Test whether all array elements are factors of a number
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I have the following question:
Write a function that returns true if all integers in an array are factors of a number, and false otherwise.
I tried the code below:
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
else {
return true
}
}
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.
javascript arrays loops if-statement
add a comment |
I have the following question:
Write a function that returns true if all integers in an array are factors of a number, and false otherwise.
I tried the code below:
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
else {
return true
}
}
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.
javascript arrays loops if-statement
2
You should get thereturn true
out of the loop ;) and leave thereturn false
inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument
– Icepickle
Apr 6 at 13:07
2
FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)
– T.J. Crowder
Apr 6 at 13:09
1
@t.j.crowder yet we don't have a good dupetarget for it.
– Jonas Wilms
Apr 6 at 13:14
2
@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)
– T.J. Crowder
Apr 6 at 13:18
@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.
– Barmar
Apr 6 at 18:43
add a comment |
I have the following question:
Write a function that returns true if all integers in an array are factors of a number, and false otherwise.
I tried the code below:
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
else {
return true
}
}
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.
javascript arrays loops if-statement
I have the following question:
Write a function that returns true if all integers in an array are factors of a number, and false otherwise.
I tried the code below:
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
else {
return true
}
}
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
else {
return true
}
}
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
else {
return true
}
}
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
javascript arrays loops if-statement
javascript arrays loops if-statement
edited Apr 6 at 16:50
isanae
2,55711437
2,55711437
asked Apr 6 at 13:04
PineNuts0PineNuts0
87631432
87631432
2
You should get thereturn true
out of the loop ;) and leave thereturn false
inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument
– Icepickle
Apr 6 at 13:07
2
FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)
– T.J. Crowder
Apr 6 at 13:09
1
@t.j.crowder yet we don't have a good dupetarget for it.
– Jonas Wilms
Apr 6 at 13:14
2
@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)
– T.J. Crowder
Apr 6 at 13:18
@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.
– Barmar
Apr 6 at 18:43
add a comment |
2
You should get thereturn true
out of the loop ;) and leave thereturn false
inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument
– Icepickle
Apr 6 at 13:07
2
FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)
– T.J. Crowder
Apr 6 at 13:09
1
@t.j.crowder yet we don't have a good dupetarget for it.
– Jonas Wilms
Apr 6 at 13:14
2
@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)
– T.J. Crowder
Apr 6 at 13:18
@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.
– Barmar
Apr 6 at 18:43
2
2
You should get the
return true
out of the loop ;) and leave the return false
inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument– Icepickle
Apr 6 at 13:07
You should get the
return true
out of the loop ;) and leave the return false
inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument– Icepickle
Apr 6 at 13:07
2
2
FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)
– T.J. Crowder
Apr 6 at 13:09
FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)
– T.J. Crowder
Apr 6 at 13:09
1
1
@t.j.crowder yet we don't have a good dupetarget for it.
– Jonas Wilms
Apr 6 at 13:14
@t.j.crowder yet we don't have a good dupetarget for it.
– Jonas Wilms
Apr 6 at 13:14
2
2
@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)
– T.J. Crowder
Apr 6 at 13:18
@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)
– T.J. Crowder
Apr 6 at 13:18
@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.
– Barmar
Apr 6 at 18:43
@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.
– Barmar
Apr 6 at 18:43
add a comment |
5 Answers
5
active
oldest
votes
Just place return true out of for loop,
If you keep return true
in else part
as soon as any of value which does not satisfies num % element !== 0
your code will return true
which should not happen in this case as you're checking for all the values in array should be factor of given number
Let's understand by 1st example
- On first element in array
1
it will check if conditionnum % element !== 0
which turns out false, so it will go to else condition andreturn true
from function and will not check for rest of values. - So you need to keep
return true
at the end so if any of the value in loop doesn't satisfy the if condition than only control will go toreturn true
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))
In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like
- keep the
failing case
return value inside for loop - keep the
passing case
return value at the end of function
JS have a inbuilt method Array.every for such cases
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
add a comment |
You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:
Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹
You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.
