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Custom functional interface as method argument

How does a method “reference to an instance method of an arbitrary object of a particular type” resolve the arbitrary object?Sort ArrayList of custom Objects by propertyCompile all files in src?Clone a single tone objectMethod override returns nullExecutorService workStealingPool and cancel methodJava - Method executed prior to Default ConstructorWhen use java regular-expression pattern.matcher(), source does not match regex.But, my hope result is ,source matches regexabstract class using its own abstract methodWould it make any difference giving arguments using scanner class instead of command line arguments?java exception - why does it catch?

3

I have the code below that does produce expected result. Could somebody explain what's happening behind the curtains? I don't understand how does the compiler/JVM knows that it's needed to invoke store(String str) on the Storer object or how does it define the doSomething(Storer s, String str) implementation.

Storer.java

public class Storer 
 private List<String> storer = new ArrayList<>();

 public void store(String str) 
 storer.add(str);
 

 @Override
 public String toString() 
 return "Storer [storer=" + storer + "]";
 

MyInterface.java

@FunctionalInterface
public interface MyInterface 
 public abstract void doSomething(Storer s, String str);

Executor.java

public class Executor 
 public void doStore(Storer storer, String s, MyInterface p) 
 p.doSomething(storer, s);
 

TestFunctionalInterfaces.java

public class TestFunctionalInterfaces 
 public static void main(String[] args) 
 Storer storer = new Storer();

 Executor test = new Executor();
 test.doStore(storer, "I've got added", Storer::store);

 System.out.println(storer);
 

An output is:

Storer [storer=[I've got added]]

Thanks in advance.

java functional-interface

share|improve this question

asked Mar 7 at 23:45

AntonAnton

224

add a comment | 

3

I have the code below that does produce expected result. Could somebody explain what's happening behind the curtains? I don't understand how does the compiler/JVM knows that it's needed to invoke store(String str) on the Storer object or how does it define the doSomething(Storer s, String str) implementation.

Storer.java

public class Storer 
 private List<String> storer = new ArrayList<>();

 public void store(String str) 
 storer.add(str);
 

 @Override
 public String toString() 
 return "Storer [storer=" + storer + "]";
 

MyInterface.java

@FunctionalInterface
public interface MyInterface 
 public abstract void doSomething(Storer s, String str);

Executor.java

public class Executor 
 public void doStore(Storer storer, String s, MyInterface p) 
 p.doSomething(storer, s);
 

TestFunctionalInterfaces.java

public class TestFunctionalInterfaces 
 public static void main(String[] args) 
 Storer storer = new Storer();

 Executor test = new Executor();
 test.doStore(storer, "I've got added", Storer::store);

 System.out.println(storer);
 

An output is:

Storer [storer=[I've got added]]

Thanks in advance.

java functional-interface

share|improve this question

asked Mar 7 at 23:45

AntonAnton

224

add a comment | 

I have the code below that does produce expected result. Could somebody explain what's happening behind the curtains? I don't understand how does the compiler/JVM knows that it's needed to invoke store(String str) on the Storer object or how does it define the doSomething(Storer s, String str) implementation.

Storer.java

public class Storer 
 private List<String> storer = new ArrayList<>();

 public void store(String str) 
 storer.add(str);
 

 @Override
 public String toString() 
 return "Storer [storer=" + storer + "]";
 

MyInterface.java

@FunctionalInterface
public interface MyInterface 
 public abstract void doSomething(Storer s, String str);

Executor.java

public class Executor 
 public void doStore(Storer storer, String s, MyInterface p) 
 p.doSomething(storer, s);
 

TestFunctionalInterfaces.java

public class TestFunctionalInterfaces 
 public static void main(String[] args) 
 Storer storer = new Storer();

 Executor test = new Executor();
 test.doStore(storer, "I've got added", Storer::store);

 System.out.println(storer);
 

An output is:

Storer [storer=[I've got added]]

Thanks in advance.

java functional-interface

share|improve this question

asked Mar 7 at 23:45

AntonAnton

224

public class Storer 
 private List<String> storer = new ArrayList<>();

 public void store(String str) 
 storer.add(str);
 

 @Override
 public String toString() 
 return "Storer [storer=" + storer + "]";
 

@FunctionalInterface
public interface MyInterface 
 public abstract void doSomething(Storer s, String str);

public class Executor 
 public void doStore(Storer storer, String s, MyInterface p) 
 p.doSomething(storer, s);
 

public class TestFunctionalInterfaces 
 public static void main(String[] args) 
 Storer storer = new Storer();

 Executor test = new Executor();
 test.doStore(storer, "I've got added", Storer::store);

 System.out.println(storer);
 

