Java - What do constructor type arguments mean when placed *before* the type?
I've recently come across this unusual (to me) Java syntax...here's an example of it:
List list = new <String, Long>ArrayList();
Notice the positioning of the <String, Long>
type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.
Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?
Why is it legal to have 2 type arguments when ArrayList
only has 1?
I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.
java grammar
add a comment |
I've recently come across this unusual (to me) Java syntax...here's an example of it:
List list = new <String, Long>ArrayList();
Notice the positioning of the <String, Long>
type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.
Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?
Why is it legal to have 2 type arguments when ArrayList
only has 1?
I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.
java grammar
Yeah so do I, I'm not asking how to create a list lol
– Nathan Adams
1 hour ago
2
A constructor may have type arguments that are placed there (this particular constructor hasn’t, so<String, Long>
is just ignored). See Generics Constructor.
– Ole V.V.
1 hour ago
1
OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor
– Nathan Adams
1 hour ago
1
No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator.List<String> list = new ArrayList<>();
– Elliott Frisch
1 hour ago
@OleV.V. if you wanna put your comment and link as an answer I'll accept it
– Nathan Adams
1 hour ago
add a comment |
I've recently come across this unusual (to me) Java syntax...here's an example of it:
List list = new <String, Long>ArrayList();
Notice the positioning of the <String, Long>
type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.
Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?
Why is it legal to have 2 type arguments when ArrayList
only has 1?
I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.
java grammar
I've recently come across this unusual (to me) Java syntax...here's an example of it:
List list = new <String, Long>ArrayList();
Notice the positioning of the <String, Long>
type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.
Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?
Why is it legal to have 2 type arguments when ArrayList
only has 1?
I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.
java grammar
java grammar
asked 1 hour ago
Nathan AdamsNathan Adams
1638
1638
Yeah so do I, I'm not asking how to create a list lol
– Nathan Adams
1 hour ago
2
A constructor may have type arguments that are placed there (this particular constructor hasn’t, so<String, Long>
is just ignored). See Generics Constructor.
– Ole V.V.
1 hour ago
1
OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor
– Nathan Adams
1 hour ago
1
No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator.List<String> list = new ArrayList<>();
– Elliott Frisch
1 hour ago
@OleV.V. if you wanna put your comment and link as an answer I'll accept it
– Nathan Adams
1 hour ago
add a comment |
Yeah so do I, I'm not asking how to create a list lol
– Nathan Adams
1 hour ago
2
A constructor may have type arguments that are placed there (this particular constructor hasn’t, so<String, Long>
is just ignored). See Generics Constructor.
– Ole V.V.
1 hour ago
1
OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor
– Nathan Adams
1 hour ago
1
No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator.List<String> list = new ArrayList<>();
– Elliott Frisch
1 hour ago
@OleV.V. if you wanna put your comment and link as an answer I'll accept it
– Nathan Adams
1 hour ago
Yeah so do I, I'm not asking how to create a list lol
– Nathan Adams
1 hour ago
Yeah so do I, I'm not asking how to create a list lol
– Nathan Adams
1 hour ago
2
2
A constructor may have type arguments that are placed there (this particular constructor hasn’t, so
<String, Long>
is just ignored). See Generics Constructor.– Ole V.V.
1 hour ago
A constructor may have type arguments that are placed there (this particular constructor hasn’t, so
<String, Long>
is just ignored). See Generics Constructor.– Ole V.V.
1 hour ago
1
1
OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor
– Nathan Adams
1 hour ago
OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor
– Nathan Adams
1 hour ago
1
1
No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator.
List<String> list = new ArrayList<>();
– Elliott Frisch
1 hour ago
No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator.
List<String> list = new ArrayList<>();
– Elliott Frisch
1 hour ago
@OleV.V. if you wanna put your comment and link as an answer I'll accept it
– Nathan Adams
1 hour ago
@OleV.V. if you wanna put your comment and link as an answer I'll accept it
– Nathan Adams
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:
public class TypeWithGenericConstructor {
public <T> TypeWithGenericConstructor(T arg) {
// TODO Auto-generated constructor stub
}
}
I suppose that more often than not we don’t need to make the type argument explicit. For example:
new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
Now T
is clearly LocalDate
. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:
new <LocalDate>TypeWithGenericConstructor(null);
Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:
new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
In your question you seem to be calling the java.util.ArrayList
constructor. That constructor is not generic (only the ArrayList
class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:
Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments
But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List
and ArrayList
, but that again is a different story).
Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?
No, it’s different. The usual type argument/s after the type (ArrayList<Integer>
) are for the generic class. The type arguments before are for the * constructor*.
The two forms may also be combined:
List<Integer> list = new <String, Long>ArrayList<Integer>();
I would consider this a bit more correct since we can now see that the list stores Integer
objects (I’d still prefer to leave out the meaningless <String, Long>
, of course).
Why is it legal to have 2 type arguments when ArrayList only has 1?
First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList
class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).
But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?
– jaspreet
53 mins ago
Thanks for asking, @jaspreet. Please see my edit.
– Ole V.V.
42 mins ago
add a comment |
Here:
List list = new <String, Long>ArrayList();
You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.
A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.
add a comment |
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2 Answers
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This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:
public class TypeWithGenericConstructor {
public <T> TypeWithGenericConstructor(T arg) {
// TODO Auto-generated constructor stub
}
}
I suppose that more often than not we don’t need to make the type argument explicit. For example:
new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
Now T
is clearly LocalDate
. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:
new <LocalDate>TypeWithGenericConstructor(null);
Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:
new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
In your question you seem to be calling the java.util.ArrayList
constructor. That constructor is not generic (only the ArrayList
class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:
Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments
But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List
and ArrayList
, but that again is a different story).
Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?
No, it’s different. The usual type argument/s after the type (ArrayList<Integer>
) are for the generic class. The type arguments before are for the * constructor*.
The two forms may also be combined:
List<Integer> list = new <String, Long>ArrayList<Integer>();
I would consider this a bit more correct since we can now see that the list stores Integer
objects (I’d still prefer to leave out the meaningless <String, Long>
, of course).
Why is it legal to have 2 type arguments when ArrayList only has 1?
First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList
class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).
But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?
– jaspreet
53 mins ago
Thanks for asking, @jaspreet. Please see my edit.
– Ole V.V.
42 mins ago
add a comment |
This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:
public class TypeWithGenericConstructor {
public <T> TypeWithGenericConstructor(T arg) {
// TODO Auto-generated constructor stub
}
}
I suppose that more often than not we don’t need to make the type argument explicit. For example:
new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
Now T
is clearly LocalDate
. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:
new <LocalDate>TypeWithGenericConstructor(null);
Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:
new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
In your question you seem to be calling the java.util.ArrayList
constructor. That constructor is not generic (only the ArrayList
class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:
Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments
But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List
and ArrayList
, but that again is a different story).
Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?
No, it’s different. The usual type argument/s after the type (ArrayList<Integer>
) are for the generic class. The type arguments before are for the * constructor*.
The two forms may also be combined:
List<Integer> list = new <String, Long>ArrayList<Integer>();
I would consider this a bit more correct since we can now see that the list stores Integer
objects (I’d still prefer to leave out the meaningless <String, Long>
, of course).
Why is it legal to have 2 type arguments when ArrayList only has 1?
First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList
class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).
But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?
– jaspreet
53 mins ago
Thanks for asking, @jaspreet. Please see my edit.
– Ole V.V.
42 mins ago
add a comment |
This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:
public class TypeWithGenericConstructor {
public <T> TypeWithGenericConstructor(T arg) {
// TODO Auto-generated constructor stub
}
}
I suppose that more often than not we don’t need to make the type argument explicit. For example:
new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
Now T
is clearly LocalDate
. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:
new <LocalDate>TypeWithGenericConstructor(null);
Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:
new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
In your question you seem to be calling the java.util.ArrayList
constructor. That constructor is not generic (only the ArrayList
class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:
Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments
But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List
and ArrayList
, but that again is a different story).
Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?
No, it’s different. The usual type argument/s after the type (ArrayList<Integer>
) are for the generic class. The type arguments before are for the * constructor*.
The two forms may also be combined:
List<Integer> list = new <String, Long>ArrayList<Integer>();
I would consider this a bit more correct since we can now see that the list stores Integer
objects (I’d still prefer to leave out the meaningless <String, Long>
, of course).
Why is it legal to have 2 type arguments when ArrayList only has 1?
First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList
class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).
This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:
public class TypeWithGenericConstructor {
public <T> TypeWithGenericConstructor(T arg) {
// TODO Auto-generated constructor stub
}
}
I suppose that more often than not we don’t need to make the type argument explicit. For example:
new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
Now T
is clearly LocalDate
. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:
new <LocalDate>TypeWithGenericConstructor(null);
Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:
new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));
In your question you seem to be calling the java.util.ArrayList
constructor. That constructor is not generic (only the ArrayList
class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:
Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments
But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List
and ArrayList
, but that again is a different story).
Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?
No, it’s different. The usual type argument/s after the type (ArrayList<Integer>
) are for the generic class. The type arguments before are for the * constructor*.
The two forms may also be combined:
List<Integer> list = new <String, Long>ArrayList<Integer>();
I would consider this a bit more correct since we can now see that the list stores Integer
objects (I’d still prefer to leave out the meaningless <String, Long>
, of course).
Why is it legal to have 2 type arguments when ArrayList only has 1?
First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList
class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).
edited 44 mins ago
answered 1 hour ago
Ole V.V.Ole V.V.
31.3k63956
31.3k63956
But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?
– jaspreet
53 mins ago
Thanks for asking, @jaspreet. Please see my edit.
– Ole V.V.
42 mins ago
add a comment |
But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?
– jaspreet
53 mins ago
Thanks for asking, @jaspreet. Please see my edit.
– Ole V.V.
42 mins ago
But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?
– jaspreet
53 mins ago
But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?
– jaspreet
53 mins ago
Thanks for asking, @jaspreet. Please see my edit.
– Ole V.V.
42 mins ago
Thanks for asking, @jaspreet. Please see my edit.
– Ole V.V.
42 mins ago
add a comment |
Here:
List list = new <String, Long>ArrayList();
You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.
A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.
add a comment |
Here:
List list = new <String, Long>ArrayList();
You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.
A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.
add a comment |
Here:
List list = new <String, Long>ArrayList();
You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.
A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.
Here:
List list = new <String, Long>ArrayList();
You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.
A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.
answered 9 mins ago
GhostCatGhostCat
95.1k1793156
95.1k1793156
add a comment |
add a comment |
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Yeah so do I, I'm not asking how to create a list lol
– Nathan Adams
1 hour ago
2
A constructor may have type arguments that are placed there (this particular constructor hasn’t, so
<String, Long>
is just ignored). See Generics Constructor.– Ole V.V.
1 hour ago
1
OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor
– Nathan Adams
1 hour ago
1
No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator.
List<String> list = new ArrayList<>();
– Elliott Frisch
1 hour ago
@OleV.V. if you wanna put your comment and link as an answer I'll accept it
– Nathan Adams
1 hour ago