Why does the SHA-1 algorithm produce different values for the same input value?

Question

Here I attach my code regards "SHA-1" algorithm based Java code for same input value given manually:

import java.security.MessageDigest;    
import java.security.NoSuchAlgorithmException;    
import sun.misc.BASE64Encoder;

public class NewClass {       
 public static void main(String args[]) throws NoSuchAlgorithmException        
 {       
    MessageDigest digest = MessageDigest.getInstance("SHA-1");    
    System.out.println("Algorithm :"+digest.getAlgorithm());    
    digest.update("welcome".getBytes());    
    byte[] result = digest.digest();     
    hash = (new BASE64Encoder()).encode(result);     
    MessageDigest digest1 = MessageDigest.getInstance("SHA-1");    
    digest1.update("welcome".getBytes());    
    byte[] result1 = digest1.digest();    
    System.out.println(result);    
    System.out.println(result1);    
    String hash1 = (new BASE64Encoder()).encode(result);    
    System.out.println("Digest value"+hash);    
    System.out.println("Digest value"+hash1);       
 }

}

output:

Algorithm :SHA-1

[B@42e816

[B@9304b1

Digest value wLE3/i15JFnyb/djzORFdKW1qwM=

Digest value wLE3/i15JFnyb/djzORFdKW1qwM=

The above algorithm produces the same hash code, but the intermediate values are not same. Why is that?

I don't answer your question, but I think you have miss take String hash1 = (new BASE64Encoder()).encode(result);, it should be result1. — secmask
– secmask, Commented Jun 2, 2012 at 12:40

Oliver Charlesworth · Accepted Answer · 2012-06-02 11:00:57Z

9

[B@42e816 is what you get when you try to print an array in Java. It just prints its memory address, not the contents. Obviously, two different arrays will have different addresses.

answered Jun 2, 2012 at 11:00

Oliver Charlesworth

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Comments

Codo · Accepted Answer · 2012-06-02 11:01:05Z

5

These two lines make no sense:

System.out.println(result);
System.out.println(result1);

The don't print the contents of the two byte arrays as you obviously expect. Instead, the print the type ([B stands for byte array) and a number which can be thought of as the memory address of the array.

To print the contents of the byte array, use the Base 64 encoding and print the resulting string. I'm confident they will be the same.

answered Jun 2, 2012 at 11:01

Codo

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