If comparefn is not
undefined
and is not a consistent comparison function for the elements of this array (see below), the behaviour ofsort
is implementation-defined.[…]
A function comparefn is a consistent comparison function for a set of values S if all of the requirements below are met for all values a, b, and c (possibly the same value) in the set S: The notation a <CF b means comparefn(a,b) < 0; a =CF b means comparefn(a,b) = 0 (of either sign); and a >CF b means comparefn(a,b) > 0.
- Calling comparefn(a,b) always returns the same value v when given a specific pair of values a and b as its two arguments. Furthermore, Type(v) is Number, and v is not
NaN
. Note that this implies that exactly one of a <CF b, a =CF b, and a >CF b will be true for a given pair of a and b.- Calling comparefn(a,b) does not modify the
this
object.- a =CF a (reflexivity)
- If a =CF b, then b =CF a (symmetry)
- If a =CF b and b =CF c, then a =CF c (transitivity of =CF)
- If a <CF b and b <CF c, then a <CF c (transitivity of <CF)
- If a >CF b and b >CF c, then a >CF c (transitivity of >CF)
28.02.10
ECMA-262 quote of the day
08.02.10
Brief talk on ES5 and Mozilla support for it
I gave a three-minute not-actually-lightning-talk-but-let’s-call-it-that-anyway on ECMA-262 5th edition, what’s in it, and the state of Mozilla’s support for it at the Mozilla weekly meeting this week. It’s probably old hat if you’ve been following the standard closely, but if you haven’t it gives a short and sweet overview of what’s new; there’s a three-minute video of the actual talk on the meeting page (start at around 7:00 into the complete video). If you’re strapped for time, view the slides and turn off stylesheets (View > Page Style > No Style in Firefox) to see notes on what roughly accompanied each slide.
15.01.10
More ES5 backwards-incompatible changes: regular expressions now evaluate to a new object, not the same object, each time they’re encountered
(preemptive clarification: coming in Firefox 3.7 and not Firefox 3.6, which is to say, a good half year away from now rather than Real Soon Now)
Disjunction: is /foo/
the same object, or a new object, each time it’s evaluated in ES3?
According to ECMA-262 3rd edition, what should this code print?
function getRegEx() { return /regex/; } print("getRegEx() === getRegEx(): " + (getRegEx() === getRegEx()));
The answer depends upon this question: when a JavaScript regular expression literal is evaluated, does it create a new RegExp
object each time, or does it evaluate to the exact same RegExp
object each time it’s evaluated? Let’s look at a few examples and make a guess.
I sense a pattern
var tests = [ function getNull() { return null; }, function getNumber() { return 1; }, function getString() { return "a"; }, function getBoolean() { return false; }, function getObject() { return {}; }, function getArray() { return []; }, function getFunction() { return function f(){}; }, ]; for (var i = 0, sz = tests.length; i < sz; i++) { var t = tests[i]; print(t.name + "() === " + t.name + "(): " + (t() === t())); }
If you test that code, you’ll see that the first four results are true, and the rest are false, all per ECMA-262 3rd edition. (Okay, technically, and bizarrely, ES3 permitted either result for the function case, but no browser ever implemented a result of true; ES5 acknowledges reality and mandates that the result be false.) The first four functions return primitive values; the last three return objects. There’s only a single instance of any primitive value — or, alternately, you might say, equality doesn’t distinguish between different instances of the same primitive. Therefore it doesn’t really matter whether primitive literals evaluate to new instances or the same instance. On the other hand, objects compare equal only if they’re the same object. Since the object cases didn’t compare identically, they must be new objects each time. This makes sense: if this were not the case, what would happen in the following example?
function makePoint(x, y) { var pt = {}; pt.x = x; pt.y = y; return pt; } var pt1 = makePoint(1, 2); var pt2 = makePoint(3, 4);
It would be complete nonsense if the object literal above evaluated to the same object every time it were encountered; the next two lines would blow away the previous point, and we would have pt1.x ===3 && pt1.y === 4
.
Plausible assertion: regular expression literals evaluate to new objects when encountered?
Returning to the original question, then, what does ES3 say this code should print?
function getRegEx() { return /regex/; } print("getRegEx() === getRegEx(): " + (getRegEx() === getRegEx()));
A regular expression is an object. If you don’t want to get weird property-poisoning of the sort just suggested, regular expression literals must evaluate to different objects each time they’re encountered, right?
Alternative: ES3 says /foo/
is the same object every time
Wrong. According to ES3, there’s only a single object for each regular expression literal that’s returned each time the literal is encountered:
A regular expression literal is an input element that is converted to a RegExp object (section 15.10) when it is scanned. The object is created before evaluation of the containing program or function begins. Evaluation of the literal produces a reference to that object; it does not create a new object.
This was originally a dubious optimization in the standard to avoid the “costly” creation of a regular expression object every time a literal would be encountered. It’s perhaps a little surprising that the same object is returned each time, but does it make a difference in real programs not written to demonstrate the quirk? Often it doesn’t matter. As a simple example, if (/^\d+$/.test(str)) { /* ... */ }
executes identically either way, assuming RegExp.prototype.test
is unmodified. The RegExp
never escapes, and its use doesn’t depend on mutable state, so creating new objects each time doesn’t make a difference (other than negligibly, in speed).
