CODEWITHSUNDEEP
javascriptcompare
hassan yousefi
hassan yousefi

Reputation: 267

relational comparison in JavaScript

I am confuse for these comparison and I don't know how can be happen. in bottom code when I compare x and y for less than or great than, it return false but when I compare them with equals they return true! do you know why this happen?

var x = { y: 10 };
var y = { y: 11 };

x < y;  // false
x == y; // false
x > y;  // false

x <= y; // true
x >= y; // true

Upvotes: 2

Views: 129

Answers (2)

hassan yousefi
hassan yousefi

Reputation: 267

now I understand how are x <= y and y <= x resulting in true however x < y and x == y and x > y are all false. in continue of accepted answer I can say these happen because the spec says for x <= y , it will actually evaluate y < x first, and then negate that result. Since y < x is also false, the result of x <= y is true.

Upvotes: 1

Yousaf
Yousaf

Reputation: 29312

x < y;  // false
x > y;  // false

When any operand of a relational operator is an object, it is converted to a primitive value. So in this case, both x and y are first converted to primitive values and then compared with each other.

Both x and y will be converted to primitive values by eventually calling .toString() method on each object which will return a string of the form: [object Object] and since both objects will be converted in this string form, x < y and x > y both evaluate to false because both strings, i.e. [object Object] are equal.

x == y; // false

This evaluates to false because when types of both operands of == operator are same, javascript performs strict equality comparison ===, and as both objects are two separate objects in memory, they are not equal.

x <= y; // true
x >= y; // true

These both expressions evaluate to true because the following comparison

"[object Object]" == "[object Object]"

evaluates to true.

Upvotes: 2

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