Storer [storer=[I've got added]]

share|improve this question

asked Mar 7 at 23:45

AntonAnton

224

share|improve this question

asked Mar 7 at 23:45

AntonAnton

224

asked Mar 7 at 23:45

AntonAnton

224

asked Mar 7 at 23:45

AntonAnton

224

2 Answers
2

active

oldest

votes

1

The method reference Store::storer is equivalent to the lambda (s, str) -> s.store(str). In general, given a functional interface that expects args a1, a2, a3, ..., a method reference of this type is equivalent to a lambda that calls a1.namedMethod(a2, a3, ...). See this answer.

share|improve this answer

answered Mar 8 at 0:11

MathiasMathias

511

add a comment | 

3

Storer::store is a method reference which can be used in place of any @FunctionalInterfaces. What you've got here is essentially a shorthand of an implementation of My Interface. The equivalent being :

public class MyInterfaceImpl implements MyInterface 
 public void doSomething(Storer storer, String str) 
 storer.store(str);
 

This implementation (specified through your method reference) is why storer.store() is executed... because of the implementation of MyInterface you've specified, and passed to, your Executor. The compiler is smart enough to match the implemention/method reference to the arguments you supplied

share|improve this answer

answered Mar 7 at 23:53

tomgeraghty3tomgeraghty3

43128

add a comment | 

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1

The method reference Store::storer is equivalent to the lambda (s, str) -> s.store(str). In general, given a functional interface that expects args a1, a2, a3, ..., a method reference of this type is equivalent to a lambda that calls a1.namedMethod(a2, a3, ...). See this answer.

share|improve this answer

answered Mar 8 at 0:11

MathiasMathias

511

add a comment | 

1

The method reference Store::storer is equivalent to the lambda (s, str) -> s.store(str). In general, given a functional interface that expects args a1, a2, a3, ..., a method reference of this type is equivalent to a lambda that calls a1.namedMethod(a2, a3, ...). See this answer.

share|improve this answer

answered Mar 8 at 0:11

MathiasMathias

511

add a comment | 

The method reference Store::storer is equivalent to the lambda (s, str) -> s.store(str). In general, given a functional interface that expects args a1, a2, a3, ..., a method reference of this type is equivalent to a lambda that calls a1.namedMethod(a2, a3, ...). See this answer.

share|improve this answer

answered Mar 8 at 0:11

MathiasMathias

511

share|improve this answer

answered Mar 8 at 0:11

MathiasMathias

511

answered Mar 8 at 0:11

MathiasMathias

511

answered Mar 8 at 0:11

MathiasMathias

511

3

Storer::store is a method reference which can be used in place of any @FunctionalInterfaces. What you've got here is essentially a shorthand of an implementation of My Interface. The equivalent being :

public class MyInterfaceImpl implements MyInterface 
 public void doSomething(Storer storer, String str) 
 storer.store(str);
 

This implementation (specified through your method reference) is why storer.store() is executed... because of the implementation of MyInterface you've specified, and passed to, your Executor. The compiler is smart enough to match the implemention/method reference to the arguments you supplied

share|improve this answer

answered Mar 7 at 23:53

tomgeraghty3tomgeraghty3

43128

add a comment | 

3

Storer::store is a method reference which can be used in place of any @FunctionalInterfaces. What you've got here is essentially a shorthand of an implementation of My Interface. The equivalent being :

public class MyInterfaceImpl implements MyInterface 
 public void doSomething(Storer storer, String str) 
 storer.store(str);
 

This implementation (specified through your method reference) is why storer.store() is executed... because of the implementation of MyInterface you've specified, and passed to, your Executor. The compiler is smart enough to match the implemention/method reference to the arguments you supplied

share|improve this answer

answered Mar 7 at 23:53

tomgeraghty3tomgeraghty3

43128

add a comment | 

Storer::store is a method reference which can be used in place of any @FunctionalInterfaces. What you've got here is essentially a shorthand of an implementation of My Interface. The equivalent being :

public class MyInterfaceImpl implements MyInterface 
 public void doSomething(Storer storer, String str) 
 storer.store(str);
 

This implementation (specified through your method reference) is why storer.store() is executed... because of the implementation of MyInterface you've specified, and passed to, your Executor. The compiler is smart enough to match the implemention/method reference to the arguments you supplied

share|improve this answer

answered Mar 7 at 23:53

tomgeraghty3tomgeraghty3

43128

public class MyInterfaceImpl implements MyInterface 
 public void doSomething(Storer storer, String str) 
 storer.store(str);
 

share|improve this answer

answered Mar 7 at 23:53

tomgeraghty3tomgeraghty3

43128

answered Mar 7 at 23:53

tomgeraghty3tomgeraghty3

43128

answered Mar 7 at 23:53

tomgeraghty3tomgeraghty3

43128