Dynamic array of Integer type in processing

Question

How can I create a Dynamic array that holds Integer values (not int) in PROCESSING.

I have stored the String into a text file (String Str= "12,13,14,15"). Now i need to split and convert them to Integer type after loading the text file.

Okay, you've said what you need to do - but you haven't said how you've tried to achieve it. What have you tried, and what went wrong? Have you got as far as splitting the numbers? Trying to parse each part in turn? — Jon Skeet, Sep 03 '13 at 10:27

score 2 · Answer 1 · edited May 23 '17 at 12:20

2

Since code is reading a file, I would use a Scanner instead:

    String str = "2,3,4,5,6,7";
    List<Integer> intList = new ArrayList<Integer>();
    Scanner sc = new Scanner(new File(/*yourFile*/));
    //Scanner sc = new Scanner(str);
    while(sc.hasNext()) {
        sc.useDelimiter(",");
        intList.add(sc.nextInt());
    }

    Integer[] wrapperArray = new Integer[intList.size()];
    intList.toArray(wrapperArray);

Scanner: How-to

edited May 23 '17 at 12:20

Community

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1

answered Sep 03 '13 at 10:43

rocketboy

8,990
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there is no keyword Scanner in processing 2 :( – MELWIN Sep 03 '13 at 10:57

score 2 · Answer 2 · answered Sep 03 '13 at 11:23

You can try this code.. It works fine for me

        try {
        FileReader fr = new FileReader("data.txt");
        BufferedReader br = new BufferedReader(fr);

        String str = br.readLine();

        String strArray[] = str.split(",");
        Integer intArray[] = new Integer[strArray.length];

        for (int i = 0; i < strArray.length; i++) {
            intArray[i] = Integer.parseInt(strArray[i]);
            System.out.println(intArray[i]);
        }

    } catch (Exception e) {
       // TODO: handle exception
       e.printStackTrace();
    }

I hope this will help you.

Sajal Dutta · Accepted Answer · 2013-09-03T10:28:45.730

1

String str = "12,13,14,15";
String[] strArray = str.split(",");

int[] intArray = new int[strArray.length];

for (int i = 0; i < strArray.length; i++) {
    try {
        intArray[i] = Integer.parseInt(strArray[i]);
    } catch (NumberFormatException e) {
        // Handle the exception properly as noted by Jon
        e.printStackTrace();
    }
}

edited Sep 03 '13 at 10:28

answered Sep 03 '13 at 10:27

Sajal Dutta

7,258
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Note that just printing the stack trace is almost *always* the wrong way of handling exceptions. It's what IDEs tend to suggest automatically, but the code should pretty much never be left that way. – Jon Skeet Sep 03 '13 at 10:28
@JonSkeet Agreed but this should be handled by the designer how to handle it. I left it there so OP can handle it depending on the spec/design. – Sajal Dutta Sep 03 '13 at 10:30
But without mentioning it (or indeed writing *any* explanatory text) you're basically inviting the OP to copy and paste your code without change. IMO you should have *at least* added a TODO comment... and ideally written some text in your answer as well as the code. (Code-only answers are almost always less useful than those with an explanation.) – Jon Skeet Sep 03 '13 at 10:31
@JonSkeet Done but didn't know that what you find in the net is good for direct "copy paste" in your project. – Sajal Dutta Sep 03 '13 at 10:32
No, it's *not* a good idea - but that doesn't mean it doesn't happen :( – Jon Skeet Sep 03 '13 at 10:34
@JonSkeet well that sucks. – Sajal Dutta Sep 03 '13 at 10:36
I was able to convert the Strings into the int format using the spilt function. But the code needed was to suit the code in one.add("Triangle",new Integer[]{157,274,157,266,157,259,157,253,157,247,157,241,157,235};one.bind("ka","detected"); Its at the place of 157,274,157,... ,235 where i should be able to use the values that i have loaded. – MELWIN Sep 03 '13 at 10:48
@MELWIN one.add("Triangle", intArray); – Sajal Dutta Sep 03 '13 at 11:16
@SajalDutta thank you very much. Your idea helped me to solve my task... thank you – MELWIN Sep 03 '13 at 11:56

Dynamic array of Integer type in processing

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