => a
return curry
}
console.log(add(1))
console.log(add(1)(2))
console.log(add(1)(2)(3))
console.log(add(1)(2)(3)(4))
Another slightly shorter approach:
const sum = a => b => b? sum(a + b) : a;
console.log(
sum(1)(2)(),
sum(3)(4)(5)()
);
Here's a solution with a generic variadic curry function in ES6 Javascript, with the caveat that a final () is needed to invoke the arguments:
const curry = (f) =>
(...args) => args.length? curry(f.bind(0, ...args)): f();
const sum = (...values) => values.reduce((total, current) => total + current, 0)
curry(sum)(2)(2)(1)() == 5 // true
Here's another one that doesn't need (), using valueOf as in #rafael's answer. I feel like using valueOf in this way (or perhaps at all) is very confusing to people reading your code, but each to their own.
The toString in that answer is unnecessary. Internally, when javascript performs a type coersion it always calls valueOf() before calling toString().
// invokes a function if it is used as a value
const autoInvoke = (f) => Object.assign(f, { valueOf: f } );
const curry = autoInvoke((f) =>
(...args) => args.length? autoInvoke(curry(f.bind(0, ...args))): f());
const sum = (...values) => values.reduce((total, current) => total + current, 0)
curry(sum)(2)(2)(1) + 0 == 5 // true
Try this
function sum (...args) {
return Object.assign(
sum.bind(null, ...args),
{ valueOf: () => args.reduce((a, c) => a + c, 0) }
)
}
console.log(+sum(1)(2)(3,2,1)(16))
Here you can see a medium post about carried functions with unlimited arguments
https://medium.com/#seenarowhani95/infinite-currying-in-javascript-38400827e581
Try this, this is more flexible to handle any type of input. You can pass any number of params and any number of paranthesis.
function add(...args) {
function b(...arg) {
if (arg.length > 0) {
return add(...[...arg, ...args]);
}
return [...args, ...arg].reduce((prev,next)=>prev + next);
}
b.toString = function() {
return [...args].reduce((prev,next)=>prev + next);
}
return b;
}
// Examples
console.log(add(1)(2)(3, 3)());
console.log(+add(1)(2)(3)); // 6
console.log(+add(1)(2, 3)(4)(5, 6, 7)); // 28
console.log(+add(2, 3, 4, 5)(1)()); // 15
Here's a more generic solution that would work for non-unary params as well:
const sum = function (...args) {
let total = args.reduce((acc, arg) => acc+arg, 0)
function add (...args2) {
if (args2.length) {
total = args2.reduce((acc, arg) => acc+arg, total)
return add
}
return total
}
return add
}
document.write( sum(1)(2)() , '<br/>') // with unary params
document.write( sum(1,2)() , '<br/>') // with binary params
document.write( sum(1)(2)(3)() , '<br/>') // with unary params
document.write( sum(1)(2,3)() , '<br/>') // with binary params
document.write( sum(1)(2)(3)(4)() , '<br/>') // with unary params
document.write( sum(1)(2,3,4)() , '<br/>') // with ternary params
ES6 way to solve the infinite currying. Here the function sum will return the sum of all the numbers passed in the params:
const sum = a => b => b ? sum(a + b) : a
sum(1)(2)(3)(4)(5)() // 15
function add(a) {
let curry = (b) => {
a += b
return curry;
}
curry[Symbol.toPrimitive] = (hint) => {
return a;
}
return curry
}
console.log(+add(1)(2)(3)(4)(5)); // 15
console.log(+add(6)(6)(6)); // 18
console.log(+add(7)(0)); // 7
console.log(+add(0)); // 0
Here is another functional way using an iterative process
const sum = (num, acc = 0) => {
if !(typeof num === 'number') return acc;
return x => sum(x, acc + num)
}
sum(1)(2)(3)()
and one-line
const sum = (num, acc = 0) => !(typeof num === 'number') ? acc : x => sum(x, acc + num)
sum(1)(2)(3)()
You can make use of the below function
function add(num){
add.sum || (add.sum = 0) // make sure add.sum exists if not assign it to 0
add.sum += num; // increment it
return add.toString = add.valueOf = function(){
var rtn = add.sum; // we save the value
return add.sum = 0, rtn // return it before we reset add.sum to 0
}, add; // return the function
}
Since functions are objects, we can add properties to it, which we are resetting when it's been accessed.
