`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?

• Felix Kling: FWIW, arguments is not a keyword. It's a normal variable (name).
• Kevin Meredith: So root.name is called first in root [name] = Batou? I don't understand
• Felix Kling: No, root is the object that is passed as first argument to merge, i.e. {name: "Batou"}. It already contains the property name.
• Kevin Meredith: 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: I updated the title - thanks, Felix King.

Answer 1

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

• cookie monster: IMO, target would have been a more descriptive name than root.

Answer 2

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