Arrays.sort() is not accepting the comparator type object

Question

Am I missing something? It gives an error on sort method (doesn't identify the comparator method) but if I use Integer type array then this error goes away. Can't we do this comparison with primitive type of arrays?

public class Test {
    public static void main(String[] args) {
        int a[]={21,58,89,45,73,24};
        Arrays.sort(a,new Comparator<Integer>(){
            @Override
            public int compare(Integer o1, Integer o2) {
                if(o1%10>o2%10) return -1;
                return 1;
            }   
        });

Madplay · Accepted Answer · 2019-04-11 06:44:03Z

5

You can't pass a primitive type to sort method. Can you try this?

// Your array
int a[]={21, 58, 89, 45, 73, 24};

int[] ints = Arrays.stream(a).boxed().sorted(new Comparator<Integer>() {
    @Override
    public int compare(Integer o1, Integer o2) {
        if (o1 % 10 > o2 % 10) return -1;
        return 1;
    }
}).mapToInt(Integer::intValue).toArray();

edited Apr 11, 2019 at 6:44

answered Apr 11, 2019 at 6:18

Madplay

1,0471 gold badge13 silver badges26 bronze badges

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

4 Comments

Ferrybig Over a year ago

After you box the stream, you can also call . sort, before converting the stream back. Streams are make for performant chaining of operations

Mehrose Over a year ago

Why can't we pass the primitive type array to sort? or did you meant we can't do this with Comparator argument?

Madplay Over a year ago

@Ferrybig Thakns! I edited the code. Is this what you're trying to say?

Madplay Over a year ago

@Mehrose Yes. Comparator and the primitive type array can not be used together.

Rafał Sokalski · Accepted Answer · 2019-04-11 06:09:02Z

2

Signature of method is: public static <T> void sort(T[] a, Comparator<? super T> c) so the first argument is generic and you cannot put there primitive type that's why you get an error. For such sorting you have predefined sort method which can take primitive arrays

answered Apr 11, 2019 at 6:09

Rafał Sokalski

2,0072 gold badges20 silver badges35 bronze badges

2 Comments

Anant Shekhar Over a year ago

Then how does this work? ` int[] arr = {3,4,2,1}; Arrays.sort(arr); ` Here arr is clearly a primitive type array. Thanks!

Rafał Sokalski Over a year ago

public static void sort(int[] a) . Arrays class has overloaded sort method to take primite type

Eran · Accepted Answer · 2019-04-11 06:24:33Z

2

No you cannot. Primitive arrays can only be sorted with the methods that accept primitive arrays, such as static void sort(int[] a).

You can't pass a primitive array to public static <T> void sort(T[] a, Comparator<? super T> c), since the generic type parameter can only be substituted with a reference type, not a primitive type.

Besides, your Comparator makes no sense and violates the general contract of the Comparator interface. Therefore you have no reason to use that Comparator (not even with Integer arrays).

Based on your comments, you need something like this:

Integer a[]={21,58,89,45,73,24};
Arrays.sort(a,new Comparator<Integer>(){
    @Override
    public int compare(Integer o1, Integer o2) {
        int l1 = o1 % 10;
        int l2 = o2 % 10;
        if (l1 != l2)
            return Integer.compare(l1,l2); // sort by last digit
        else
            return Integer.compare(o1,o2); // sort by natural order
    }   
});

edited Apr 11, 2019 at 6:24

answered Apr 11, 2019 at 6:07

Eran

395k57 gold badges726 silver badges793 bronze badges

7 Comments

Eran Over a year ago

@Mehrose can you explain first what is the purpose of this code?

Mehrose Over a year ago

i have this array mentioned in the code and now I want to sort the numbers in ascending order based on the last(unit digit). Expected output: 21,73,24,45,58,89.

Eran Over a year ago

@Mehrose what should happen if two numbers have the same last digit?

Mehrose Over a year ago

the it'll be sort by default nature sorting

Mehrose Over a year ago

Thanks buddy but we don't necessarily type cast the value to Integer as we already defined the Generics with Comparator. The code works fine if I just change the primitive type to Integer object

|

Misha Betekhtin · Accepted Answer · 2019-04-11 07:56:03Z

You can perform sorting with Comparator on Integer because Integer is implementing Comparable Interface. primitive can be sorted only by the native way - int for example by number order.

You can do one of the following:

Convert int to Integer int a[]={21,58,89,45,73,24}; to Integer a[]={21,58,89,45,73,24};
if using int don't use Comparator Arrays.sort(a);
create your class, implement Comparable Interface and then manipulate int array inside - Harder approach and not recommended

public class MyInt implements Comparable<Integer> {

    private final int value;

    public MyInt(int value) {
        this.value = value;
    }

    @Override
    public int compareTo(Integer o) {
        return compare(this.value, o);
    }

    public static int compare(Integer o1, Integer o2) {
        if(o1%10>o2%10) return -1;
        return 1;
    }
}

Faseela Thayattuchira · Accepted Answer · 2024-03-05 13:09:27Z

0

 Arrays.sort(array, Comparator.comparingInt(interval -> interval[0]));

answered Mar 5, 2024 at 13:09

Faseela Thayattuchira

5577 silver badges16 bronze badges

Collectives™ on Stack Overflow

Arrays.sort() is not accepting the comparator type object

5 Answers 5

4 Comments

2 Comments

7 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

4 Comments

2 Comments

7 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related