Sometimes, however, the shared-object misoptimization does matter meaningfully: when a RegExp
with mutable state is used in ways that depend on that state. Most regular expressions don’t store any state, so if the same RegExp
object is used twice it’s no big deal. However, it can matter a lot for regular expressions specified with the global
flag:
var s = "abcddeeefffffgggggggghhhhhhhhhhhhh"; function next(s) { var r = /(.)\1*/g; r.exec(s); return r.lastIndex; } var r = []; for (var i =0; i < 8; i++) r.push(next(s)); print(r.join(", "));
Each time a regular expression with the global
flag is used, its lastIndex
property is updated with the index of the location in the matched string where matching should resume when the regular expression is next used. Thus, in this example we have mutable state, and if next
is called multiple times we have uses which will depend on that mutable state. Let’s see what happens in engines which implemented regular expression literals per ES3. If you download the Firefox 3.6 release candidate and test the above code in it (adjusting the implied print
to alert
), the printed result will be this:
1, 2, 3, 5, 8, 13, 21, 34
ES5: an escape to sanity
Is ES3’s behavior what you’d expect? No, it isn’t. In fact, ES3’s behavior, which Mozilla and SpiderMonkey implement, is the second-most duplicated bug filed against Mozilla’s JavaScript engine. SpiderMonkey and (strangely enough) v8 are the only notable JavaScript engines out there that implement ES3’s behavior. ES3’s behavior is rarely what web developers expect, and it doesn’t provide any real value, so ES5 is changing to the behavior you’d expect: evaluating a regular expression literal creates a new object every time.
Starting with Firefox 3.7, Firefox will implement what ES5 specifies. Download a Firefox nightly from nightly.mozilla.org and test it out as above (use the profile manager if you want to keep your current Firefox settings and install untouched). Instead of the Fibonacci sequence you’ll get this:
1, 1, 1, 1, 1, 1, 1, 1
The bottom line
Starting with Firefox 3.7, evaluating a regular expression literal like /foo/
will create a new RegExp
object, just as evaluating {}
or []
currently creates a new object or array. The optimization ES3 specified has resulted in clear developer confusion and was misguided and inconsistent with respect to other object literal syntax in JavaScript.
Again, as with my previous post, we doubt this change will affect many scripts (in this case, except for the better). The fact that few browsers implemented ES3’s semantics means that most sites have to cope with either choice of semantics, so the semantics in ES5, implemented by Mozilla for Firefox 3.7, are likely already handled. Still, it’s possible that this change might break some sites (particularly those which include browser-specific code), so we’re giving a heads-up as early as possible.
12.01.10
More ES5 backwards-incompatible changes: the global properties undefined
, NaN
, and Infinity
are now immutable
(preemptive clarification: coming in Firefox 3.7 and not Firefox 3.6, which is to say, a good half year away from now rather than Real Soon Now)
JavaScript and the undefined
, Infinity
, and NaN
“keywords”
Consider the following JavaScript program: what do you think it does?
print("undefined before: " + undefined); undefined = 17; print("undefined after: " + undefined);
The above program will print this output:
undefined before: undefined undefined after: 17
Surely you can’t be serious!
A sane person might think that this program isn’t even a program. Doesn’t undefined
always refer to the primitive value undefined
? After all, this “program” isn’t one, nor would be the same one for true
or false
, mutatis mutandis:
print("null before: " + null); null = 17; // !!! NullLiteral is not a LeftHandSideExpression print("null after: " + null);
I am serious…and don’t call me Shirley
Curiously, the program that assigns to undefined
is a valid JavaScript program, but programs that assign to null
, true
, and false
are not. Why not? The latter are all keywords with intrinsic meaning within the language; undefined
, on the other hand, is just a normal property of the global object. According to ECMA-262 3rd edition, if you assign a different value to undefined
, that different value becomes the new value of undefined
.
This is a clear botch in ES3. undefined
should have been a keyword in JavaScript from the beginning; similarly, the global properties Infinity
and NaN
probably should have been keywords as well (or perhaps the properties should not have existed, given that Math.Infinity
and Math.NaN
exist and are immutable). ECMA-262 5th edition doesn’t quite go so far as to change these three properties into keywords due to backwards compatibility concerns (making that change would be guaranteed to break any programs that even tried to assign to those names, regardless whether the program relied on that assignment for correctness). Instead, it changes these properties to be read-only, in the same way that the various numeric properties on the Math
object are read-only. Assigning to these properties in ES5 won’t do anything (unless you opt into strict mode, in which case a TypeError
exception will be thrown after we fix bug 537873), but at least it won’t definitely and completely break existing programs that relied on this.
We’ve made this change in SpiderMonkey, and it is now in trunk builds of Firefox, slated for the eventual Firefox 3.7 release. Download a nightly build from nightly.mozilla.org and test out the change for yourself (use the profile manager if you want to keep your current Firefox settings and install untouched). This change should have no effect on the vast, vast majority of web developers who don’t try to change the values of these properties; as for the [civility and my religion require I redact this description] developers who did change the value of the global undefined
, NaN
, or Infinity
properties, well…you had it coming.
The bottom line
The global properties undefined
, Infinity
, and NaN
will be read-only and immutable in Firefox 3.7. Assigning to these properties will do nothing (except in strict mode where a TypeError
exception will be thrown once we fix a bug) rather than changing their values. This shouldn’t break the vast, vast, vast majority of scripts out there — but there’s no way to guarantee it will break no one, so we think it’s worth announcing this backwards-incompatible change as proactively as possible.