Reputation: 29
class BubbleSort<T> extends ArraySort<T>
{
public void iSort(T[] inArray) {
int n = inArray.length;
int k;
for (int m = n; m >= 0; m--) {
for (int i = 0; i < n - 1; i++) {
k = i + 1;
if (compare(inArray[i], inArray[k])) {
T temp;
temp = inArray[i];
inArray[i] = inArray[k];
inArray[k] = temp;
}
}
}
print(inArray);
}
public static <T extends Comparable<T>> boolean compare(T a, T b) {
if (a.compareTo(b) > 0) {
return true;
} return false;
}
I'm getting (T extends comparable < T >, T extends comparable < T >) in the type bubblesort< T > is not applicable for the arguments (T,T)
error!
The quick fix is telling me to change method compare(T,T)< T >
to compare(T,T)
, but that wouldn't resolve my problem. It works perfectly fine when I enter in actual value of elements, for example, compare(3, 5)
or compare("hi", "hello")
instead of compare(inArray[i], inArray[k])
.
I would appreciate it a lot if someone could explain why it's doing that and give me a solution.
Upvotes: 1
Views: 50
Reputation: 29680
You've already defined your class with a generic type T
, so there's no point in declaring a separate one for the method:
class BubbleSort<T extends Comparable<T>> extends ArraySort<T> {
public void iSort(T[] inArray) {
int n = inArray.length;
int k;
for (int m = n; m >= 0; m--) {
for (int i = 0; i < n - 1; i++) {
k = i + 1;
if (compare(inArray[i], inArray[k])) {
T temp;
temp = inArray[i];
inArray[i] = inArray[k];
inArray[k] = temp;
}
}
}
print(inArray);
}
public static boolean compare(T a, T b) {
if (a.compareTo(b) > 0) {
return true;
}
return false;
}
}
Anyway, you should denote T extends Comparable<T>
in the class declaration rather than in the method.
Upvotes: 1