Runnable::new vs new Runnable()
Clash Royale CLAN TAG#URR8PPP
Why doesn't the first of the following examples work?
run(R::new);
methodR.run
is not called.run(new R());
methodR.run
is called.
Both examples are compiled-able.
public class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
static class R implements Runnable
R()
System.out.println("R constructor runs");
@Override
public void run()
System.out.println("R.run runs");
The output is:
R constructor runs
-----------------------
R constructor runs
R.run runs
In the first example, the R
constructor is called, it returns lambda (which is not object):
But then how is it possible, that the example is compiled successfully?
java java-8 runnable constructor-reference
add a comment |
Why doesn't the first of the following examples work?
run(R::new);
methodR.run
is not called.run(new R());
methodR.run
is called.
Both examples are compiled-able.
public class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
static class R implements Runnable
R()
System.out.println("R constructor runs");
@Override
public void run()
System.out.println("R.run runs");
The output is:
R constructor runs
-----------------------
R constructor runs
R.run runs
In the first example, the R
constructor is called, it returns lambda (which is not object):
But then how is it possible, that the example is compiled successfully?
java java-8 runnable constructor-reference
2
Note thatRunnable runnable = R::new;
runnable instanceof R
-> false
– Prasad Karunagoda
Jan 7 at 10:48
I don't know about Java specifically butnew
is usually an indicator that you want to allocate some memory that you promise you will clean up yourself.R::new
just sounds like a factory method, a static function in R that creates and returns an instance of Runnable. If this instance is not captured by assigning it to a variable it might be cleaned up the moment it goes out of scope.
– kevin
Jan 8 at 14:22
add a comment |
Why doesn't the first of the following examples work?
run(R::new);
methodR.run
is not called.run(new R());
methodR.run
is called.
Both examples are compiled-able.
public class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
static class R implements Runnable
R()
System.out.println("R constructor runs");
@Override
public void run()
System.out.println("R.run runs");
The output is:
R constructor runs
-----------------------
R constructor runs
R.run runs
In the first example, the R
constructor is called, it returns lambda (which is not object):
But then how is it possible, that the example is compiled successfully?
java java-8 runnable constructor-reference
Why doesn't the first of the following examples work?
run(R::new);
methodR.run
is not called.run(new R());
methodR.run
is called.
Both examples are compiled-able.
public class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
static class R implements Runnable
R()
System.out.println("R constructor runs");
@Override
public void run()
System.out.println("R.run runs");
The output is:
R constructor runs
-----------------------
R constructor runs
R.run runs
In the first example, the R
constructor is called, it returns lambda (which is not object):
But then how is it possible, that the example is compiled successfully?
java java-8 runnable constructor-reference
java java-8 runnable constructor-reference
edited Jan 9 at 12:35
double-beep
1,7132724
1,7132724
asked Jan 7 at 10:15
user1722245user1722245
810722
810722
2
Note thatRunnable runnable = R::new;
runnable instanceof R
-> false
– Prasad Karunagoda
Jan 7 at 10:48
I don't know about Java specifically butnew
is usually an indicator that you want to allocate some memory that you promise you will clean up yourself.R::new
just sounds like a factory method, a static function in R that creates and returns an instance of Runnable. If this instance is not captured by assigning it to a variable it might be cleaned up the moment it goes out of scope.
– kevin
Jan 8 at 14:22
add a comment |
2
Note thatRunnable runnable = R::new;
runnable instanceof R
-> false
– Prasad Karunagoda
Jan 7 at 10:48
I don't know about Java specifically butnew
is usually an indicator that you want to allocate some memory that you promise you will clean up yourself.R::new
just sounds like a factory method, a static function in R that creates and returns an instance of Runnable. If this instance is not captured by assigning it to a variable it might be cleaned up the moment it goes out of scope.
