Null in functional interface with different type return
I write this code, but I don't know why it compiles.
The UnaryOperator takes a specific type argument and returns the result with the same type of its argument.
My question: if I put an if-statement with the return null, isn't there a compiler error?
null isn't the type of the its argument (in my case is Doll)?
Can a built-in functional interface (like Consumer, UnaryOperator, Function) return null instead of its standard return?
This is my code:
import java.util.function.*;
public class Doll {
private int layer;
public Doll(int layer) {
super();
this.layer = layer;
}
public static void open(UnaryOperator<Doll> task, Doll doll) {
while ((doll = task.apply(doll)) != null) {
System.out.println("X");
}
}
public static void main(String args) {
open(s -> {
if (s.layer <= 0)
return null;
else
return new Doll(s.layer--);
}, new Doll(5));
}
}
Thanks a lot!
java if-statement return-type unary-operator
edited Nov 12 '18 at 15:19
deHaar
2,15631527
asked Nov 12 '18 at 15:01
Adryr83
889
add a comment |
I write this code, but I don't know why it compiles.
The UnaryOperator takes a specific type argument and returns the result with the same type of its argument.
My question: if I put an if-statement with the return null, isn't there a compiler error?
null isn't the type of the its argument (in my case is Doll)?
Can a built-in functional interface (like Consumer, UnaryOperator, Function) return null instead of its standard return?
This is my code:
import java.util.function.*;
public class Doll {
private int layer;
public Doll(int layer) {
super();
this.layer = layer;
}
public static void open(UnaryOperator<Doll> task, Doll doll) {
while ((doll = task.apply(doll)) != null) {
System.out.println("X");
}
}
public static void main(String args) {
open(s -> {
if (s.layer <= 0)
return null;
else
return new Doll(s.layer--);
}, new Doll(5));
}
}
Thanks a lot!
java if-statement return-type unary-operator
edited Nov 12 '18 at 15:19
deHaar
2,15631527
asked Nov 12 '18 at 15:01
Adryr83
889
add a comment |
I write this code, but I don't know why it compiles.
The UnaryOperator takes a specific type argument and returns the result with the same type of its argument.
My question: if I put an if-statement with the return null, isn't there a compiler error?
null isn't the type of the its argument (in my case is Doll)?
Can a built-in functional interface (like Consumer, UnaryOperator, Function) return null instead of its standard return?
This is my code:
import java.util.function.*;
public class Doll {
private int layer;
public Doll(int layer) {
super();
this.layer = layer;
}
public static void open(UnaryOperator<Doll> task, Doll doll) {
while ((doll = task.apply(doll)) != null) {
System.out.println("X");
}
}
public static void main(String args) {
open(s -> {
if (s.layer <= 0)
return null;
else
return new Doll(s.layer--);
}, new Doll(5));
}
}
Thanks a lot!
java if-statement return-type unary-operator
edited Nov 12 '18 at 15:19
deHaar
2,15631527
asked Nov 12 '18 at 15:01
Adryr83
889
I write this code, but I don't know why it compiles.
The UnaryOperator takes a specific type argument and returns the result with the same type of its argument.
My question: if I put an if-statement with the return null, isn't there a compiler error?
null isn't the type of the its argument (in my case is Doll)?
Can a built-in functional interface (like Consumer, UnaryOperator, Function) return null instead of its standard return?
This is my code:
import java.util.function.*;
public class Doll {
private int layer;
public Doll(int layer) {
super();
this.layer = layer;
}
public static void open(UnaryOperator<Doll> task, Doll doll) {
while ((doll = task.apply(doll)) != null) {
System.out.println("X");
}
}
public static void main(String args) {
open(s -> {
if (s.layer <= 0)
return null;
else
return new Doll(s.layer--);
}, new Doll(5));
}
}
Thanks a lot!
java if-statement return-type unary-operator
java if-statement return-type unary-operator
edited Nov 12 '18 at 15:19
deHaar
2,15631527
asked Nov 12 '18 at 15:01
Adryr83
889
edited Nov 12 '18 at 15:19
deHaar
2,15631527
asked Nov 12 '18 at 15:01
Adryr83
889
edited Nov 12 '18 at 15:19
deHaar
2,15631527
edited Nov 12 '18 at 15:19
deHaar
2,15631527
edited Nov 12 '18 at 15:19
deHaar
2,15631527
2,15631527
asked Nov 12 '18 at 15:01
Adryr83
889
asked Nov 12 '18 at 15:01
Adryr83
889
asked Nov 12 '18 at 15:01
Adryr83
889
889
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Think of it this way:
Doll d = null;
null is a valid reference for any object, and isn't different for functional interfaces.
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
add a comment |
There's nothing special about built-in functional interfaces. Any method that returns a reference type can return null. This includes the lambda expression implementation of the method of a functional interface.
answered Nov 12 '18 at 15:05
Eran
280k37452538
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
Think of it this way:
Doll d = null;
null is a valid reference for any object, and isn't different for functional interfaces.
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
add a comment |
Think of it this way:
Doll d = null;
null is a valid reference for any object, and isn't different for functional interfaces.
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
add a comment |
Think of it this way:
Doll d = null;
null is a valid reference for any object, and isn't different for functional interfaces.
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
Think of it this way:
Doll d = null;
null is a valid reference for any object, and isn't different for functional interfaces.
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
answered Nov 12 '18 at 15:05
Jacob G.
15.2k52162
15.2k52162
add a comment |
add a comment |
There's nothing special about built-in functional interfaces. Any method that returns a reference type can return null. This includes the lambda expression implementation of the method of a functional interface.
answered Nov 12 '18 at 15:05
Eran
280k37452538
add a comment |
There's nothing special about built-in functional interfaces. Any method that returns a reference type can return null. This includes the lambda expression implementation of the method of a functional interface.
answered Nov 12 '18 at 15:05
Eran
280k37452538
add a comment |
There's nothing special about built-in functional interfaces. Any method that returns a reference type can return null. This includes the lambda expression implementation of the method of a functional interface.
answered Nov 12 '18 at 15:05
Eran
280k37452538
There's nothing special about built-in functional interfaces. Any method that returns a reference type can return null. This includes the lambda expression implementation of the method of a functional interface.
answered Nov 12 '18 at 15:05
Eran
280k37452538
answered Nov 12 '18 at 15:05
Eran
280k37452538
answered Nov 12 '18 at 15:05
Eran
280k37452538
answered Nov 12 '18 at 15:05
Eran
280k37452538
280k37452538
add a comment |
add a comment |
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