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Java always passes arguments by value, NOT by reference.
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public class Main { public static void main(String[] args) { Foo f = new Foo("f"); changeReference(f); // It won't change the reference! modifyReference(f); // It will modify the object that the reference variable "f" refers to! } public static void changeReference(Foo a) { Foo b = new Foo("b"); a = b; } public static void modifyReference(Foo c) { c.setAttribute("c"); } }
I will explain this in steps:-
Declaring a reference named
f
of typeFoo
and assign it a new object of typeFoo
with an attribute"f"
.Foo f = new Foo("f");
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From the method side, a reference of type
Foo
with a namea
is declared and it's initially assignednull
.public static void changeReference(Foo a)
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As you call the method
changeReference
, the referencea
will be assigned the object which is passed as an argument.changeReference(f);
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Declaring a reference named
b
of typeFoo
and assign it a new object of typeFoo
with an attribute"b"
.Foo b = new Foo("b");
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a = b
makes a new assignment to the referencea
, notf
, of the object whose attribute is"b"
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As you call
modifyReference(Foo c)
method, a referencec
is created and assigned the object with attribute"f"
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c.setAttribute("c");
will change the attribute of the object that referencec
points to it, and it's the same object that referencef
points to it.
I hope you understand now how passing objects as arguments works in Java 🎉
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