Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.
It goes like this:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
// we pass the object to foo
foo(aDog);
// aDog variable is still pointing to the "Max" dog when foo(...) returns
aDog.getName().equals("Max"); // true
aDog.getName().equals("Fifi"); // false
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// change d inside of foo() to point to a new Dog instance "Fifi"
d = new Dog("Fifi");
d.getName().equals("Fifi"); // true
}
In the example above aDog.getName()
will still return "Max"
. The value aDog
within main
is not changed in the function foo
with the Dog
"Fifi"
as the object reference is passed by value. If it were passed by reference, then the aDog.getName()
in main
would return "Fifi"
after the call to foo
.
Likewise:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
foo(aDog);
// when foo(...) returns, the name of the dog has been changed to "Fifi"
aDog.getName().equals("Fifi"); // true
// but it is still the same dog:
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// this changes the name of d to be "Fifi"
d.setName("Fifi");
}
In the above example, Fifi
is the dog's name after call to foo(aDog)
because the object's name was set inside of foo(...)
. Any operations that foo
performs on d
are such that, for all practical purposes, they are performed on aDog
, but it is not possible to change the value of the variable aDog
itself.
For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.
Whereas one approach is to implement the ICloneable
interface (described here, so I won't regurgitate), here's a nice deep clone object copier I found on The Code Project a while ago and incorporated it into our code.
As mentioned elsewhere, it requires your objects to be serializable.
using System;
using System.IO;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Formatters.Binary;
/// <summary>
/// Reference Article http://www.codeproject.com/KB/tips/SerializedObjectCloner.aspx
/// Provides a method for performing a deep copy of an object.
/// Binary Serialization is used to perform the copy.
/// </summary>
public static class ObjectCopier
{
/// <summary>
/// Perform a deep copy of the object via serialization.
/// </summary>
/// <typeparam name="T">The type of object being copied.</typeparam>
/// <param name="source">The object instance to copy.</param>
/// <returns>A deep copy of the object.</returns>
public static T Clone<T>(T source)
{
if (!typeof(T).IsSerializable)
{
throw new ArgumentException("The type must be serializable.", nameof(source));
}
// Don't serialize a null object, simply return the default for that object
if (ReferenceEquals(source, null)) return default;
using var Stream stream = new MemoryStream();
IFormatter formatter = new BinaryFormatter();
formatter.Serialize(stream, source);
stream.Seek(0, SeekOrigin.Begin);
return (T)formatter.Deserialize(stream);
}
}
The idea is that it serializes your object and then deserializes it into a fresh object. The benefit is that you don't have to concern yourself about cloning everything when an object gets too complex.
In case of you prefer to use the new extension methods of C# 3.0, change the method to have the following signature:
public static T Clone<T>(this T source)
{
// ...
}
Now the method call simply becomes objectBeingCloned.Clone();
.
EDIT (January 10 2015) Thought I'd revisit this, to mention I recently started using (Newtonsoft) Json to do this, it should be lighter, and avoids the overhead of [Serializable] tags. (NB @atconway has pointed out in the comments that private members are not cloned using the JSON method)
/// <summary>
/// Perform a deep Copy of the object, using Json as a serialization method. NOTE: Private members are not cloned using this method.
/// </summary>
/// <typeparam name="T">The type of object being copied.</typeparam>
/// <param name="source">The object instance to copy.</param>
/// <returns>The copied object.</returns>
public static T CloneJson<T>(this T source)
{
// Don't serialize a null object, simply return the default for that object
if (ReferenceEquals(source, null)) return default;
// initialize inner objects individually
// for example in default constructor some list property initialized with some values,
// but in 'source' these items are cleaned -
// without ObjectCreationHandling.Replace default constructor values will be added to result
var deserializeSettings = new JsonSerializerSettings {ObjectCreationHandling = ObjectCreationHandling.Replace};
return JsonConvert.DeserializeObject<T>(JsonConvert.SerializeObject(source), deserializeSettings);
}
Best Answer
Objects aren't passed at all. By default, the argument is evaluated and its value is passed, by value, as the initial value of the parameter of the method you're calling. Now the important point is that the value is a reference for reference types - a way of getting to an object (or null). Changes to that object will be visible from the caller. However, changing the value of the parameter to refer to a different object will not be visible when you're using pass by value, which is the default for all types.
If you want to use pass-by-reference, you must use
out
orref
, whether the parameter type is a value type or a reference type. In that case, effectively the variable itself is passed by reference, so the parameter uses the same storage location as the argument - and changes to the parameter itself are seen by the caller.So:
I have an article which goes into a lot more detail in this. Basically, "pass by reference" doesn't mean what you think it means.