I have the following code which has
a mutable Person class, String and a method to modify the instances of String and Person
class Person{
int a = 8;
public int getA() {
return a;
}
public void setA(int a) {
this.a = a;
}
@Override
public String toString() {
return "Person [a=" + a + "]";
}
}
—
public class TestMutable {
public static void main(String[] args)
{
Person p = new Person();
p.setA(34);
String s = "bar";
modifyObject(s, p); //Call to modify objects
System.out.println(s);
System.out.println(p);
}
private static void modifyObject(String str, Person p)
{
str = "foo";
p.setA(45);
}
}
The output is as expected. It prints
bar
Person [a=45]
Now, my question is
What is happening at the place you say str=”foo” ?
Initially let’s assume that s=’bar’ and the data resides in 0x100 memory
Now the reference of string is passed to another method, the other method tries to change the contents of the memory location(0x100) to ‘foo’ using s=”foo”. Is this what is happening, or is ‘foo’ is created in differennt memory location ?
Does java pass references by value ?
Java always passes arguments by value NOT by reference.
Let me explain this through an example:
I will explain this in steps:
1- Declaring a reference named
fof typeFooand assign it to a new object of typeFoowith an attribute"f".2- From the method side, a reference of type
Foowith a nameais declared and it’s initially assigned tonull.3- As you call the method
changeReference, the referenceawill be assigned to the object which is passed as an argument.4- Declaring a reference named
bof typeFooand assign it to a new object of typeFoowith an attribute"b".5-
a = bis re-assigning the referenceaNOTfto the object whose its attribute is"b".6- As you call
modifyReference(Foo c)method, a referencecis created and assigned to the object with attribute"f".7-
c.setAttribute("c");will change the attribute of the object that referencecpoints to it, and it’s same object that referencefpoints to it.I hope you understand now how passing objects as arguments works in Java 🙂