CODEWITHSUNDEEP
javascript
evilcman
evilcman

Reputation: 123

Confusing behavior with JavaScript Arguments object

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?

Upvotes: 0

Views: 29

Answers (1)

BenM
BenM

Reputation: 53198

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));

Upvotes: 2

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