检查两个java对象的Deep Equal的最快捷有效的方法是什么？

Question

I have got two java objects with byte[] field of the size of the order of millions . What is the fastest and efficient way to check Deep Equal for these java objects?

Sample Entity:

@Entity
public class NormalBook
{

  @Id
  private String bookId;

  @Column
  private String title;

  @Column
  private byte[] pdfFile;

  //setters and getters

  }

Note: I am doing it for an ORM tool basically I am checking an object ( which is in managed state ) with an object present in Persistence Context.

Answer 1

Override equals() or have a *helper method (bad option!) and do it in 5 steps :

1. Check for *not null*.
2. Check for same *type*.
3. Check for *size of byte[]*.
4. Check for `==` (*reference equality* of byte[]) 
5. Start comparing byte values 

Answer 2

Use the following in the definition of equals() on your object's class:

java.util.Arrays.equals(bs1, bs2)

You might also want to check if they are the same array (instance) first. Though this method may well do that anyway.

For example (and making some assumptions about your class that contains the arrays):

public boolean equals(Object obj) {
    if(this == obj)
        return true;
    if(!(obj instanceof MyObject)) // covers case where obj null, too.
        return false;
    return Arrays.equals(this.bytes, ((MyObject)obj).bytes);
}

If there are other fields in your class, your equals() should take those into account too.

(There might be a better answer to the question if you could provide more information about what kind of data is stored in the arrays.)

Answer 3

if your class have fields like byte[] you can use something like:

public class MyClass {


    byte[] a;

    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        MyClass other = (MyClass) obj;
        if (!Arrays.equals(a, other.a))
            return false;
        return true;
    }


}

If you are concern about performance and can assure a unique hascode (this is important hascode need to be unique ) then you can just compare the hascode.