incompatible type wile returning generic method of statictype
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I tried to implement a BST class was like
class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
}
I want to get the successor node in a method which takes a Node as a parameter and returns a successor node
private <T> Node<T> get(Node<T> current node){
Node<T> successor = root;
}
here I'm getting incompatible types . if I remove the method bound
private Node<T> get(Node<T> current node) {
Node<T> successor = root;
}
now it is compiling fine. What is the reason?
java generics
edited Jan 4 at 4:01
Naman
45.9k11102204
asked Jan 4 at 2:58
user3878073user3878073
3015
add a comment |
I tried to implement a BST class was like
class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
}
I want to get the successor node in a method which takes a Node as a parameter and returns a successor node
private <T> Node<T> get(Node<T> current node){
Node<T> successor = root;
}
here I'm getting incompatible types . if I remove the method bound
private Node<T> get(Node<T> current node) {
Node<T> successor = root;
}
now it is compiling fine. What is the reason?
java generics
edited Jan 4 at 4:01
Naman
45.9k11102204
asked Jan 4 at 2:58
user3878073user3878073
3015
add a comment |
I tried to implement a BST class was like
class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
}
I want to get the successor node in a method which takes a Node as a parameter and returns a successor node
private <T> Node<T> get(Node<T> current node){
Node<T> successor = root;
}
here I'm getting incompatible types . if I remove the method bound
private Node<T> get(Node<T> current node) {
Node<T> successor = root;
}
now it is compiling fine. What is the reason?
java generics
edited Jan 4 at 4:01
Naman
45.9k11102204
asked Jan 4 at 2:58
user3878073user3878073
3015
I tried to implement a BST class was like
class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
}
I want to get the successor node in a method which takes a Node as a parameter and returns a successor node
private <T> Node<T> get(Node<T> current node){
Node<T> successor = root;
}
here I'm getting incompatible types . if I remove the method bound
private Node<T> get(Node<T> current node) {
Node<T> successor = root;
}
now it is compiling fine. What is the reason?
java generics
java generics
edited Jan 4 at 4:01
Naman
45.9k11102204
asked Jan 4 at 2:58
user3878073user3878073
3015
edited Jan 4 at 4:01
Naman
45.9k11102204
asked Jan 4 at 2:58
user3878073user3878073
3015
edited Jan 4 at 4:01
Naman
45.9k11102204
edited Jan 4 at 4:01
Naman
45.9k11102204
edited Jan 4 at 4:01
Naman
45.9k11102204
45.9k11102204
asked Jan 4 at 2:58
user3878073user3878073
3015
asked Jan 4 at 2:58
user3878073user3878073
3015
asked Jan 4 at 2:58
user3878073user3878073
3015
3015
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
reason is T has already been defined at class level BST<T>. which means that you have to use BST<String>.method(); If you remove <T> from BST, the type will then have to be specified at the method level and you will need to add <T> at your method declaration. And calling the method will be BST.method<String>();
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
add a comment |
Your current code(which doesn't compile) reads as :
static class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
//
private <S> Node<T> get(Node<T> currentNode) {
Node<S> successor = root;
return null;
}
}
now you could easily relate by Node<S> couldn't be compatible with Node<T>, right?
Because of the presence of the method bound, the compiler treats both the method bound and the successor bound to be the same in this case. But the bound represented in the class BST<T> is not the same as bound represented by <T> Node<T> get method further still and that's where the incompatibility arises from.
If you wish to use the same bound, you could simply change the above code to get rid of your method bound, which is where you don't see any compile errors. Since then the bound of successor is inferred as the same as the bound of the class BST.
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
1
The scope of declaration could be worth reading for the specifications related to this.
