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How does the sort() method of the Collection class call the Comparable's compareTo()?

Suppose I want to sort a list of Employee objects:

Employee emp1 = new Employee("Abhijit", 10);
Employee emp2 = new Employee("Aniket", 5);
Employee emp3 = new Employee("Chirag", 15);

List<Employee> employees = new ArrayList<Employee>();
employees.add(emp1);
employees.add(emp2);
employees.add(emp3);
Collections.sort(employees);
System.out.println("sorted List is: "+employees);

And my Employee class implements Comparable, therefore it must override the compareTo() method.
And I have to write my sorting logic in the compareTo() method.

class Employee implements Comparable<Employee>
{
    String name;
    int empId;
    public Employee(String name, int empId)
    {
        this.name= name;
        this.empId = empId;
    }
    public String getName()
    {
        return name;
    }
    public void setName(String name)
    {
        this.name = name;
    }
    public int getEmpId()
    {
        return empId;
    }
    public void setEmpId(int empId)
    {
        this.empId = empId;
    }
    @Override
    public int compareTo(Employee e)
    {
        // TODO Auto-generated method stub
        return this.empId > e.empId ? 1 : (this.empId < e.empId ? -1 : 0);
        //return 0;
    }
    @Override
    public String toString()
    {
        return String.valueOf(empId);
    }
}

How does sort() call the compareTo() method internally?

like image 498
Abhijit Thorat Avatar asked Jul 23 '15 12:07

Abhijit Thorat


1 Answers

Please see open jdk source codes. I guess it helps.

java.util.Collections:

132     public static <T extends Comparable<? super T>> void More     ...sort(List<T> list) {
133         Object[] a = list.toArray();
134         Arrays.sort(a);
135         ListIterator<T> i = list.listIterator();
136         for (int j=0; j<a.length; j++) {
137             i.next();
138             i.set((T)a[j]);
139         }
140     }

Collections sort calls Arrays.sort

java.util.Arrays:

1218    public static <T> void More ...sort(T[] a, Comparator<? super T> c)   {
1219        T[] aux = (T[])a.clone();
1220        if (c==null)
1221            mergeSort(aux, a, 0, a.length, 0);
1222        else
1223            mergeSort(aux, a, 0, a.length, 0, c);
1224    }

Arrays.sort calls Arrays.mergeSort and your answer is on the line 1157.

1145
1146    private static void More ...mergeSort(Object[] src,
1147                                  Object[] dest,
1148                                  int low,
1149                                  int high,
1150                                  int off) {
1151        int length = high - low;
1152
1153        // Insertion sort on smallest arrays
1154        if (length < INSERTIONSORT_THRESHOLD) {
1155            for (int i=low; i<high; i++)
1156                for (int j=i; j>low &&
1157                         ((Comparable) dest[j-1]).compareTo(dest[j])>0; j--)
1158                    swap(dest, j, j-1);
1159            return;
1160        }
1161
1162        // Recursively sort halves of dest into src
1163        int destLow  = low;
1164        int destHigh = high;
1165        low  += off;
1166        high += off;
1167        int mid = (low + high) >>> 1;
1168        mergeSort(dest, src, low, mid, -off);
1169        mergeSort(dest, src, mid, high, -off);
1170
1171        // If list is already sorted, just copy from src to dest.  This is an
1172        // optimization that results in faster sorts for nearly ordered lists.
1173        if (((Comparable)src[mid-1]).compareTo(src[mid]) <= 0) {
1174            System.arraycopy(src, low, dest, destLow, length);
1175            return;
1176        }
1177
1178        // Merge sorted halves (now in src) into dest
1179        for(int i = destLow, p = low, q = mid; i < destHigh; i++) {
1180            if (q >= high || p < mid && ((Comparable)src[p]).compareTo(src[q])<=0)
1181                dest[i] = src[p++];
1182            else
1183                dest[i] = src[q++];
1184        }
1185    }
like image 164
jeton Avatar answered Sep 20 '22 21:09

jeton