– kevin
Jan 8 at 14:22
2
2
Note that
Runnable runnable = R::new;
runnable instanceof R
-> false– Prasad Karunagoda
Jan 7 at 10:48
Note that
Runnable runnable = R::new;
runnable instanceof R
-> false– Prasad Karunagoda
Jan 7 at 10:48
I don't know about Java specifically but
new
is usually an indicator that you want to allocate some memory that you promise you will clean up yourself. R::new
just sounds like a factory method, a static function in R that creates and returns an instance of Runnable. If this instance is not captured by assigning it to a variable it might be cleaned up the moment it goes out of scope.– kevin
Jan 8 at 14:22
I don't know about Java specifically but
new
is usually an indicator that you want to allocate some memory that you promise you will clean up yourself. R::new
just sounds like a factory method, a static function in R that creates and returns an instance of Runnable. If this instance is not captured by assigning it to a variable it might be cleaned up the moment it goes out of scope.– kevin
Jan 8 at 14:22
add a comment |
4 Answers
4
active
oldest
votes
Your run
method takes a Runnable
instance, and that explains why run(new R())
works with the R
implementation.
R::new
is not equivalent to new R()
. It can fit the signature of a Supplier<Runnable>
(or similar functional interfaces), but R::new
cannot be used as a Runnable
implemented with your R
class.
A version of your run
method that can takeR::new
could look like this (but this would be unnecessarily complex):
void run(Supplier<Runnable> r)
r.get().run();
Why does it compile?
Because the compiler can make a Runnable
out of the constructor call, and that would be equivalent to this lambda expression version:
new ConstructorRefVsNew().run(() ->
new R(); //discarded result, but this is the run() body
);
The same applies to these statements:
Runnable runnable = () -> new R();
new ConstructorRefVsNew().run(runnable);
Runnable runnable2 = R::new;
new ConstructorRefVsNew().run(runnable2);
But, as you can notice, the Runnable
created with R::new
does just call new R()
in its run
method body.
A valid use of a method reference to execute R#run
could use an instance, like this (but you'd surely rather use the r
instance directly, in this case):
R r = new R();
new ConstructorRefVsNew().run(r::run);
Should we assume, that java compiler created from my method run(Runnable r) -> run(Supplier<Runnable> r)?
– user1722245
Jan 7 at 10:21
1
@user1722245 I edited the answer. The compiler made() -> new R();
out ofR::new
in that context.
– ernest_k
Jan 7 at 10:36
add a comment |
The first example:
new ConstructorRefVsNew().run(R::new);
is more or less equivalent to:
new ConstructorRefVsNew().run( () -> new R(); );
The effect is you just create an instance of R but do not call its run
method.
2
It means, that I have two nested runnable. On the first method run is called only.
– user1722245
Jan 7 at 10:29
add a comment |
The run
method expects a Runnable
.
The easy case is new R()
. In this case you know the result is an object of type R
. R
itself is a runnable, it has a run
method, and that's how Java sees it.
But when you pass R::new
something else is happening. What you tell it is to create an anonymous object compatible with a Runnable
whose run
method runs the operation you passed it.
The operation you passed it is not R
's run
method. The operation is the costructor of R
. Thus, it's like you have passed it an anonymous class like:
new Runnable()
public void run()
new R();
(Not all the details are the same, but this is the closest "classical" Java construct ).
R::new
, when called, calls new R()
. Nothing more, nothing less.
add a comment |
Compare two calls:
((Runnable)() -> new R()).run();
new R().run();
By ((Runnable)() -> new R())
or ((Runnable) R::new)
, you create a new Runnable
which does nothing1.
By new R()
, you create an instance of the R
class where the run
method is well-defined.
1 Actually, it creates an object of R
which has no impact on execution.
I was thinking of treating 2 invocations identically without modifying the main
method. We would need to overload run(Runnable)
with run(Supplier<Runnable>)
.
class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
void run(Supplier<Runnable> s)
run(s.get());
static class R implements Runnable ...
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your run
method takes a Runnable
instance, and that explains why run(new R())
works with the R
implementation.
R::new
is not equivalent to new R()
. It can fit the signature of a Supplier<Runnable>
(or similar functional interfaces), but R::new
cannot be used as a Runnable
implemented with your R
class.
A version of your run
method that can takeR::new
could look like this (but this would be unnecessarily complex):
void run(Supplier<Runnable> r)
r.get().run();
Why does it compile?
Because the compiler can make a Runnable
out of the constructor call, and that would be equivalent to this lambda expression version:
new ConstructorRefVsNew().run(() ->
new R(); //discarded result, but this is the run() body
);
The same applies to these statements:
Runnable runnable = () -> new R();
new ConstructorRefVsNew().run(runnable);
Runnable runnable2 = R::new;
new ConstructorRefVsNew().run(runnable2);
But, as you can notice, the Runnable
created with R::new
does just call new R()
in its run
method body.
