0

Consider the following:

public class Foo {

private String name;

public Foo(String name) {
this.name = name;
}

public String get() {
return name;
}

public void set(String name) {
this.name = name;
}
}

import java.util.ArrayList;

A container class:

public class FooContainer {

private ArrayList<Foo> foos;

public FooContainer(int nbrOfFoos) {
foos = new ArrayList<Foo>(nbrOfFoos);
}

public void add(Foo foo) {
foos.add(foo);
}

public void printFoos() {
for(Foo foo: foos) {
    System.out.println(foo.get());
}
}
}

A simple test:

public class FooTest {

public static void main(String[] args) {

Foo foo = new Foo("Long");
FooContainer cont = new FooContainer(1);
cont.add(foo);
cont.printFoos();
foo.set("John");
cont.printFoos();
}
}

Of course, when executed, the test program prints "Long" and then "John".

Is it poor design to have a program where this is not desirable, i.e. where one would not like the name of the Foo to change?

Is the only workaround to copy/clone the object when passed to the add method?

Alexandre Vandermonde
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2 Answers2

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That's the way Java works: Objects are passed as reference and not as copies. In most cases, this seems to make sense. Copying an object can be expensive. If you want to put a copy into the container, you need to explicitly create a new object - I would not consider this a workaround.

If your point is mainly about preventing an existing object from being modified, you could make it immutable, i.e. make 'name' final and remove the 'set' method.

Frank Puffer
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  • Alright, thanks! I tried to make my question as general as possible, but in the example that prompted the question, I was actually passing BitSet:s. Perhaps it makes better sense there to make a copy of the BitSet, as I will essentially be using it as a more memory efficient boolean array, and would not like it to be modified from outside. – Alexandre Vandermonde Feb 14 '16 at 15:04
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Just for clarification, Java always pass parameters as values. The misunderstanding comes as Java pass a copy of the reference thus it appears that an object is passed by reference.

Read this question to learn more: Is Java "pass-by-reference" or "pass-by-value"?

Community
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Manuel Vera Silvestre
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