`arguments` Object in JavaScript

Question

Secrets of the JavaScript Ninja explains the arguments keyword with the merge() function:

function merge(root){
 for (var i = 1; i < arguments.length; i++) { // starts at i = 1, not 0
   for (var key in arguments[i]) {
     root[key] = arguments[i][key];
   }
 }
 return root;
}

var merged = merge(
 {name: "Batou"},
 {city: "Niihama"});

Note the assertions:

assert(merged.name == "Batou", 
    "The original name is intact.");
assert(merged.city == "Niihama", 
    "And the city has been copied over.");

Why does merged.name equal Batou rather than undefined?

Since, as I understand, merge() does not look at the first argument in the outer for-loop, how does the name: Batou get added to root?

FWIW, arguments is not a keyword. It's a normal variable (name). — Felix Kling
– Felix Kling, Commented Dec 26, 2013 at 23:47
So root.name is called first in root [name] = Batou? I don't understand — Kevin Meredith
– Kevin Meredith, Commented Dec 27, 2013 at 1:58
No, root is the object that is passed as first argument to merge, i.e. {name: "Batou"}. It already contains the property name. — Felix Kling
– Felix Kling, Commented Dec 27, 2013 at 1:58
perhaps it's due to inexperience with arguments, but the merge function above seems confusing to me. Is it an acceptable/passes code review practice in JS? — Kevin Meredith
– Kevin Meredith, Commented Dec 27, 2013 at 2:42

Sebastian Paaske Tørholm · Accepted Answer · 2013-12-26 23:01:58Z

4

root itself is the first argument, thus we don't need to add whatever members is in it to itself.

answered Dec 26, 2013 at 23:01

Sebastian Paaske Tørholm

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1 Comment

cookie monster Over a year ago

IMO, target would have been a more descriptive name than root.

Stephen Thomas · Accepted Answer · 2013-12-26 23:02:07Z

1

It's already in root. That's the first parameter

answered Dec 26, 2013 at 23:02

Stephen Thomas

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