Page 198 - JavaScript

false != 0; // false
false != "0"; // false

The results start to make sense if you consider how JavaScript converts empty strings to numbers.

Number(""); // 0
Number("0"); // 0
Number(false); // 0

The Solution

In the statement false B, both the operands are strings ("" and "0"), hence there will be no type
conversion and since "" and "0" are not the same value, "" == "0" is false as expected.

One way to eliminate unexpected behavior here is making sure that you always compare
operands of the same type. For example, if you want the results of numerical comparison use
explicit conversion:

var test = (a,b) => Number(a) == Number(b);
test("", 0); // true;
test("0", 0); // true
test("", "0"); // true;
test("abc", "abc"); // false as operands are not numbers

Or, if you want string comparison:

var test = (a,b) => String(a) == String(b);
test("", 0); // false;
test("0", 0); // true
test("", "0"); // false;

Side-note: Number("0") and new Number("0") isn't the same thing! While the former performs a type
conversion, the latter will create a new object. Objects are compared by reference and not by
value which explains the results below.

Number("0") == Number("0"); // true;
new Number("0") == new Number("0"); // false

Finally, you have the option to use strict equality and inequality operators which will not perform
any implicit type conversions.

"" === 0; // false
0 === "0"; // false
"" === "0"; // false

Further reference to this topic can be found here:

Which equals operator (== vs ===) should be used in JavaScript comparisons?.

Abstract Equality (==)

https://riptutorial.com/ 155