You realize that your boss actually wanted you to do:
Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²
Or in code:
// ¹
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
return true;
} else {
return false;
}
}
}
// ²
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
continue; // this could be omitted, as a loop keeps looping nevertheless
} else {
return false;
}
}
return true;
}
As this is a very common task in programming, there is already a shorter way to express this:
if(chocolates.every(isTasty)) {
alert("all chocolates are fine");
} else {
alert("Oh, that doesnt taste good");
}
whereas isTasty
is a function taking a chocolate and returning either true or false.
If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)
1
A tiny quibble - it's "chocolate", not "choclate".
– Wai Ha Lee
Apr 6 at 13:25
1
@waiHaLee oh, pronounciation tricked me ...
– Jonas Wilms
Apr 6 at 13:28
add a comment |
Inside the loop the input num
was was tested for divisibility, if the num
was divisible the control was going in the else
block from where the function returned true
.
The loop was not checking for all numbers of the input array it was returning true
when the first number was divisible.
Just use a flag variable to see if all of the elements are divisible by the input number num
, if any one is not divisible the flag
will set to false
and then we can break
out of the loop and return it as there is no point to check the other numbers.
function checkFactors(factors, num) {
let flag = true;
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
flag = false;
break;
}
}
return flag;
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
You can also use Array.every
to check the same in a concise way:
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
add a comment |
Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
}
return true;
}
1
He is looking for an explanation as well though
– Icepickle
Apr 6 at 13:10
I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.
– javapedia.net
Apr 6 at 13:12
add a comment |
Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.
New contributor
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Just place return true out of for loop,
If you keep return true
in else part
as soon as any of value which does not satisfies num % element !== 0
your code will return true
which should not happen in this case as you're checking for all the values in array should be factor of given number
Let's understand by 1st example
- On first element in array
1
it will check if conditionnum % element !== 0
which turns out false, so it will go to else condition andreturn true
from function and will not check for rest of values. - So you need to keep
return true
at the end so if any of the value in loop doesn't satisfy the if condition than only control will go toreturn true
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))
In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like
- keep the
failing case
return value inside for loop - keep the
passing case
return value at the end of function
JS have a inbuilt method Array.every for such cases
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
add a comment |
Just place return true out of for loop,
If you keep return true
in else part
as soon as any of value which does not satisfies num % element !== 0
your code will return true
which should not happen in this case as you're checking for all the values in array should be factor of given number
Let's understand by 1st example
- On first element in array
1
it will check if conditionnum % element !== 0
which turns out false, so it will go to else condition andreturn true
from function and will not check for rest of values. - So you need to keep
return true
at the end so if any of the value in loop doesn't satisfy the if condition than only control will go toreturn true
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))
In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like
- keep the
failing case
return value inside for loop - keep the
passing case
return value at the end of function
JS have a inbuilt method Array.every for such cases
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
add a comment |
Just place return true out of for loop,
If you keep return true
in else part
as soon as any of value which does not satisfies num % element !== 0
your code will return true
which should not happen in this case as you're checking for all the values in array should be factor of given number
Let's understand by 1st example
- On first element in array
1
it will check if conditionnum % element !== 0
which turns out false, so it will go to else condition andreturn true
from function and will not check for rest of values. - So you need to keep
return true
at the end so if any of the value in loop doesn't satisfy the if condition than only control will go toreturn true
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))
In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like
- keep the
failing case
return value inside for loop - keep the
passing case
return value at the end of function
JS have a inbuilt method Array.every for such cases
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
Just place return true out of for loop,
If you keep return true
in else part
as soon as any of value which does not satisfies num % element !== 0
your code will return true
which should not happen in this case as you're checking for all the values in array should be factor of given number
Let's understand by 1st example
- On first element in array
1
it will check if conditionnum % element !== 0
which turns out false, so it will go to else condition andreturn true
from function and will not check for rest of values. - So you need to keep
return true
at the end so if any of the value in loop doesn't satisfy the if condition than only control will go toreturn true
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))
In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like
- keep the
failing case
return value inside for loop - keep the
passing case
return value at the end of function
JS have a inbuilt method Array.every for such cases
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}
console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
edited Apr 6 at 16:22
answered Apr 6 at 13:07
Code ManiacCode Maniac
11.8k2834
11.8k2834
add a comment |
add a comment |
You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:
Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹
You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.
