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I am not getting proper output with nested for-loop in javascript [closed]

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Closed 3 years ago.
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I'm supposed to get the multiplication value of the multi-dimensional array. but I am getting '1' as output whatever values being changed in array.
function arrayMultiplyer(arr){
var multi = 1;
for(var i=0;i < arr.length; i++){
for(var j =0; j<arr[i];j++){
multi *= arr[i][j];
}
}
return multi;
}
var multi = arrayMultiplyer([[2,33],[33,2],[5,6,9]]);
console.log(multi);

You need to check arr[i].length in the j loop.
function arrayMultiplyer(arr){
var multi = 1;
for(var i=0;i < arr.length; i++){
for(var j =0; j<arr[i].length;j++){ // you need to check arr[i].length here
multi *= arr[i][j];
}
}
return multi;
}
var multi = arrayMultiplyer([[2,33],[33,2],[5,6,9]]);
console.log(multi);

You can simply do this in two lines.
function arrayMultiplyer(arr){
let flattenedArray = arr.flat();
return flattenedArray.reduce((x, y) => x * y);
}
var multi = arrayMultiplyer([[2,33],[33,2],[5,6,9]]);
console.log(multi);

An alternative solution using .reduce()
function arrayMultiplier(arr) {
return arr.reduce((tot,arr2) =>
arr2.reduce((subTot, n) =>
subTot * n
, tot)
, 1);
}

Javascript Question: Sum of all odd numbers to 100 [closed]

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I am trying to find out the sum off all odd numbers from 1 to 100 using a for loop. Here is my code so far:
var sum = 0;
for (var i = 1; i <= 100; i++) {
if (i % 2 == 1);
sum += i;
}
console.log(sum);
I thought maybe using a while loop maybe the solution, but whenever I try it, I always get an infinite loop error. So I tried switching back to a for loop, but the answer does not come out right. I believe I am looping through each number instead of every odd number. Is my "if" condition not correct? Any help is appreciated.
Thank You

You have a ; just after your if statement, so the next line is not going to get executed.
Just removing the ; should work:
var sum = 0;
for (var i = 1; i <= 100; i++) {
if (i % 2 == 1)
sum += i;
}
console.log(sum);

use step 2
var sum = 0;
for (var i = 1; i < 100; i+=2) {
sum += i;
}
console.log(sum);

i want to create an empty array to fill it later [closed]

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Closed 4 years ago.
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hi guys i tryed to create an empty array to fill it later. I post my code so you can help me. XD
"use strict"
var array = new array(6);
for(var i = 0; i <= 5; i++){
do {
var number = prompt("Put the element" + (i+1), 0);
}while(isNaN(number));
arr.push(number);
};
write.data(array);

It appears you may have copy and pasted some code without understanding what it is doing.
You are trying to push to an array, but you've declared your array as array, but trying to push to arr. Which is why it's not working.
var arr = new Array();
for(var i = 0; i <= 5; i++){
arr.push(i + 1);
}
console.log(arr);
alert(arr);
EDIT * You actually don't even need to declare the new Array(6), you can just use new Array() to push. However, if you would like to declare the size, you can do this instead.
var arr = new Array(6);
for(var i = 0; i <= 5; i++){
arr[i] = i + 1;
}
console.log(arr);
alert(arr);

You can also use this:
var arr = []; // create an empty object
for(var i = 0; i <= 5; i++){
arr.push(i + 1); // fill the object
}
console.log(arr);
alert(arr); // shows 1,2,3,4,5,6
This does what you are trying to achieve.

You can do it like
var a = new Array(6);
for(var i = 0;i < 6;i++){
a.push(i+1);
}
console.log(a);

Javascript nested array and setting the variable [closed]

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for (i = 0; i < length1; i++) {
for (j = 0; j < length2; j++) {
for (k = j; k < length2; k++) {}
}
}
I just wondered... in the last loop where I initialize k to be equal to j, would it cause problems for me later on?
I'm assuming every time j increments by 1, k is set to the new value of j and then increments up to length2

You are correct that for each iteration of the j-for-loop, the innermost loop will begin at J's current value and iterate up to length2-1. If this is what you want, shouldn't be a problem.

Javascript substring not working as expected [closed]

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var test = "abcdefghijklmnopqrstuvwxyz";
for(i = 0; i < test.length; i++) {
alert(test.substring(i,1));
}
I expected each alert to return each letter of the alphabet individually.
Instead, the first 5 alerts displayed as follows. Why?
a
b
bc
bcd
bcde

var test = "abcdefghijklmnopqrstuvwxyz";
for(i = 0; i < test.length; i++) {
console.log(test.substring(i,i+1));
}
actually, it's
substring(start, end)
not
substring(start, length)
unlike substr, which is indeed, substr(start, length)

If "start" is greater than "end", this method (substring) will swap the two arguments, meaning str.substring(1,4) == str.substring(4,1).
Use:
for(i = 0; i < test.length; i++) {
alert(test[i]);
}