Undesirable behavior of javascript operator

问题

I do not understand why this is happening,

console.info("") //(an empty string)

console.info(0)  //0

console.info(typeof "") //string

console.info(typeof 0) //number

console.info(""==0) //true

console.info(0=="") //true

console.info (0>="") //true

console.info ("">=0) //true

Shouldn't it return false for last four statements.

I did all these in FireBug, a firefox addons.

回答1:

The ECMAScript Equals Operator (==) uses the Abstract Equality Comparison Algorithm, see ECMA-262 §11.9.3.

回答2:

The last 4 statements involve implicit type casting via the logical comparison operators. Because a String and Number are being compared, the String is converted to a Number. "" turns into 0 which is equal to 0.

You can find it in the specification _{which is not an easy read IMO}:

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), go to step 14.
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.
14. If x is null and y is undefined, return true.
15. If x is undefined and y is null, return true.
16. If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).
17. If Type(x) is String and Type(y) is Number, return the result of the comparison ToNumber(x)== y.
18. If Type(x) is Boolean, return the result of the comparison ToNumber(x)== y.
19. If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).
20. If Type(x) is either String or Number and Type(y) is Object, return the result of the comparison x == ToPrimitive(y).
21. If Type(x) is Object and Type(y) is either String or Number, return the result of the comparison ToPrimitive(x)== y.
22. Return false.

回答3:

well you have run into javascripts falsey and truthy value concept. In javascript

Truthy: Something which evaluates to TRUE.
Falsey: Something which evaluates to FALSE.

It’s mostly logical. One (1) is truthy, Zero (0) is falsey. An object of any kind (including functions, arrays, RegExp objects, etc.) is always truthy. The easiest way to determine if something is truthy is to determine that it’s not falsey. There are only five falsey values in JavaScript:

undefined, null, NaN, 0, "" (empty string), and false.

Thus when you end up comparing two falsey values using the Abstract Equality Comparer it returns true.

To get around the problem use the typesafe comparison by using the === operator

来源：https://stackoverflow.com/questions/8265401/undesirable-behavior-of-javascript-operator