– Naman
Jan 4 at 4:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
reason is T has already been defined at class level BST<T>. which means that you have to use BST<String>.method(); If you remove <T> from BST, the type will then have to be specified at the method level and you will need to add <T> at your method declaration. And calling the method will be BST.method<String>();
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
add a comment |
reason is T has already been defined at class level BST<T>. which means that you have to use BST<String>.method(); If you remove <T> from BST, the type will then have to be specified at the method level and you will need to add <T> at your method declaration. And calling the method will be BST.method<String>();
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
add a comment |
reason is T has already been defined at class level BST<T>. which means that you have to use BST<String>.method(); If you remove <T> from BST, the type will then have to be specified at the method level and you will need to add <T> at your method declaration. And calling the method will be BST.method<String>();
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
reason is T has already been defined at class level BST<T>. which means that you have to use BST<String>.method(); If you remove <T> from BST, the type will then have to be specified at the method level and you will need to add <T> at your method declaration. And calling the method will be BST.method<String>();
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
answered Jan 4 at 3:02
mkjhmkjh
1,1751023
1,1751023
add a comment |
add a comment |
Your current code(which doesn't compile) reads as :
static class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
//
private <S> Node<T> get(Node<T> currentNode) {
Node<S> successor = root;
return null;
}
}
now you could easily relate by Node<S> couldn't be compatible with Node<T>, right?
Because of the presence of the method bound, the compiler treats both the method bound and the successor bound to be the same in this case. But the bound represented in the class BST<T> is not the same as bound represented by <T> Node<T> get method further still and that's where the incompatibility arises from.
If you wish to use the same bound, you could simply change the above code to get rid of your method bound, which is where you don't see any compile errors. Since then the bound of successor is inferred as the same as the bound of the class BST.
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
1
The scope of declaration could be worth reading for the specifications related to this.
– Naman
Jan 4 at 4:02
add a comment |
Your current code(which doesn't compile) reads as :
static class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
//
private <S> Node<T> get(Node<T> currentNode) {
Node<S> successor = root;
return null;
}
}
now you could easily relate by Node<S> couldn't be compatible with Node<T>, right?
Because of the presence of the method bound, the compiler treats both the method bound and the successor bound to be the same in this case. But the bound represented in the class BST<T> is not the same as bound represented by <T> Node<T> get method further still and that's where the incompatibility arises from.
If you wish to use the same bound, you could simply change the above code to get rid of your method bound, which is where you don't see any compile errors. Since then the bound of successor is inferred as the same as the bound of the class BST.
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
1
The scope of declaration could be worth reading for the specifications related to this.
– Naman
Jan 4 at 4:02
add a comment |
Your current code(which doesn't compile) reads as :
static class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
//
private <S> Node<T> get(Node<T> currentNode) {
Node<S> successor = root;
return null;
}
}
now you could easily relate by Node<S> couldn't be compatible with Node<T>, right?
Because of the presence of the method bound, the compiler treats both the method bound and the successor bound to be the same in this case. But the bound represented in the class BST<T> is not the same as bound represented by <T> Node<T> get method further still and that's where the incompatibility arises from.
If you wish to use the same bound, you could simply change the above code to get rid of your method bound, which is where you don't see any compile errors. Since then the bound of successor is inferred as the same as the bound of the class BST.
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
Your current code(which doesn't compile) reads as :
static class BST<T> {
private Node<T> root;
private static class Node<T> {
T element;
Node<T> left;
Node<T> right;
}
//
private <S> Node<T> get(Node<T> currentNode) {
Node<S> successor = root;
return null;
}
}
now you could easily relate by Node<S> couldn't be compatible with Node<T>, right?
Because of the presence of the method bound, the compiler treats both the method bound and the successor bound to be the same in this case. But the bound represented in the class BST<T> is not the same as bound represented by <T> Node<T> get method further still and that's where the incompatibility arises from.
If you wish to use the same bound, you could simply change the above code to get rid of your method bound, which is where you don't see any compile errors. Since then the bound of successor is inferred as the same as the bound of the class BST.
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
answered Jan 4 at 3:55
NamanNaman
45.9k11102204
45.9k11102204
1
The scope of declaration could be worth reading for the specifications related to this.
– Naman
Jan 4 at 4:02
add a comment |
1
The scope of declaration could be worth reading for the specifications related to this.
– Naman
Jan 4 at 4:02
1
1
The scope of declaration could be worth reading for the specifications related to this.
– Naman
Jan 4 at 4:02
The scope of declaration could be worth reading for the specifications related to this.
– Naman
Jan 4 at 4:02
add a comment |
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