var items = ["a", "b", false, "c", null];
for (var i = 0, item; typeof (item = items[i]) != ‘undefined’; i++) {
alert(item);
}

but u still have problem if the array contains undefined value, like:

var items = ["a", undefined, "b"];

“for in” is not so safe…it loops on properties, so the array instance must be “pure array” to work correctly:

Array.prototype.bar=”bar”;
var items = ["a", "b", "c"];
items.foo=’foo’;

for (var p in items) {
alert(items[p]);
};
//shows a, b, c, foo, bar

Perhaps because it says: “Warning: Never use a loop like this on arrays” on https://developer.mozilla.org/En/Core_JavaScript_1.5_Reference:Statements:for_each…in

The ‘for (each) in’ loops over all the properties of an object. Not just the index of an array.

for each (item in items)
{
alert(item);
}

Is any reasonable environment still stuck with JS < 1.6?

Things are getting out of control!

#2 iterates over the array’s properties, which are not necessarily only its values. It should be avoided unless you have a good reason to do it this way.
#3 fails for the reason Robert pointed out, and also for sparse arrays. For example:

var items = ["a", "b", "c"];
items[47] = “d”;

items.length will return 48, so method #1 will find [47] without trouble. Method #3 will stop at [3], because it returns undefined.

It’ll stop before it gets to “c”

The first solution will correctly iterate over all of them.

@Femi
The variable is still manually assigned in the header of the loop, you can’t avoid it. In this simple case, it would have been easier to simply write alert(items[i]) but when you have more code than that, it’s confusing to always write it completly instead of its shorter version, item.