Compare x ='' and x = 0 in Javascript

Question

I have the following:

var a = 0;

if (Number(a) != '')
{
    alert('hi');
}

I never get the alert even though a is not empty, it is 0 and I need this value for numeric calculations.enter code here

What's going on here?

Answer 1

try the following

var a = 0;

if (Number(a) !== '') {
  alert('hi');
}

If you compare two values using == or != operator, javascript type-coerces both of them and then compares. On the contrary, if you use === instead of == (or !== instead of !=), type coercion does not happen.

Answer 2

The != operator does type conversion if needed, so '' is converted to the numeric value 0.

You can use the !== operator to avoid the type conversion:

var a = 0;

if (Number(a) !== '')
{
    alert('hi');
}

Answer 3

You are comparing 0 != 0 which will always be false.

The main issue is that the implicit cast here is causing '' to become 0. This can be easily verified.

alert(Number(''));//0

But why does this happen? It is defined behavior in the Language Specification as seen at

• 9.3.1 ToNumber Applied to the String Type
  http://www.ecma-international.org/ecma-262/5.1/#sec-9.3.1

  The conversion of a String to a Number value is similar overall to the determination of the Number value for a numeric literal (see 7.8.3), but some of the details are different, so the process for converting a String numeric literal to a value of Number type is given here in full. This value is determined in two steps: first, a mathematical value (MV) is derived from the String numeric literal; second, this mathematical value is rounded as described below.

  The MV of StringNumericLiteral ::: [empty] is 0.

This is why the value of implicitly converting an empty string to a number is 0.

Going forward, it would make more sense to check for a number by using isNaN to ensure that a number was entered by the user or is present in the variable.

var a = 0;

if (!isNaN(a))
{
    alert('a is a number');
}

Answer 4

use the !== operator to avoid type conversion