we can also use this easy way.
function sum(a) {
return function(b){
if(b) return sum(a+b);
return a;
}
}
console.log(sum(1)(2)(3)(4)(5)());
To make sum(1) callable as sum(1)(2), it must return a function.
The function can be either called or converted to a number with valueOf.
function sum(a) {
var sum = a;
function f(b) {
sum += b;
return f;
}
f.toString = function() { return sum }
return f
}
function sum(a){
let res = 0;
function getarrSum(arr){
return arr.reduce( (e, sum=0) => { sum += e ; return sum ;} )
}
function calculateSumPerArgument(arguments){
let res = 0;
if(arguments.length >0){
for ( let i = 0 ; i < arguments.length ; i++){
if(Array.isArray(arguments[i])){
res += getarrSum( arguments[i]);
}
else{
res += arguments[i];
}
}
}
return res;
}
res += calculateSumPerArgument(arguments);
return function f(b){
if(b == undefined){
return res;
}
else{
res += calculateSumPerArgument(arguments);
return f;
}
}
}
let add = (a) => {
let sum = a;
funct = function(b) {
sum += b;
return funct;
};
Object.defineProperty(funct, 'valueOf', {
value: function() {
return sum;
}
});
return funct;
};
console.log(+add(1)(2)(3))
After looking over some of the other solutions on here, I would like to provide my two solutions to this problem.
Currying two items using ES6:
const sum = x => y => (y !== undefined ) ? +x + +y : +x
sum(2)(2) // 4
Here we are specifying two parameters, if the second one doesnt exist we just return the first parameter.
For three or more items, it gets a bit trickier; here is my solution. For any additional parameters you can add them in as a third
const sum = x => (y=0) => (...z) => +x + +y + +z.reduce((prev,curr)=>prev+curr,0)
sum(2)()()//2
sum(2)(2)()//4
sum(2)(2)(2)//6
sum(2)(2)(2,2)//8
I hope this helped someone
Related
first part sums two numbers, second part checks whether the sum of the numbers is odd or even and the final part prints out the message but in this case it prints out 8 instead of even
let sum = (a, b) => {
return a + b};
let evaluate = (c) => {
if (c % 2 == 0) {
console.log("Even")
} else {
console.log("odd")
}
return c;
};
let Hello = (name, status) => {
let message = `Hi I am ${name} , and the number is ${status}`;
return message;
};
console.log( Hello("David", evaluate(sum(5, 3))));
As Sergey mentions above, returning a console.log is not the same as returning a string.
I've re-written your code to return a string and it should now behave as expected.
let sum = (a, b) => {
return a + b};
let evaluate = (c) => {
if (c % 2 == 0) {
return "Even";
}
return "Odd";
};
let Hello = (name, status) => {
let message = `Hi I am ${name} , and the number is ${status}`;
return message;
};
console.log( Hello("David", evaluate(sum(5, 3))));
I am a beginner of JavaScript. I am trying to understand the functions and promises in JS. I wrote a small code.
code Purpose: I wrote this code to understand the working of promise function(then) on function with a return value.
input: a,b
output: if the sum of a & b is 0, then show the alert statement else show sum in the alert value.
const func = (a, b) => {
let operand1 = a * 10;
let operand2 = b * 10;
return operand1 + operand2
};
const funcwrapper = (a, b) => {
func(a, b).then((sum) => {
if (sum == 0) {
window.alert("value is zero");
} else {
window.alert(("sum is " + sum));
}
})
};
funcwrapper(5, 5);
I am still confused with this promise on functions with some return value after going through lot of investigation. So finally tried to implement the code to understand and got stuck with this error.
