Confusing behavior with JavaScript Arguments object

Question

I have the following piece of code:

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

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

It removes the elements from an array which equal the optional arguments. This code gives [1,1] as I expect.

But if I change the 4th line to

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

I get [1,2,3,1,2,3] instead, when I expect [1,1]. Why is that the case?

In the second case you use the arguments object of the inner function and not the outer one. — Sirko
– Sirko, Commented Sep 9, 2017 at 8:16
you could use arrow notation arr = arr.filter(x => x != arguments[i]); — Jaromanda X
– Jaromanda X, Commented Sep 9, 2017 at 8:20

BenM · Accepted Answer · 2017-09-09 08:20:11Z

2

Because when you use arguments within the anonymous function, you're accessing arguments of that function, not destroyer().

You will need to copy the arguments of destroyer(), preferably before your loop as follows:

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

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

Alternatively, an arrow function can also be used to achieve the same functionality:

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

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

answered Sep 9, 2017 at 8:20

BenM

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1 Comment

Jaromanda X Over a year ago

of course, if you're going to use ES6 ... function destroyer(arr, ...args) { return arr.filter(x => !args.includes(x)); }

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