A valid use of a method reference to execute R#run
could use an instance, like this (but you'd surely rather use the r
instance directly, in this case):
R r = new R();
new ConstructorRefVsNew().run(r::run);
Should we assume, that java compiler created from my method run(Runnable r) -> run(Supplier<Runnable> r)?
– user1722245
Jan 7 at 10:21
1
@user1722245 I edited the answer. The compiler made() -> new R();
out ofR::new
in that context.
– ernest_k
Jan 7 at 10:36
add a comment |
Your run
method takes a Runnable
instance, and that explains why run(new R())
works with the R
implementation.
R::new
is not equivalent to new R()
. It can fit the signature of a Supplier<Runnable>
(or similar functional interfaces), but R::new
cannot be used as a Runnable
implemented with your R
class.
A version of your run
method that can takeR::new
could look like this (but this would be unnecessarily complex):
void run(Supplier<Runnable> r)
r.get().run();
Why does it compile?
Because the compiler can make a Runnable
out of the constructor call, and that would be equivalent to this lambda expression version:
new ConstructorRefVsNew().run(() ->
new R(); //discarded result, but this is the run() body
);
The same applies to these statements:
Runnable runnable = () -> new R();
new ConstructorRefVsNew().run(runnable);
Runnable runnable2 = R::new;
new ConstructorRefVsNew().run(runnable2);
But, as you can notice, the Runnable
created with R::new
does just call new R()
in its run
method body.
A valid use of a method reference to execute R#run
could use an instance, like this (but you'd surely rather use the r
instance directly, in this case):
R r = new R();
new ConstructorRefVsNew().run(r::run);
Should we assume, that java compiler created from my method run(Runnable r) -> run(Supplier<Runnable> r)?
– user1722245
Jan 7 at 10:21
1
@user1722245 I edited the answer. The compiler made() -> new R();
out ofR::new
in that context.
– ernest_k
Jan 7 at 10:36
add a comment |
Your run
method takes a Runnable
instance, and that explains why run(new R())
works with the R
implementation.
R::new
is not equivalent to new R()
. It can fit the signature of a Supplier<Runnable>
(or similar functional interfaces), but R::new
cannot be used as a Runnable
implemented with your R
class.
A version of your run
method that can takeR::new
could look like this (but this would be unnecessarily complex):
void run(Supplier<Runnable> r)
r.get().run();
Why does it compile?
Because the compiler can make a Runnable
out of the constructor call, and that would be equivalent to this lambda expression version:
new ConstructorRefVsNew().run(() ->
new R(); //discarded result, but this is the run() body
);
The same applies to these statements:
Runnable runnable = () -> new R();
new ConstructorRefVsNew().run(runnable);
Runnable runnable2 = R::new;
new ConstructorRefVsNew().run(runnable2);
But, as you can notice, the Runnable
created with R::new
does just call new R()
in its run
method body.
A valid use of a method reference to execute R#run
could use an instance, like this (but you'd surely rather use the r
instance directly, in this case):
R r = new R();
new ConstructorRefVsNew().run(r::run);
Your run
method takes a Runnable
instance, and that explains why run(new R())
works with the R
implementation.
R::new
is not equivalent to new R()
. It can fit the signature of a Supplier<Runnable>
(or similar functional interfaces), but R::new
cannot be used as a Runnable
implemented with your R
class.
A version of your run
method that can takeR::new
could look like this (but this would be unnecessarily complex):
void run(Supplier<Runnable> r)
r.get().run();
Why does it compile?
Because the compiler can make a Runnable
out of the constructor call, and that would be equivalent to this lambda expression version:
new ConstructorRefVsNew().run(() ->
new R(); //discarded result, but this is the run() body
);
The same applies to these statements:
Runnable runnable = () -> new R();
new ConstructorRefVsNew().run(runnable);
Runnable runnable2 = R::new;
new ConstructorRefVsNew().run(runnable2);
But, as you can notice, the Runnable
created with R::new
does just call new R()
in its run
method body.
A valid use of a method reference to execute R#run
could use an instance, like this (but you'd surely rather use the r
instance directly, in this case):
R r = new R();
new ConstructorRefVsNew().run(r::run);
edited Jan 9 at 8:28
Bakudan
13.5k84264
13.5k84264
answered Jan 7 at 10:18
ernest_kernest_k
21k42344
21k42344
Should we assume, that java compiler created from my method run(Runnable r) -> run(Supplier<Runnable> r)?
