comparing arrays in java
Use Arrays.equals method. Example: boolean b = Arrays.equals(nir1, nir2); //prints true in this case
Use Arrays.equals method. Example: boolean b = Arrays.equals(nir1, nir2); //prints true in this case
Domain-Driven Design makes the distinction between Entities and Value Objects. This is a good distinction to observe since it guides how you implement Equals. Entities are equal if their IDs equal each other. Value Objects are equal if all their (important) constituent elements are equal to each other. In any case, the implementation of GetHashCode … Read more
From the article Override equals and hashCode in Java: Default implementation of equals() class provided by java.lang.Object compares memory location and only return true if two reference variable are pointing to same memory location i.e. essentially they are same object. Java recommends to override equals and hashCode method if equality is going to be defined … Read more
A difference is that “foo”.equals((String)null) returns false while “foo”.compareTo((String)null) == 0 throws a NullPointerException. So they are not always interchangeable even for Strings.
You should almost always use equals. You can be certain that string1 == string2 will work if: You’ve already made sure you’ve got distinct values in some other way (e.g. you’re using string values fetched from a set, but comparing them for some other reason) You know you’re dealing with compile-time string constants You’ve manually … Read more
Unitils has this functionality: Equality assertion through reflection, with different options like ignoring Java default/null values and ignoring order of collections
The answer is in the JavaDoc of the equals() method: Unlike compareTo, this method considers two BigDecimal objects equal only if they are equal in value and scale (thus 2.0 is not equal to 2.00 when compared by this method). In other words: equals() checks if the BigDecimal objects are exactly the same in every … Read more
It doesn’t say the hashcode for an object has to be completely unique, only that the hashcode for two equal objects returns the same hashcode. It’s entirely legal to have two non-equal objects return the same hashcode. However, the more unique a hashcode distribution is over a set of objects, the better performance you’ll get … Read more
The key to the answer is called object interning. Java interns small numbers (less than 128), so all instances of Integer(n) with n in the interned range are the same. Numbers greater than or equal to 128 are not interned, hence Integer(1000) objects are not equal to each other.
How about some reflection, perhaps using Expression.Compile() for performance? (note the static ctor here ensures we only compile it once per T): using System; using System.Linq.Expressions; public class Report { public int Id { get; set; } public int ProjectId { get; set; } static void Main() { Report a = new Report { Id … Read more