Please help me in understanding this topic.
Your help is appreciated.
Thanks in advance
func() needs to return a promise:
const func = (a, b) => {
let operand1 = a * 10;
let operand2 = b * 10;
return new Promise((resolve, reject) => resolve(operand1 + operand2));
};
const funcwrapper = (a, b) => {
func(a, b).then((sum) => {
if (sum == 0) {
window.alert("value is zero");
} else {
window.alert(("sum is " + sum));
}
})
};
funcwrapper(5, 5);
I faced this question in one interview. I did not get how to solve this.
Question: Write a sum function which will add 2 numbers, but numbers can be passed to a function in following ways:
sum(3)(4) // answer should be 7
sum(3)()(4)//answer should be 7
sum(3)()()()()(4) //answer should b 7
I can solve first function using closure, in fact for the second function also I can check the arguments and if arguments length is zero I can again make a call to sum to except next parameter.
But how to make it generic ? Means even your first parameter and last parameter has 'N' number of calls & those can be empty or parameterized, it should return sum.
Recorded a video how to solve it:
https://youtu.be/7hnYMIOVEg0
Text answer:
function sum(numberOne) {
return function innerSum(numberTwo) {
if (typeof(numberTwo) === 'number') {
return numberOne + numberTwo;
}
return innerSum;
}
}
Output:
sum(3)(4); => 7
sum(5)()()(10); => 15
Basically, you need to return inner function (innerSum) up until you receive a value - then you return number.
You could also choose another name - like _sum(), or addToFirstNumber() for your method.
You can always return a function from within a function:
let a;
function sum(value) {
if (typeof value !== "number") {
return sum;
}
if (typeof a !== "number") {
a = value;
return sum;
}
let result = a + value;
a = null;
return result;
}
see https://jsfiddle.net/d9tLh11k/1/
function sum(num1) {
return function sum2(num2) {
if(num2 === undefined) {
return sum2;
}
return num1 + num2;
}
}
console.log(sum(4)()()()(3)); // -> 7
Or in ES6:
const add = num1 => num2 => num2 === undefined ? add(num1) : num1 + num2;
console.log(add(4)()()()()()(3)); // -> 7
function add (n) {
var func = function (x) {
if(typeof x==="undefined"){
x=0;
}
return add (n + x);
};
func.valueOf = func.toString = function () {
return n;
};
return func;
}
console.log(+add(1)(2)(3)()()(6));
I have already given answer of this question Here
but according to your question I have modified that
function add (n) {
var func = function (x) {
if(typeof x==="undefined"){
x=0;
}
return add (n + x);
};
func.valueOf = func.toString = function () {
return n;
};
return func;
}
console.log(+add(1)(2)(3)()()(6));
Run code like that
console.log(+add(1)(2)(3)()()(6));
This should do it
function sum(num1) {
if (arguments.length === 0){
return sum;
}
return function innerSum(num2) {
if (arguments.length > 0){
return num1 + num2;
}else{
return innerSum;
}
}
}
You can do this in a number of ways, but mostly you'll want named recursion. That is, you can have something like:
sum = start => (...args) => args.length? args[0] + start : sum(start)
but it might look cleaner to write this as:
function sum(start) {
function out(...args) {
return args.length? start + args[0] : out
}
return out;
}
function add(n) {
return function(x) {
this.value = n;
return this.value + x;
}
}
var add3 = add(3);
var add4 = add(4);
I'm trying to figure out if I can re-write this function above to:
Allow to modify the first parameter like this:
add3.value(1);
console.log(add3(4))
// 5
Also so that the value method returns the current value if no parameter is passed:
add3.value(1);
console.log(add3.value())
// 1
Functions are just objects, so you can assign a property value to it, whose value is a function that modifies n:
function add(n) {
var result = function(x) {
return n + x;
};
result.value = function(x) {
if (typeof x === 'undefined') {
return n;
}
n = x;
};
return result;
}
Why you'd want to do that I don't know, but it's possible.