– user1722245
Jan 7 at 10:21
1
@user1722245 I edited the answer. The compiler made() -> new R();
out ofR::new
in that context.
– ernest_k
Jan 7 at 10:36
add a comment |
Should we assume, that java compiler created from my method run(Runnable r) -> run(Supplier<Runnable> r)?
– user1722245
Jan 7 at 10:21
1
@user1722245 I edited the answer. The compiler made() -> new R();
out ofR::new
in that context.
– ernest_k
Jan 7 at 10:36
Should we assume, that java compiler created from my method run(Runnable r) -> run(Supplier<Runnable> r)?
– user1722245
Jan 7 at 10:21
Should we assume, that java compiler created from my method run(Runnable r) -> run(Supplier<Runnable> r)?
– user1722245
Jan 7 at 10:21
1
1
@user1722245 I edited the answer. The compiler made
() -> new R();
out of R::new
in that context.– ernest_k
Jan 7 at 10:36
@user1722245 I edited the answer. The compiler made
() -> new R();
out of R::new
in that context.– ernest_k
Jan 7 at 10:36
add a comment |
The first example:
new ConstructorRefVsNew().run(R::new);
is more or less equivalent to:
new ConstructorRefVsNew().run( () -> new R(); );
The effect is you just create an instance of R but do not call its run
method.
2
It means, that I have two nested runnable. On the first method run is called only.
– user1722245
Jan 7 at 10:29
add a comment |
The first example:
new ConstructorRefVsNew().run(R::new);
is more or less equivalent to:
new ConstructorRefVsNew().run( () -> new R(); );
The effect is you just create an instance of R but do not call its run
method.
2
It means, that I have two nested runnable. On the first method run is called only.
– user1722245
Jan 7 at 10:29
add a comment |
The first example:
new ConstructorRefVsNew().run(R::new);
is more or less equivalent to:
new ConstructorRefVsNew().run( () -> new R(); );
The effect is you just create an instance of R but do not call its run
method.
The first example:
new ConstructorRefVsNew().run(R::new);
is more or less equivalent to:
new ConstructorRefVsNew().run( () -> new R(); );
The effect is you just create an instance of R but do not call its run
method.
edited Jan 7 at 10:48
answered Jan 7 at 10:23
HenryHenry
33.6k54260
33.6k54260
2
It means, that I have two nested runnable. On the first method run is called only.
– user1722245
Jan 7 at 10:29
add a comment |
2
It means, that I have two nested runnable. On the first method run is called only.
– user1722245
Jan 7 at 10:29
2
2
It means, that I have two nested runnable. On the first method run is called only.
– user1722245
Jan 7 at 10:29
It means, that I have two nested runnable. On the first method run is called only.
– user1722245
Jan 7 at 10:29
add a comment |
The run
method expects a Runnable
.
The easy case is new R()
. In this case you know the result is an object of type R
. R
itself is a runnable, it has a run
method, and that's how Java sees it.
But when you pass R::new
something else is happening. What you tell it is to create an anonymous object compatible with a Runnable
whose run
method runs the operation you passed it.
The operation you passed it is not R
's run
method. The operation is the costructor of R
. Thus, it's like you have passed it an anonymous class like:
new Runnable()
public void run()
new R();
(Not all the details are the same, but this is the closest "classical" Java construct ).
R::new
, when called, calls new R()
. Nothing more, nothing less.
add a comment |
The run
method expects a Runnable
.
The easy case is new R()
. In this case you know the result is an object of type R
. R
itself is a runnable, it has a run
method, and that's how Java sees it.
But when you pass R::new
something else is happening. What you tell it is to create an anonymous object compatible with a Runnable
whose run
method runs the operation you passed it.
The operation you passed it is not R
's run
method. The operation is the costructor of R
. Thus, it's like you have passed it an anonymous class like:
new Runnable()
public void run()
new R();
(Not all the details are the same, but this is the closest "classical" Java construct ).
R::new
, when called, calls new R()
. Nothing more, nothing less.
add a comment |
The run
method expects a Runnable
.
The easy case is new R()
. In this case you know the result is an object of type R
. R
itself is a runnable, it has a run
method, and that's how Java sees it.