You realize that your boss actually wanted you to do:
Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²
Or in code:
// ¹
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
return true;
} else {
return false;
}
}
}
// ²
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
continue; // this could be omitted, as a loop keeps looping nevertheless
} else {
return false;
}
}
return true;
}
As this is a very common task in programming, there is already a shorter way to express this:
if(chocolates.every(isTasty)) {
alert("all chocolates are fine");
} else {
alert("Oh, that doesnt taste good");
}
whereas isTasty
is a function taking a chocolate and returning either true or false.
If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)
1
A tiny quibble - it's "chocolate", not "choclate".
– Wai Ha Lee
Apr 6 at 13:25
1
@waiHaLee oh, pronounciation tricked me ...
– Jonas Wilms
Apr 6 at 13:28
add a comment |
You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:
Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹
You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.
You realize that your boss actually wanted you to do:
Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²
Or in code:
// ¹
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
return true;
} else {
return false;
}
}
}
// ²
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
continue; // this could be omitted, as a loop keeps looping nevertheless
} else {
return false;
}
}
return true;
}
As this is a very common task in programming, there is already a shorter way to express this:
if(chocolates.every(isTasty)) {
alert("all chocolates are fine");
} else {
alert("Oh, that doesnt taste good");
}
whereas isTasty
is a function taking a chocolate and returning either true or false.
If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)
1
A tiny quibble - it's "chocolate", not "choclate".
– Wai Ha Lee
Apr 6 at 13:25
1
@waiHaLee oh, pronounciation tricked me ...
– Jonas Wilms
Apr 6 at 13:28
add a comment |
You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:
Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹
You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.
You realize that your boss actually wanted you to do:
Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²
Or in code:
// ¹
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
return true;
} else {
return false;
}
}
}
// ²
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
continue; // this could be omitted, as a loop keeps looping nevertheless
} else {
return false;
}
}
return true;
}
As this is a very common task in programming, there is already a shorter way to express this:
if(chocolates.every(isTasty)) {
alert("all chocolates are fine");
} else {
alert("Oh, that doesnt taste good");
}
whereas isTasty
is a function taking a chocolate and returning either true or false.
If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)
You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:
Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹
You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.
You realize that your boss actually wanted you to do:
Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²
Or in code:
// ¹
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
return true;
} else {
return false;
}
}
}
// ²
function checkChocolates(chocolates) {
for(const chocolate of chocolates) {
if(isTasty(chocolate)) {
continue; // this could be omitted, as a loop keeps looping nevertheless
} else {
return false;
}
}
return true;
}
As this is a very common task in programming, there is already a shorter way to express this:
if(chocolates.every(isTasty)) {
alert("all chocolates are fine");
} else {
alert("Oh, that doesnt taste good");
}
whereas isTasty
is a function taking a chocolate and returning either true or false.
If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)
edited Apr 6 at 13:31
answered Apr 6 at 13:10
Jonas WilmsJonas Wilms
65.6k53660
65.6k53660
1
A tiny quibble - it's "chocolate", not "choclate".
– Wai Ha Lee
Apr 6 at 13:25
1
@waiHaLee oh, pronounciation tricked me ...
– Jonas Wilms
Apr 6 at 13:28
add a comment |
1
A tiny quibble - it's "chocolate", not "choclate".
– Wai Ha Lee
Apr 6 at 13:25
1
@waiHaLee oh, pronounciation tricked me ...
– Jonas Wilms
Apr 6 at 13:28
1
1
A tiny quibble - it's "chocolate", not "choclate".
– Wai Ha Lee
Apr 6 at 13:25
A tiny quibble - it's "chocolate", not "choclate".
– Wai Ha Lee
Apr 6 at 13:25
1
1
@waiHaLee oh, pronounciation tricked me ...
– Jonas Wilms
Apr 6 at 13:28
@waiHaLee oh, pronounciation tricked me ...
– Jonas Wilms
Apr 6 at 13:28
add a comment |
Inside the loop the input num
was was tested for divisibility, if the num
was divisible the control was going in the else
block from where the function returned true
.
The loop was not checking for all numbers of the input array it was returning true
when the first number was divisible.
Just use a flag variable to see if all of the elements are divisible by the input number num
, if any one is not divisible the flag
will set to false
and then we can break
out of the loop and return it as there is no point to check the other numbers.
function checkFactors(factors, num) {
let flag = true;
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
flag = false;
break;
}
}
return flag;
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
You can also use Array.every
to check the same in a concise way:
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
add a comment |
Inside the loop the input num
was was tested for divisibility, if the num
was divisible the control was going in the else
block from where the function returned true
.
