Does it matter which equals operator (== vs ===) I use in JavaScript comparisons?(待翻译)
发布在stackoverflow 热门 javascript 问答2013年12月10日view:4782
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问题 http://stackoverflow.com/q/359494/343194

I’m using JSLint to go through some horrific JavaScript at work and it’s returning a huge number of suggestions to replace == (two equals signs) with === (three equals signs) when doing things like comparing idSele_UNVEHtype.value.length == 0 inside of an if statement.

Is there a performance benefit to replacing == with ===?

Any performance improvement would probably be welcomed as there are hundreds (if not thousands) of these comparison operators being used throughout the file.

Would I be correct in assuming that if no type conversion takes place, there would be a small (probably extremely small) performance gain over ==?

回答 http://stackoverflow.com/a/359509/343194

The identity (===) operator behaves identically to the equality (==) operator except no type conversion is done, and the types must be the same to be considered equal.

Reference: Javascript Tutorial: Comparison Operators

The == operator will compare for equality after doing any necessary type conversions. The === operator will not do the conversion, so if two values are not the same type === will simply return false. It’s this case where === will be faster, and may return a different result than ==. In all other cases performance will be the same.

To quote Douglas Crockford’s excellent JavaScript: The Good Parts,

JavaScript has two sets of equality operators: === and !==, and their evil twins == and !=. The good ones work the way you would expect. If the two operands are of the same type and have the same value, then === produces true and !== produces false. The evil twins do the right thing when the operands are of the same type, but if they are of different types, they attempt to coerce the values. the rules by which they do that are complicated and unmemorable. These are some of the interesting cases:

'' == '0'           // false
0 == ''             // true
0 == '0'            // true

false == 'false'    // false
false == '0'        // true

false == undefined  // false
false == null       // false
null == undefined   // true

' \t\r\n ' == 0     // true

The lack of transitivity is alarming. My advice is to never use the evil twins. Instead, always use === and !==. All of the comparisons just shown produce false with the === operator.


Update:

A good point was brought up by @Casebash in the comments and in @Phillipe Laybaert’s answer concerning reference types. For reference types == and === act consistently with one another (except in a special case).

var a = [1,2,3];
var b = [1,2,3];

var c = { x: 1, y: 2 };
var d = { x: 1, y: 2 };

var e = "text";
var f = "te" + "xt";

a == b            // false
a === b           // false

c == d            // false
c === d           // false

e == f            // true
e === f           // true

The special case is when you compare a literal with an object that evaluates to the same literal, due to its toString or valueOf method. For example, consider the comparison of a string literal with a string object created by the String constructor.

"abc" == new String("abc")    // true
"abc" === new String("abc")   // false

Here the == operator is checking the values of the two objects and returning true, but the === is seeing that they’re not the same type and returning false. Which one is correct? That really depends on what you’re trying to compare. My advice is to bypass the question entirely and just don’t use the String constructor to create string objects.

回答二 http://stackoverflow.com/a/1813267/343194

This question is more than a year old, but please let me add this counsel:

If in doubt, read the specification!

ECMA-262 is the specification for a scripting language of which Javascript is a dialect. Of course in the practice it matters more how the most important browsers behave than an esoteric definition how something is supposed to be handled. But it is helpful to understand why new String(”a”) !== “a”.

Please let me explain how to read the specification to clarify this question. I see that in this very old topic nobody had an answer for the very strange effect. So, if you can read a specification, this will help you in your profession tremendously. It is an acquired skill. So, let’s continue.

Searching the PDF file for === brings me to page 56 of the specification: 11.9.4. The Strict Equals Operator ( === ), and after wading through the specificationalese I find:

11.9.6 The Strict Equality Comparison Algorithm
The comparison x === y, where x and y are values, produces true or false. Such a comparison is performed as follows:
  1. If Type(x) is different from Type(y), return false.
  2. If Type(x) is Undefined, return true.
  3. If Type(x) is Null, return true.
  4. If Type(x) is not Number, go to step 11.
  5. If x is NaN, return false.
  6. If y is NaN, return false.
  7. If x is the same number value as y, return true.
  8. If x is +0 and y is −0, return true.
  9. If x is −0 and y is +0, return true.
  10. Return false.
  11. If Type(x) is String, then return true if x and y are exactly the same sequence of characters (same length and same characters in corresponding positions); otherwise, return false.
  12. If Type(x) is Boolean, return true if x and y are both true or both false; otherwise, return false.   13. Return true if x and y refer to the same object or if they refer to objects joined to each other (see 13.1.2). Otherwise, return false.

Interesting ist step 11. Yes, strings are treated as value types. But this does not explain why new String(”a”) !== “a”. Do we have a browser not conforming to ECMA-262?

Not so fast!

Let’s check the types of the operands. Try it out for yourself by wrapping them in typeof(). I find that new String(”a”) is an object, and step 1 is used: return false if the types are different.

If you wonder why new String(”a”) does not return a string, how about some exercise reading a specification? Have fun!


Aidiakapi wrote this in a comment below:

From the specification

11.2.2 The new Operator:

If Type(constructor) is not Object, throw a TypeError exception.

With other words, if String wouldn’t be of type Object it couldn’t be used with the new operator.

new always returns an Object, even for String constructors, too. And alas! The value semantics for strings (see step 11) is lost.

And this finally means: new String(”a”) !== “a”.

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