But when you pass R::new
something else is happening. What you tell it is to create an anonymous object compatible with a Runnable
whose run
method runs the operation you passed it.
The operation you passed it is not R
's run
method. The operation is the costructor of R
. Thus, it's like you have passed it an anonymous class like:
new Runnable()
public void run()
new R();
(Not all the details are the same, but this is the closest "classical" Java construct ).
R::new
, when called, calls new R()
. Nothing more, nothing less.
The run
method expects a Runnable
.
The easy case is new R()
. In this case you know the result is an object of type R
. R
itself is a runnable, it has a run
method, and that's how Java sees it.
But when you pass R::new
something else is happening. What you tell it is to create an anonymous object compatible with a Runnable
whose run
method runs the operation you passed it.
The operation you passed it is not R
's run
method. The operation is the costructor of R
. Thus, it's like you have passed it an anonymous class like:
new Runnable()
public void run()
new R();
(Not all the details are the same, but this is the closest "classical" Java construct ).
R::new
, when called, calls new R()
. Nothing more, nothing less.
answered Jan 7 at 10:24
RealSkepticRealSkeptic
27.9k63261
27.9k63261
add a comment |
add a comment |
Compare two calls:
((Runnable)() -> new R()).run();
new R().run();
By ((Runnable)() -> new R())
or ((Runnable) R::new)
, you create a new Runnable
which does nothing1.
By new R()
, you create an instance of the R
class where the run
method is well-defined.
1 Actually, it creates an object of R
which has no impact on execution.
I was thinking of treating 2 invocations identically without modifying the main
method. We would need to overload run(Runnable)
with run(Supplier<Runnable>)
.
class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
void run(Supplier<Runnable> s)
run(s.get());
static class R implements Runnable ...
add a comment |
Compare two calls:
((Runnable)() -> new R()).run();
new R().run();
By ((Runnable)() -> new R())
or ((Runnable) R::new)
, you create a new Runnable
which does nothing1.
By new R()
, you create an instance of the R
class where the run
method is well-defined.
1 Actually, it creates an object of R
which has no impact on execution.
I was thinking of treating 2 invocations identically without modifying the main
method. We would need to overload run(Runnable)
with run(Supplier<Runnable>)
.
class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
void run(Supplier<Runnable> s)
run(s.get());
static class R implements Runnable ...
add a comment |
Compare two calls:
((Runnable)() -> new R()).run();
new R().run();
By ((Runnable)() -> new R())
or ((Runnable) R::new)
, you create a new Runnable
which does nothing1.
By new R()
, you create an instance of the R
class where the run
method is well-defined.
1 Actually, it creates an object of R
which has no impact on execution.
I was thinking of treating 2 invocations identically without modifying the main
method. We would need to overload run(Runnable)
with run(Supplier<Runnable>)
.
class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
void run(Supplier<Runnable> s)
run(s.get());
static class R implements Runnable ...
Compare two calls:
((Runnable)() -> new R()).run();
new R().run();
By ((Runnable)() -> new R())
or ((Runnable) R::new)
, you create a new Runnable
which does nothing1.
By new R()
, you create an instance of the R
class where the run
method is well-defined.
1 Actually, it creates an object of R
which has no impact on execution.
I was thinking of treating 2 invocations identically without modifying the main
method. We would need to overload run(Runnable)
with run(Supplier<Runnable>)
.
class ConstructorRefVsNew
public static void main(String args)
new ConstructorRefVsNew().run(R::new);
System.out.println("-----------------------");
new ConstructorRefVsNew().run(new R());
void run(Runnable r)
r.run();
void run(Supplier<Runnable> s)
run(s.get());
static class R implements Runnable ...
edited Jan 7 at 15:07
answered Jan 7 at 10:28
Andrew TobilkoAndrew Tobilko
27.1k104285
27.1k104285
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2
Note that
Runnable runnable = R::new;
runnable instanceof R
-> false– Prasad Karunagoda
Jan 7 at 10:48
I don't know about Java specifically but
new
is usually an indicator that you want to allocate some memory that you promise you will clean up yourself.R::new
just sounds like a factory method, a static function in R that creates and returns an instance of Runnable. If this instance is not captured by assigning it to a variable it might be cleaned up the moment it goes out of scope.– kevin
Jan 8 at 14:22