To use the interface you've described, your inner function will need to return a object with a 'value' method.
eg.
function add (n) {
return {
value: function (x) {
x = x || 0;
return n + x;
}
};
}
But, what it sounds like you're trying to do is currying & partial application. In which case, returning a function is probably simpler.
function add (n) {
return function (x) {
x = x || 0;
return n + x;
};
}
This would be used like so:
var add3 = add(3);
console.log(add3(4)); // output: 7
or, with the default value:
var add4 = add(4);
console.log(add4()); // output: 4
Similar to Felix's answer, but permitting empty inputs, remembering last values and also giving it a valueOf so it can be used like a number itself.
function adder(i) {
i || (i = 0);
var add = function (x) {
return i += (+x || 0); // += modifies i
};
add.value = function (x) {
return !x && x !== 0 ? i : i = (+x || 0); // if x falsy but not zero
};
add.valueOf = function () {
return i;
};
return add;
}
var add = adder();
add.value(1); // 1
add(4); // 5
add(5); // 10
add + 2; // 12, via valueOf
add(); // 10
Here is the problem. I have to implement make function:
var sum = function (a, b) { return a + b; }
var mult = function (a, b) { return a * b; }
//'make' function goes here
var res = make(1)(2)(3)(4);
console.log(res(sum)); //OUTPUT: 10
console.log(res(mult)); //OUTPUT: 24
I have implemented it, but I feel like a little better way still exists. :)
So, here is my solution:
function make(a, arr) {
if (a instanceof Function) { return arr.reduce(a); }
arr = arr || [];
arr.push(a);
return function (b) { return make(b, arr); };
}
You want functions? You can have functions!
function sum(a, b) { return a + b; }
function mult(a, b) { return a * b; }
function diff(a, b) { return a - b; }
function make(x, f) {
if (typeof x === 'function') return f(x);
return function(y) {
return make(y, function(a) {
return f ? a(f(a), x) : x;
});
};
}
console.log(make(1)(2)(3)(4)(sum)); // -> 10
console.log(make(1)(2)(3)(4)(mult)); // -> 24
console.log(make(4)(3)(2)(1)(diff)); // -> -2
Instead of building up an array, this builds up a function that, when given a function, reduces all the elements using that function, from left to right :)
I'm pretty sure this wouldn't qualify as code that anyone would ever want to see in their codebase.
function make(inp, arr) {
if (typeof inp == 'function') {
while(arr.length > 1)
arr.push(inp(arr.pop(), arr.pop()));
return arr[0];
}
if (!arr) arr = [];
arr.push(inp);
return function (inp1) {
return make(inp1, arr);
};
}
Here's a better JavaScript specific solution:
var test = make(1)(2)(3)(4);
alert(test(add)); // ((1 + 2) + 3) + 4 = 10
alert(test(mul)); // ((1 * 2) * 3) * 4 = 24
alert(test(sub)); // ((1 - 2) - 3) - 4 = -8
function make(x) {
if (typeof x !== "function") return foldl([x]);
throw new TypeError("Initial value can't be a function");
}
function foldl(array) {
return function (fx) {
return typeof fx !== "function" ?
foldl(array.concat(fx)) :
array.reduce(fx);
};
}
function add(a, b) { return a + b; }
function mul(a, b) { return a * b; }
function sub(a, b) { return a - b; }
There are several advantages to this method:
Better error reporting when you write incorrect code like make(add).
Simple functional definition which is easy to read and understand.
If you want foldr then just rename reduce to reduceRight.
Using concat instead of push to preserve original array.
I can't think of a better functional implementation for make.