I have the code below. I am purposefully trying to use forEach in this instance.
function check(arr, el) {
arr.forEach((element) => {
console.log(element)
if (element === el) {
return true
}
})
}
check([1, 2, 3, 4, 5], 3)
I am expecting the code to return true because the el value of 3 is in the array. But instead it returns undefined. What am I doing wrong?
forEach don't return anything ( means undefined ), you can use some
function check(arr, el) {
return arr.some( element => element === el)
}
console.log(check([1, 2, 3, 4, 5], 3))If you want to use forEach than use a variable to store value and than return later from function
function check(arr, el) {
let found = false
arr.forEach((element) => {
if (element === el && !found){
found = true
}
})
return found
}
console.log(check([1, 2, 3, 4, 5], 3))Cannot use return inside a forEach statement.
NOTE: This answer is only because you need to use forEach.
Ordinarily you would always use some().
function check(arr, el) {
let found = false;
arr.forEach((element) => {
console.log(element)
if (element === el) {
found = true;
}
});
return found;
}
console.log( check([1, 2, 3, 4, 5], 3));
Just to use OP's context. since one has to use forEach.
function check(arr, el) {
let found = false;
arr.forEach((element) => {
console.log(element)
if (element === el){
found = true;
}
})
return found;
}
If you want to use forEach you need to have a variable being updated when you find a match. Array.forEach by default returns undefined. There is no build in way to break out of the forEach.
Since you are just looking for a simple element match simply use Array.includes:
let check = (arr, el) => arr.includes(el)
console.log(check([1, 2, 3, 4, 5], 3))Array.some gives you an iterator function which in this case you really do not need.
With Array.forEach:
function check(arr, el) {
let result = false
arr.forEach((element) => {
if (element === el) {
result = true // <-- update the result on match
}
})
return result // <-- return the result
}
console.log(check([1, 2, 3, 4, 5], 3))