Java camparing two numbers, equality for Primitive

Question

Question maybe very simple. Checked the equality of two Primitive types, then got some mistakes. One of the double, second one long.

public class TesterPrimitive {

public static void main(String[] args) {
    System.out.println("Equality - " + (5.0 == 5)); // Return true
    System.out.println("Equality - " + (5.000000000000001D == 5L)); // Return false
    System.out.println("Equality - " + (5.0000000000000001D == 5L)); // Return true
}}

Why for third equality, for double and long, I got true? Is it means, long 0 numbers after (.), change any value to absolute 0? Or is it changing bits and then we can get 5.0 for double?

I think this might help https://stackoverflow.com/questions/31744937/how-to-compare-correctly-long-and-float-primitives — Igor Flakiewicz, Jan 21 '22 at 18:35

J Asgarov · Accepted Answer · 2022-01-21T18:44:49.427

2

As you can see here https://www.geeksforgeeks.org/difference-float-double-c-cpp/ double has exactly 15 digits precision. 16th digit gets rounded.

Another thing that is perhaps important to understand - when performing operations on different primitives like here (5.0 == 5), int gets promoted to double for the sake of equality operation. More here https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html

edited Jan 21 '22 at 18:44

answered Jan 21 '22 at 18:36

J Asgarov

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score 0 · Answer 2 · answered Aug 05 '22 at 14:56

double variables have a precision of about 15 digits, which is enough for most purposes.double data type can occupy 8 bytes of space in the computer memory and stores 15 to 16 numbers after the decimal point, you can check this article, hope this will help you

Java camparing two numbers, equality for Primitive

2 Answers2