Seek and Destroy (need help)

JavaScript
1

I am trying to loop through all the arguments in my function (except the first) and filter out the numbers that i am looping in the array (which would be the first argument). I am not sure why this is not working. Thanks.


function destroyer(arr) {
  // Remove all the values
  for(let i = 1; i < arguments.length; i++){
      arr = arr.filter(function(x){
      return x !== arguments[i];
    })
  }
  return arr;
}

console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));

User Agent is: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_13_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/76.0.3809.132 Safari/537.36.

Link to the challenge:

2

what makes you think this is not working?

3

the entire array that is in the argument was printed, instead of the modified array.

4

Emm, I’m not actually sure why this doesn’t work, but it seems to be to do with scoping and the loop because:

Edit: I’m an idiot, see my other reply

function destroyer(arr) {
  for(let i = 1; i < arguments.length; i++){
    arr = arr.filter((x) => x !== arguments[i])
  }
  return arr;
}

Works absolutely fine.

As does doing the operations manually (not that you would do that, just for comparison):

> arr = [1, 2, 3, 1, 2, 3];
Array(6) [ 1, 2, 3, 1, 2, 3 ]
> arr = arr.filter(function (x) { return x !== 2; })
Array(4) [ 1, 3, 1, 3 ]
> arr = arr.filter(function (x) { return x !== 3; })
Array(2) [ 1, 1 ]
5

don’t use arguments inside loop it is an array like object but not true array, therefore, filter function was evaluating everything true.
just modified your code:

> function destroyer(arr) {
>   // Remove all the values
>   let args = [...arguments]
>   for(let i = 1; i < arguments.length; i++){
>       arr = arr.filter(function(x){
>       return x !== args[i];
>     })
>   }

return arr;
}

console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));

6

This is not the reason, but what you’re doing does fix it and made it obvious why:

@joshfer2000 arguments refers to the arguments of a function. So if you use it like

arr.filter(function(x) { return x !== arguments[i]; })

arguments is the arguments for that function above that you’re giving to filter.

If you use an arrow function, arrow functions do not have an arguments object; arguments points at the arguments of the outer function, so it works fine. Or you copy the arguments from the outer function and use those, as @avatarfreak did.

7

please refert to The arguments object - JavaScript | MDN
and know the actual reason… `

Note: “Array-like” means that arguments has a length property and properties indexed from zero, but it doesn’t have Array 's built-in methods like forEach() and map()

please refer to the links given above

`

8

That’s irrelevant, it’s because arguments refers to the arguments for the currently scoped function, @joshfer2000’s solution is failing because of that. I’m sorry, but what you’ve posted literally contradicts what you’re saying.

1 Like