Use of ... parameters in Java

Question

I had a question regarding ... parameters in java. Here is the code example

class Foo{

    private void m1(Object... params){
        //do something with params[0]
        //do something with params[1]
        //do something with params[2]
        //do something with params[3]
    }

    public void m2(Object... params){
        Object additionalParam = new Object();
        m1(additionalParam, params);
    }

}
class Example{

    public void main(String[] args){
        Foo f = new Foo();
        Object o1 = new Object();
        Object o2 = new Object();
        Object o3 = new Object();
        f.m2(o1, o2, o3);
    }
}

Does something like this work? I didn't get a compiler error, but when doing something similar, I had problems when changing the value of one of the objects from m1().

SOLUTION: I thought that the array would be flattened, but apparently that doesn't happen.

class Foo{

    private void m1(Object... params){
        //do something with params[0]
        //do something with params[1][0]
        //do something with params[1][1]
        //do something with params[1][2]
    }

    public void m2(Object... params){
        Object additionalParam = new Object();
        m1(additionalParam, params);
    }

}
class Example{

    public void main(String[] args){
        Foo f = new Foo();
        Object o1 = new Object();
        Object o2 = new Object();
        Object o3 = new Object();
        f.m2(o1, o2, o3);
    }
}

Thanks, Sibbo for your contribution.

wouldn't m2 result in an infinite regress, due to it repeatedly calling itself with an increasingly large number of arguments? — Davis Broda, Apr 08 '15 at 13:47
m1() is private, maybe this why you were getting problems... — DAG, Apr 08 '15 at 13:48
@DavisBroda i guess he wanted to write a method call to m1, since the m1 method wouldn´t make sense otherwise — SomeJavaGuy, Apr 08 '15 at 13:49
Ok, for those people calling `...` as `params`, learn it's **actual** definition. It's *variable arguments* or [*varargs*](https://docs.oracle.com/javase/1.5.0/docs/guide/language/varargs.html) for short. — Buhake Sindi, Apr 08 '15 at 13:56
@BuhakeSindi I know, but parameters/arguments: same difference. The fact is: I know that Type... makes you able to pass any number of arguments (parameters) of the class Type, or pass an array of the class Type — Rik Schaaf, Apr 08 '15 at 14:06
@fabian It is not a direct copy of the code I used, therefore I forgot the return statement: fixed — Rik Schaaf, Apr 08 '15 at 14:07
@Coolcat, not exactly, arguments and parameters have been used in 2 different constructs, especially when referencing it using Generics Parameterized types/arguments. Don't confuse the 2. — Buhake Sindi, Apr 08 '15 at 14:16
@BuhakeSindi [Param vs args](http://stackoverflow.com/questions/1788923/parameter-vs-argument) if this is what you mean: ok then I get that there is a difference. But that means that my use of the name params isn't wrong, since it is indeed a list of the parameters. — Rik Schaaf, Apr 08 '15 at 14:26
@MarounMaroun I added a second program that might clarify my question more — Rik Schaaf, Apr 08 '15 at 14:29

Sibbo · Accepted Answer · 2015-04-08T14:44:06.357

1

You are passing two arguments to m1. One is an Object (A), and one an Object[] (B). To access A, you can use params[0], but to access any element i from B, you need to use params[1][i].

So if you expected that the resulting array will be flattened, no, it won't be.

If you want to change the values that are stored in the array you pass in the main method, just do it like you did in your second example. If you want to do it nicely, use proper types instead of Object.

edited Apr 08 '15 at 14:44

answered Apr 08 '15 at 13:51

Sibbo

3,796
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I indeed found out that was one of the problems, though it is not the solution. Thanks anyway – Rik Schaaf Apr 08 '15 at 13:55
What are you asking? – Sibbo Apr 08 '15 at 13:56
The other answer is my solution, though I cannot accept it for an other 5 minutes :P – Rik Schaaf Apr 08 '15 at 13:56
I added a second program that might clarify my question more – Rik Schaaf Apr 08 '15 at 14:29
Sorry, but I still don't get what you want to do. – Sibbo Apr 08 '15 at 14:33
I want to set the value of o1, o2 and o3 in m1 by calling m2 – Rik Schaaf Apr 08 '15 at 14:36
1

I have a feeling that this might actually be the solution after all. I think something went wrong when running (maybe I forgot to save or something) because the code I now provided does work. Thanks for the help. – Rik Schaaf Apr 27 '15 at 15:46

Bryan Reilly · Answer 2 · 2015-04-08T14:56:19.070

0

The (TYPE... NAME) parameter simply says "pass in an number of objects of type TYPE". These can then be accessed regularly via NAME[x] where x is the index of the object you passed in.

EDIT: The second part of my answer was wrong. Sorry about any confusion.

edited Apr 08 '15 at 14:56

answered Apr 08 '15 at 13:50

Bryan Reilly

114
4

The objects are not copied. Java passes objects by reference. – Sibbo Apr 08 '15 at 13:54
You are wrong partially, java creates a copy that is true, but not of the object itself, it creates a copy of the reference value which in the end references the object itself. – SomeJavaGuy Apr 08 '15 at 14:10
@KevinEsche Does it reference the variable or does it reference its value? – Rik Schaaf Apr 08 '15 at 14:19

Kelevandos · Answer 3 · 2015-04-08T19:14:17.253

0

Varargs params are treated by the compiler as an Array. So your example will have to look like this:

class Foo{

    private void m1(Object... params){
        //do something
    }

    public void m2(Object... params){
        Object additionalParam = new Object();
        Object[] newArray = /* Here you need to create an array with all param elements and the new one*/
        m2(newArray);
    }

}
class Example{

    public int main(String[] args){
        Foo f = new Foo();
        Object o1 = new Object();
        Object o2 = new Object();
        Object o3 = new Object();
        f.m2(new Object[]{o1, o2, o3});
    }
}

edited Apr 08 '15 at 19:14

answered Apr 08 '15 at 13:52

Kelevandos

7,024
2
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uhm, you know the difference between Object... and Object[], right? – Rik Schaaf Apr 08 '15 at 13:59
1

"Difference" is not the right word here. Varargs (the "...") allows you to pass any number of objects as the given parameter, so null, 0, one or an array. Whichever you pass, in the method code it will be treated as an array (which is aither null, empty, has only index 0 etc...) – Kelevandos Apr 08 '15 at 14:03

Use of ... parameters in Java

3 Answers3