JavaScript Abstract or loose Equality Comparisons
As per ECMA the comparison x == y, where x and y are values which produce true or false. And, such comparison is performed as follow.
Type(x) is same as Type(y), Then | ||
---|---|---|
If Type(x) is Undefined | Return True | |
If Type(x) is Null | Return True | |
If Type(x) is Number | If x is NaN | Return False |
If y is NaN | Return False | |
If x is same number as y | Return True | |
If x is +0 & y is -0 | Return True | |
If x is -0 & y is +0 | Return True | |
Else Return False | ||
If Type(x) is String | If x & y are exactly same, – same sequence of character – same length – same character in corresponding positions |
Return True |
Else Return False | ||
If Type(x) is Boolean | If x & y are both True or both False | Return True |
Else Return False | ||
If x & y refers to the same object ( for more detail check description below ) |
Return True | |
Else Return False |
In the case of Type of x & y is different | ||
---|---|---|
If x is Null | y is Undefined | Return True |
If x is Undefined | y is Null | Return True |
If Type(x) is Number | Type(y) is String | Return the Result of comparison x == ToNumber(y) |
If Type(x) is String | Type(y) is Number | Return the Result of comparison ToNumber(x) == y |
If Type(x) is Boolean | Type(y) is any | Return the Result of comparison ToNumber(x) == y |
If Type(x) is any | Type(y) is Boolean | Return the Result of comparison x == ToNumber(y) |
If Type(x) is either String or Number | Type(y) is Object | Return the Result of comparison x == ToPrimitive(y) |
Type(x) is Object | If Type(y) is either String or Number | Return the Result of comparison ToPrimitive(x) == y |
- For all other cases, it will Return False
Here some things to keep in note that ToNumber stands for coercing/typecasting value to a number. So in the case of Type(x) is boolean and y is any value, the JS will perform implicit coercion on value of y to make it comparable with x.
ToPrimitive stands for typecasting non-primitive value ( eg. Object including function and array ) to a primitive value ( eg. string, number, etc ).
In the case of x & y refers to the same object only returns true else return false means lets suppose,
var a = [1,2,3];
var b = [1,2,3];
var c = '1,2,3';
a == c // true.
b == c // true.
a == b // false.
In the case of a == c, a is coerced to string as c is a string and both values are same (because while coercing array to string it return a comma separated values) so returns true. and same for b == c.
While comparing a == b, coercing won’t happen as the type on both side is same. The reason it will return false is, JS now don’t check the value of the object instead it will check its reference. Even though the value of a and b is the same, but because of different reference, it will return false.
It can be true in the case of,
var a = [1,2,3];
var b = a;
a == b // true.