function dropElements(arr, func) {
let newArr = [];
// Step 1: Create a for loop that goes through everything in the arr
for (let i = 0; i < arr.length; i++){
// Step 2: If the function is not passed, then continue to loop
if (arr[i] !== func) {
i = i + 1;
}
// Step 3: Once the function is passed, return the remaining numbers in the array and remove numbers that didn't pass
else if(arr[i] == func) {
arr.shift() == newArr
return newArr;
}
return newArr;
}
}
dropElements([1, 2, 3], function(n) {return n < 3; });
Even though this is wrong, I feel like I am getting better because I was able to get to this without looking at anything. It only passed one of the six tests. Can someone help me out?
In the starter code you can see that the value function(n) {return n < 3; } is passed as the argument for func.
In your if statement condition, you’re checking the equality of numbers against that function. I don’t think the number 3 is ever going to strictly equal a function.
You need to follow this line in the instructions
when the iterated element is passed through it
In other words, you need to call the func function and pass the element from the array, and checking its return value.
This will always remove the first element of the array and not the one currently being iterated over (the one being checked with the call to func). Is that what you wanted to do?
function dropElements(arr, func) {
// Step 1: Create a for loop that checks each number in arr
for (let i = 0; i < arr.length; i++){
// Step 2: If number plugged into function is true...STOP
if (func(arr[i])) {
break;
}
// Step 3: if not true...remove that number
else {
arr.shift();
}
}
return arr;
}
dropElements([1, 2, 3], function(n) {return n < 3; });
This is what I thought I was doing, is my logic off?
No, you are not removing that particular number in the array. You are only ever removing the first element of the array. That is what the shift method does.
Ahhhh I see, do you know a method that removes any number regardless of if its at the beginning, middle, or end? Isn’t it the pop() method
function dropElements(arr, func) {
// Step 1: Create a for loop that checks each number in arr
for (let i = 0; i < arr.length; i++){
// Step 2: If number plugged into function is true...STOP
if (func(arr[0])) {
break;
}
// if not true...remove that number
else {
arr.shift();
}
}
return arr;
}
dropElements([1, 2, 3], function(n) {return n < 3; });
It’s working better now that I am specifying the index arr[0], it passes 5/6
You can look at splice, but be careful with it, because if you remove an element, you affect the index being iterated over.
To avoid mutating the original array, why not just create a new array and when the element currently being iterated over causes func to evaluate to true, push the element to the new array.
NOTE: By the way, arr[0] is not the current element being iterated over. It is just the first element of the array. You changed arr[i] to arr[0] for some reason.
function dropElements(arr, func) {
let orignalLength = arr.length
// Step 1: Create a for loop that checks each number in arr
for (let i = 0; i < orignalLength; i++){
// Step 2: If number plugged into function is true...STOP
if (func(arr[0])) {
break;
}
// if not true...remove that number
else {
arr.shift();
}
}
return arr;
}
dropElements([1, 2, 3], function(n) {return n < 3; });
Figured it out, but only because I felt close so I looked at solutions. Somehow let orignalLength = arr.length makes this work. Can you help me understand why? Isn’t creating a variable to name arr.length the same as just using arr.length?
Every time you remove a member of the array, its length becomes one less. This messes with the exit condition logic in your for loop.
The logic becomes much simpler if you avoid mutating an array while looping over it (not only for this problem, in general). For example, you could create a new array that you build up within your loop, and return that at the end of the function.