The loop was not checking for all numbers of the input array it was returning true
when the first number was divisible.
Just use a flag variable to see if all of the elements are divisible by the input number num
, if any one is not divisible the flag
will set to false
and then we can break
out of the loop and return it as there is no point to check the other numbers.
function checkFactors(factors, num) {
let flag = true;
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
flag = false;
break;
}
}
return flag;
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
You can also use Array.every
to check the same in a concise way:
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
add a comment |
Inside the loop the input num
was was tested for divisibility, if the num
was divisible the control was going in the else
block from where the function returned true
.
The loop was not checking for all numbers of the input array it was returning true
when the first number was divisible.
Just use a flag variable to see if all of the elements are divisible by the input number num
, if any one is not divisible the flag
will set to false
and then we can break
out of the loop and return it as there is no point to check the other numbers.
function checkFactors(factors, num) {
let flag = true;
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
flag = false;
break;
}
}
return flag;
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
You can also use Array.every
to check the same in a concise way:
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
Inside the loop the input num
was was tested for divisibility, if the num
was divisible the control was going in the else
block from where the function returned true
.
The loop was not checking for all numbers of the input array it was returning true
when the first number was divisible.
Just use a flag variable to see if all of the elements are divisible by the input number num
, if any one is not divisible the flag
will set to false
and then we can break
out of the loop and return it as there is no point to check the other numbers.
function checkFactors(factors, num) {
let flag = true;
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
flag = false;
break;
}
}
return flag;
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
You can also use Array.every
to check the same in a concise way:
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
function checkFactors(factors, num) {
let flag = true;
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
flag = false;
break;
}
}
return flag;
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
function checkFactors(factors, num) {
let flag = true;
for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
flag = false;
break;
}
}
return flag;
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));
console.log(checkFactors([2, 4, 3, 6, 9], 12));
console.log(checkFactors([3, 5, 2, 6, 9], 15));
console.log(checkFactors([4, 2, 8, 1], 16));
edited Apr 6 at 13:29
answered Apr 6 at 13:12
Amardeep BhowmickAmardeep Bhowmick
5,52321128
5,52321128
add a comment |
add a comment |
Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
}
return true;
}
1
He is looking for an explanation as well though
– Icepickle
Apr 6 at 13:10
I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.
– javapedia.net
Apr 6 at 13:12
add a comment |
Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
}
return true;
}
1
He is looking for an explanation as well though
– Icepickle
Apr 6 at 13:10
I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.
– javapedia.net
Apr 6 at 13:12
add a comment |
Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
}
return true;
}
Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.
function checkFactors(factors, num) {
for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)
if (num % element !== 0){
return false
}
}
return true;
}
answered Apr 6 at 13:08
javapedia.netjavapedia.net
575312
575312
1
He is looking for an explanation as well though
– Icepickle
Apr 6 at 13:10
I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.
– javapedia.net
Apr 6 at 13:12
add a comment |
1
He is looking for an explanation as well though
– Icepickle
Apr 6 at 13:10
I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.
– javapedia.net
Apr 6 at 13:12
1
1
He is looking for an explanation as well though
– Icepickle
Apr 6 at 13:10
He is looking for an explanation as well though
– Icepickle
Apr 6 at 13:10
I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.
– javapedia.net
Apr 6 at 13:12
I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.
– javapedia.net
Apr 6 at 13:12
add a comment |
Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.
New contributor
add a comment |
Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.
New contributor
add a comment |
Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.
New contributor
Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.
New contributor
New contributor
answered Apr 6 at 13:26
W.BrightW.Bright
133
133
New contributor
New contributor
add a comment |
add a comment |
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2
You should get the
return true
out of the loop ;) and leave thereturn false
inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument– Icepickle
Apr 6 at 13:07
2
FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)
– T.J. Crowder
Apr 6 at 13:09
1
@t.j.crowder yet we don't have a good dupetarget for it.
– Jonas Wilms
Apr 6 at 13:14
2
@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)
– T.J. Crowder
Apr 6 at 13:18
@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.
– Barmar
Apr 6 at 18:43