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n = 1;
while (n < 1024) {
alert(Math.pow(n, 2));
n = n + 2;
}
How can I make this display the powers of 2 from 2^0 to 2^10 inclusive i.e. (1-1024) and end once it reaches 1024 in Javascript
I would do something like this:
for(let n = 0; n <= 10; n++) {
console.log(2 ** n)
}
This logs the answer of each loop iteration. As Pointy suggested in the comment, we have to make the n value iterate from 1 to 10, since that is the variable that generates the output, whereas the 1024 is simply the output that comes from the power values. You could also use Math.pow as follows:
for(let n = 0; n <= 10; n++) {
console.log(Math.pow(2, n))
}
Implementing Secan's suggestion, here is a function which will take any number as a parameter:
function foo(x, exp) {
for(let i = 0; i <= exp; i++) {
console.log(x ** i)
}
}
So to apply this to the original answer, we would call it like this:
foo(2, 10)
Sorry for the ambuiguity with my original question. I've figured it out-
n=0;
while (n <= 10) {
alert(Math.pow(2, n));
n = n + 1;
}
Setting n=0, n<=10 in while and changing n + 1 instead of 2 resolved my issue. Thanks for your responses!
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Hello everybody can you help me get percent with steps?enter image description here
function getPlan(currentProduction, months, percent) {
// write code here
let sum = 0;
for(let i = 0; i < months; i++){
let workCalculate = currentProduction * percent / 100;
sum *= workCalculate;
}
return Math.floor(sum);
}
example:
getPlan(1000, 6, 30) === [1300, 1690, 2197, 2856, 3712, 4825]
getPlan(500, 3, 50) === [750, 1125, 1687]
Simply push each iteration to an array and return that array.
function getPlan(currentProduction, months, percent) {
// write code here
// starting at currentProduction
let sum = currentProduction;
// output
let output = [];
for(let i = 0; i < months; i++){
// progressive from sum and not from currentProduction
let workCalculate = sum * percent / 100;
sum += Math.floor(workCalculate);
output.push(sum)
};
return output
};
console.log(getPlan(1000, 6, 30))
console.log(getPlan(500, 3, 50))
Currently your method returns a number, not an array. What do you need exactly? Do you need it to return an array or you just want to see the intermediate values of the calculation done inside the loop?
In the first case, create an empty array and add the values you want to it in each step of the loop:
function getPlan(currentProduction, months, percent) {
// write code here
let sum = 0;
var result= [];
for(let i = 0; i < months; i++){
let workCalculate = currentProduction * percent / 100;
sum *= workCalculate;
result.push(sum);
}
return result;
}
In the second case, you have 2 options:
Add a console.log so the values are printed to the console.
Add a breaking point so the code stops at it and you can see the values of the variables and go step by step through the execution of the program.
This is a bit vague because your need is unclear, hope it helps though!
function getPlan(currentProduction, months, percent) {
var plan=[];
var workCalculate=currentProduction;
for(var i=0; i<months; i++) {
workCalculate*=(1+percent/100);
plan.push(Math.floor(workCalculate));
}
return plan;
}
console.log(getPlan(1000, 6, 30));
console.log(getPlan(500, 3, 50));
.as-console-wrapper { max-height: 100% !important; top: 0; }
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I would like to sum up all the elements inside the array covertedValue. Why I am getting a Nan result? whats wrong with the recursive function I have written?
function findOutlier(integers){
var covertedValue = integers.map(x => x % 2);
var total = 0;
for (i=0 ; i<covertedValue.length ; i++){
total = total + covertedValue[0] + findOutlier(integers.splice(1));
}
console.log(total);
}
findOutlier([0, 1, 2]);
because this line create a function and a private alias to itself
var factorial =
// create a variable named factorial in global scope
// it contains the function fac
function fac(n) {
// fac is a function available only in the factorial var context
return n < 2 ? // if n < 2
1 : // return 1
n * fac(n - 1); // else return n * fac(n - 1)
};
console.log(factorial(5));
// calls factorial(5) which is 5*fac(4) which is 5*(4*fac(3)) ...
