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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.
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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);
}
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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);
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I wrote a program to have that particular output in javascript:
i=2 j=1
i=3 j=2
i=1 j=3
i=2 j=1
i=3 j=2
i=1 j=3
and so on...
But the following code never results in the output I need.
c=2;
for (i=c; i<=3; i++){
alert(i);
if(i==1){
j=3;
}else{
j=j-1;
}
alert(j);
if(i==3){
c=1;
continue;
}
}
That's seems to an easy problem : you just have to count 1 by 1 with 2 variables (i and j), the differences is than you don't start at 1 for both.
You need to increment j and i at the same time.
// this for not making infinite loop
var nb_loop=0;
var max_loop=10;
var j=0;
for (var i=2; i<=3 ; i++){
nb_loop++;
j++;
console.log("i="+i+", j="+j); // or alert if you want
if (j>=3)
j=0;
if (i>=3)
i=0;
if (nb_loop>max_loop)
break;
}
Note: a while loop may be better than the "for"
I don't know if u need to use this, i reproduce your output (see console)
var c=1,
j = 2,
i = 0,
x = 2; // this a limit for loop to prevent infinite loop
for (c; c<=3; c++){
if(c == 1){
i = 2;
}else if(c == 2){
i = 3;
}else{
i = 1;
}
//alert("i="+i+" j="+c); uncomment to see alert
console.log("i="+i+" j="+c);
//you can change this condition to stop the loop where you need
if(c==3 && x != 0){
c = 1;
x--;
}
}
see this example, maybe can help:
http://jsfiddle.net/pvMby/1/
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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]);
}
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I can't work out why this only loops through the array once in JavaScript. It should log the second nested array as well.
JSFiddle below and code below;
http://jsfiddle.net/HJfbT/
b = [["one", "is"],
["two", "is"]];
for (var i = 0; i < b.length; i++) {
for (var x = 0; x < b[x].length; x++) {
console.log(b[i][x]);
}
}
Because you have a typo:
// --- should be i ---v
for (var x = 0; x < b[x].length; x++) {
DEMO: http://jsfiddle.net/HJfbT/1/
Use b[i].length in the second loop.
I think is because the inner loop have the running condition with an error.
Is:
for (var x = 0; x < b[x].length; x++)
but must be:
for (var x = 0; x < b[i].length; x++)