Passing an ArrayList of any type to a method

Question

I'm working on my first Android app, a math game for my kid, and am learning Java in the process. I have two ArrayLists, one of integers 0..9 and one strings of possible operations, at present, just + and -.

I would like to write a method that returns a random index into an ArrayList, so I can select a random element. What I'm running into is that I need two methods, one for each type of ArrayList, even though the code is identical. Is there a way to do this in a single method?

What I use now:

Random randomGenerator = new Random();

  . . .

n = randomIndexInt(possibleOperands);
int op1 = possibleOperands.get(n);
n = randomIndexInt(possibleOperands);
int op2 = possibleOperands.get(n);
n = randomIndexStr(possibleOperations);
String operation = possibleOperations.get(n);

    . . .

int randomIndexInt(ArrayList<Integer> a){
    int n = randomGenerator.nextInt(a.size());
    return n;
}

int randomIndexStr(ArrayList<String> a){
    int n = randomGenerator.nextInt(a.size());
    return n;
}

What I'd like to do is collapse randomIndexInt and randomIndexStr into a single method.

You can write a method that will take A (of type int) as an argument and return any random number between 0 and A and call this method passing your array list sizes. — Andrew Logvinov
– Andrew Logvinov, Commented Jul 31, 2012 at 7:18
Remember to accept an answer, and maybe +1 if you like someone else as well :) — Anders Metnik
– Anders Metnik, Commented Jul 31, 2012 at 7:43
Thanks, all, with the help I got here I was able to improve the method further, and my code now looks like this: int op1 = (Integer) getRandomElement(possibleOperands); int op2 = (Integer) getRandomElement(possibleOperands); String operation = (String) getRandomElement(possibleOperations); . . . Object getRandomElement(ArrayList<?> a){ int n = randomGenerator.nextInt(a.size()); return a.get(n); } Apologies, I can't get this comment to format correctly. — codingatty
– codingatty, Commented Jul 31, 2012 at 17:34

BenMorel · Accepted Answer · 2013-11-02 21:48:05Z

18

declare your method as int randomIndexInt(ArrayList<?> a)

edited Nov 2, 2013 at 21:48

BenMorel

37k52 gold badges208 silver badges339 bronze badges

answered Jul 31, 2012 at 7:20

didxga

6,1555 gold badges45 silver badges60 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Paul Bellora Over a year ago

It could just be Collection<?> since that's what declares size().

Nermeen · Accepted Answer · 2012-07-31 07:19:19Z

7

You need only the size of the array right? so do it:

int randomIndex(int size){
    int n = randomGenerator.nextInt(size);
    return n;
}

answered Jul 31, 2012 at 7:19

Nermeen

16k5 gold badges62 silver badges73 bronze badges

1 Comment

codingatty Over a year ago

Thanks. That's another approach, and one I'd done earlier, but I actually wanted to push as much common code as possible (including getting the Array size) into the method.

neworld · Accepted Answer · 2012-07-31 17:36:14Z

5

with this code you can pass any type of list

int randomIndex(List<?> list){
    int n = randomGenerator.nextInt(list.size());
    return n;
}

edited Jul 31, 2012 at 17:36

answered Jul 31, 2012 at 7:23

neworld

7,7933 gold badges42 silver badges62 bronze badges

3 Comments

codingatty Over a year ago

Thanks; that worked okay with the array, but the int array chokes it as a primitive type. I suppose I could use Integer instead of int, but the <?> approach fixed it nicely.

Paul Bellora Over a year ago

You wouldn't be able to pass a List<Integer> or a List<String> into this method.

neworld Over a year ago

Ups, I forget about generics cast. Thanks @PaulBellora.

Anders Metnik · Accepted Answer · 2012-07-31 07:20:38Z

2

just make it:

private int randomIndex(int size){
return randomGenerator(size);
}

And then call them with randomIndex(yourArray.size());

answered Jul 31, 2012 at 7:20

Anders Metnik

6,2677 gold badges44 silver badges80 bronze badges

Comments

Priyank Doshi · Accepted Answer · 2012-07-31 14:36:46Z

2

More generic is to use List than ArrayList in method signature.

int your_method(List<?> a){
//your code
}

answered Jul 31, 2012 at 14:36

Priyank Doshi

13.2k18 gold badges61 silver badges82 bronze badges

Comments

sunil · Accepted Answer · 2012-07-31 10:56:22Z

1

you can use Generics as follows

declare your function as below

private <T> int randomIndex(ArrayList<T> a){
    int n = randomGenerator.nextInt(a.size());
    return n;
}

now you can pass ArrayList<String> or ArrayList<Integer> to this function without any issue like this

ArrayList<String> strList = new ArrayList<String>();
strList.add("on1");
System.out.println(randomIndex(strList));
ArrayList<Integer> intList = new ArrayList<Integer>();
intList.add(1);
System.out.println(randomIndex(intList));

edited Jul 31, 2012 at 10:56

answered Jul 31, 2012 at 10:27

sunil

6,6541 gold badge36 silver badges44 bronze badges

Comments

Shark · Accepted Answer · 2012-07-31 08:41:07Z

0

int randomIndex1(ArrayList<?> a)
{
   int n = randomGenerator.nextInt(a.size());
   return n;
}

int randomIndex2(int size)
{
   int n = randomGenerator.nextInt(size);
   return n;
}

answered Jul 31, 2012 at 8:41

Shark

6,6183 gold badges31 silver badges51 bronze badges

Collectives™ on Stack Overflow

Passing an ArrayList of any type to a method

7 Answers 7

1 Comment

1 Comment

3 Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

7 Answers 7

1 Comment

1 Comment

3 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related