How to can i use integer in ArrayList?

Question

When i replace EmpNo(Integer) to EmpName(String) its working properly. My code is not working for Int. If i search a string in my Array its working but when i search Integer in my array, the ouput is Employee Number not Found!!

int EmpNo[]=new int[a];

switch(a){
    case 1:
        System.out.print("Enter Employee Number:");
        number = Integer.parseInt(reader.readLine());
        int y = Arrays.asList(EmpNo).indexOf(number);
        if(Arrays.asList(EmpNo).contains(number))
        {
            System.out.print("Employee Number: "+EmpNo[y]+" \nEmployee Name "+EmpName[y]+" \nSalary "+Salary[y]+"");
        }
        else
        {
            System.out.println("Employee Number not Found!!");
        }
        break;
}

is indexOf returning -1? why not just check the condition y!=-1 in the if? — SteveL, Jul 09 '14 at 07:06
There is no need to use contains() when indexOf() is already used before. Value of y can tell you if number is present in EmpNo or not. if y == -1, employee not found else found at location y. — dips, Jul 09 '14 at 07:07
I'm not sure of this - but I think the issue is that when you pass `int[]` to `asList` it's treated as a single element; because it can't be cast to `Integer[]` or `Object[]`. — Dawood ibn Kareem, Jul 09 '14 at 07:08
@DavidWallace You are right, `int[]` will be converted to a singular Object — Scary Wombat, Jul 09 '14 at 07:09
Yeah, I thought so. Which means you've got a singleton list, whose only element is an `int[]` - and you're trying to find an `Integer` in that list. Nothing will ever be found by `contains` or `indexOf`. — Dawood ibn Kareem, Jul 09 '14 at 07:10

score 0 · Answer 1 · answered Jul 09 '14 at 07:12

Change

int EmpNo[]=new int[a];

to

Integer EmpNo[]=new Integer[a];

Also there is not need to use contains() when indexOf() is already used before. Value of y can tell you if number is present in EmpNo or not. if y == -1, employee not found else found at location y.

score 0 · Accepted Answer · answered Jul 09 '14 at 07:13

0

As David Wallace has pointed out the int[] does not autobox to Integer[] but will be automatically converted to a singular Object.

Define empNo as

Integer [] empNo = new Integer [a] ();

Also if indexOf is -1 then contains will also be false.

answered Jul 09 '14 at 07:13

Scary Wombat

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score 0 · Answer 3 · edited May 23 '17 at 12:29

0

EmpNo is an array of primitive integers. A List<> cannot store primitive data types. It needs Integers instead. So when you do:

Arrays.asList(EmpNo)

it returns a List<int[]> instead of a List<Integer>. So the contains and indexOf obviously do not match for an int.

If you can change the type of EmpNo, the simplest solution is to declare it as:

Integer[] EmpNo = new Integer[a];

Notice the Integer instead of the int. This should work correctly. If you cannot change the type of EmpNo, you will need to use some other methods as given here:

Arrays.asList() not working as it should?

edited May 23 '17 at 12:29

Community

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answered Jul 09 '14 at 07:15

metacubed

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Nothing's wrong with this answer. I suspect the downvoter misread it. It happens sometimes. – Dawood ibn Kareem Jul 09 '14 at 09:21

score -2 · Answer 4 · answered Jul 09 '14 at 07:18

-2

.contains() will treat the argument inside it as string.so,try using something like

if(Arrays.asList(EmpNo).contains(new Int (number))

answered Jul 09 '14 at 07:18

user1591182

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what is a `new Int` ? – Scary Wombat Jul 09 '14 at 07:30

How to can i use integer in ArrayList?

4 Answers4