note that the fac function is not available outside of itself
var factorial = function fac(n) { return n < 2 ? 1 : n * fac(n - 1); };
console.log(fac(5));
This way of coding where a function call itself is called recursive programming it can be hard to grasp at first but can be VERY powerful in some situations
recurcive programming can easily be used when the result of func(n) depend directly on the result of func(n+x)
the cond ? then : else structure is called ternary operator, it's a way to make if in one line, can become messy very quick
as you noted in a comment : recursion can be replaced by a loop (it can be hard to do it sometime)
here's a more beginner way to write it
function factorial(n) {
if(n < 2) {
return 1
} else {
let fac = 1;
for(var i = n; i > 0; i--) {
fac = fac * i
// here fac takes the value of the precedent iteration
}
// this whole loop is the same as i*fac(i-1) in the recurcive way
return fac
}
};
console.log(factorial(5));
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I have a test, where I needed to code for multiplying two numbers without using multiplication,
the code is as follows,
function multiply(num,toNum){
var product = 0;
for(var i = 1; i <= toNum; i++){
product += num;
}
return product;
}
console.log(multiply(2,5));
The output is
rahul#rahul:~/myPractise/Algo$ node MultiplyWithoutLoop.js
10
rahul#rahul:~/myPractise/Algo$
Is the above code satisfactory or need is there a room for improvement.
Can a better logic be applied.
Hey,
I solved it using recursion,
this is the code,
function multiply01(num,toNum){
var product = num;
return (toNum >= 1) ? product + multiply01(product,--toNum) : 0;
}
Compact way:
function multiply(a, b) {
return a / (1 / b);
}
console.log(multiply(2, 5)); // 10
You could use addition for odd numbers and and bit shifting. Better known as Ancient Egyptian multiplication.
The value of b is summed, if a is odd. Then a is divided by 2 and the integer part is assigned. b is doubled.
Example:
a b sum
--- --- ---
5 4 4 add 4
2 8 4
1 16 20 add 16
0 32 20 <- result
function multiply(a, b) {
var sum = 0;
while (a) {
if (a & 1) {
sum += b;
}
a >>= 1;
b <<= 1;
}
return sum;
}
console.log(multiply(5, 4));
console.log(multiply(3, 7));
console.log(multiply(191, 7));
Please try this, it will also handle negative value
function multiply(x, y)
{
/* Add x one by one */
if(y > 0 )
{
return (x + multiply(x, y-1));
}
/* the case where y is negative */
else if(y < 0 )
{
return -multiply(x, -y);
}
return 0;
}
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var x = function (a) { return a + a/4 - 600}
I have a function that does something.
So now my goal is to repeat this function 12 times with an argument used from previous operation;
Let's say initial ammount is 5000;
So
`x(5000) = 5650;
x(5650) =6462.5;
x(6462.) =...;
and this should be repeated 12 times;
So how can this be done in code?
`
Certainly the easiest way would be to use a for loop:
var x = function(a) { return a + a/4 - 600 },
v = 5000;
for (var i = 0; i < 12; i++) {
v = x(v);
}
console.log(v);
I'd not recommend doing something like this, but instead of a loop you could use recursion:
var x = function(42.34, 12);
----
function(double a, int maxSize)
{
a = ((a + a) / (4 - 600));
if(maxSize>0){
maxSize--;
return function(a,maxSize);
}else{
return a;
}
}
(just pseudocode)
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I am trying to write a script in shorter way but it doesn't work..
It is not really clear as i see.. so: here all of the codes..
$(document).ready(function(){
$('button').click(function(){
var benen = $('[name=benen]').val();
var zadelmaat = benen * 1.89;
var kadermaat = benen * 0.69;
var remainder = zadelmaat % 2;
if(remainder % 2 == 0)
{
return zadelmaat;
}else
{
zadelmaat = zadelmaat - remainder;
}
var remainder = kadermaat % 2;
if(remainder % 2 == 0)
{
return kadermaat;
}else
{
kadermaat = kadermaat - remainder;
}
$('.zadelmaat').append(zadelmaat);
$('.kadermaat').append(kadermaat);
return false;
});
});
That works for me! But what i am trying to change is to write a simple function to get rid of the remainder(in case)!
I hope it s more clear now...
Thnx!
This will do it:
function getRemainder(x) {
var remainder = x % 2;
return x - remainder;
}
You forgot to return the result in case the remainder was not 0. And you don't need to explicitly test for that, as x - 0 == x.
You pass a value to the function, not a pointer. To make it work you must add return x; to the else clause.
Change your function to this:
you only return a different value if it is not ==0
function getRemainder(x) {
var remainder = x % 2;
if(x % 2 != 0) {
x = x - remainder;
}
return x
}
Someone edited your code which seems to be an invalid edit as it really changed the question/problem.
Simplified code for function:
function getRemainder (x) {
return x % 2 === 0 ? x : x = x - x % 2;
};
I made some assumptions from the original code in a sample fiddle: http://jsfiddle.net/eqQzM/
alerts "28:6"