In Java 8, methods can be created as Lambda expressions and can be passed by reference (with a little work under the hood). There are plenty of examples online with lambdas being created and used with methods, but no examples of how to make a method taking a lambda as a parameter. What is the syntax for that?
MyClass.method((a, b) -> a+b);
class MyClass{
//How do I define this method?
static int method(Lambda l){
return l(5, 10);
}
}
Basically to pass a lamda expression as a parameter, we need a type in which we can hold it. Just as an integer value we hold in primitive int or Integer class. Java doesn't have a separate type for lamda expression instead it uses an interface as the type to hold the argument. But that interface should be a functional interface.
There's a public Web-accessible version of the Lambda-enabled Java 8 JavaDocs, linked from http://lambdafaq.org/lambda-resources. (This should obviously be a comment on Joachim Sauer's answer, but I can't get into my SO account with the reputation points I need to add a comment.) The lambdafaq site (I maintain it) answers this and a lot of other Java-lambda questions.
NB This answer was written before the Java 8 GA documentation became publicly available. I've left in place, though, because the Lambda FAQ might still be useful to people learning about features introduced in Java 8.
Lambda is not a object but a Functional Interface. One can define as many as Functional Interfaces as they can using the @FuntionalInterface as an annotation
@FuntionalInterface
public interface SumLambdaExpression {
public int do(int a, int b);
}
public class MyClass {
public static void main(String [] args) {
SumLambdaExpression s = (a,b)->a+b;
lambdaArgFunction(s);
}
public static void lambdaArgFunction(SumLambdaExpression s) {
System.out.println("Output : "+s.do(2,5));
}
}
The Output will be as follows
Output : 7
The Basic concept of a Lambda Expression is to define your own logic but already defined Arguments. So in the above code the you can change the definition of the do function from addition to any other definition, but your arguments are limited to 2.
To me, the solution that makes the most sense is to define a Callback interface :
interface Callback {
void call();
}
and then to use it as parameter in the function you want to call :
void somewhereInYourCode() {
method(() -> {
// You've passed a lambda!
// method() is done, do whatever you want here.
});
}
void method(Callback callback) {
// Do what you have to do
// ...
// Don't forget to notify the caller once you're done
callback.call();
}
A lambda is not a special interface, class or anything else you could declare by yourself. Lambda is just the name given to the () -> {} special syntax, which allows better readability when passing single-method interfaces as parameter. It was designed to replace this :
method(new Callback() {
@Override
public void call() {
// Classic interface implementation, lot of useless boilerplate code.
// method() is done, do whatever you want here.
}
});
So in the example above, Callback is not a lambda, it's just a regular interface ; lambda is the name of the shortcut syntax you can use to implement it.
You can use functional interfaces as mentioned above. below are some of the examples
Function<Integer, Integer> f1 = num->(num*2+1);
System.out.println(f1.apply(10));
Predicate<Integer> f2= num->(num > 10);
System.out.println(f2.test(10));
System.out.println(f2.test(11));
Supplier<Integer> f3= ()-> 100;
System.out.println(f3.get());
Hope it helps
To use Lambda expression you need to either create your own functional interface or use Java functional interface for operation that require two integer and return as value. IntBinaryOperator
Using user defined functional interface
interface TwoArgInterface {
public int operation(int a, int b);
}
public class MyClass {
public static void main(String javalatte[]) {
// this is lambda expression
TwoArgInterface plusOperation = (a, b) -> a + b;
System.out.println("Sum of 10,34 : " + plusOperation.operation(10, 34));
}
}
Using Java functional interface
import java.util.function.IntBinaryOperator;
public class MyClass1 {
static void main(String javalatte[]) {
// this is lambda expression
IntBinaryOperator plusOperation = (a, b) -> a + b;
System.out.println("Sum of 10,34 : " + plusOperation.applyAsInt(10, 34));
}
}
There is flexibility in using lambda as parameter. It enables functional programming in java. The basic syntax is
param -> method_body
Following is a way, you can define a method taking functional interface (lambda is used) as parameter.
a. if you wish to define a method declared inside a functional interface,
say, the functional interface is given as an argument/parameter to a method called from main()
@FunctionalInterface
interface FInterface{
int callMeLambda(String temp);
}
class ConcreteClass{
void funcUsesAnonymousOrLambda(FInterface fi){
System.out.println("===Executing method arg instantiated with Lambda==="));
}
public static void main(){
// calls a method having FInterface as an argument.
funcUsesAnonymousOrLambda(new FInterface() {
int callMeLambda(String temp){ //define callMeLambda(){} here..
return 0;
}
}
}
/***********Can be replaced by Lambda below*********/
funcUsesAnonymousOrLambda( (x) -> {
return 0; //(1)
}
}
FInterface fi = (x) -> { return 0; };
funcUsesAnonymousOrLambda(fi);
Here above it can be seen, how a lambda expression can be replaced with an interface.
Above explains a particular usage of lambda expression, there are more. ref Java 8 lambda within a lambda can't modify variable from outer lambda
Well, that's easy. The purpose of lambda expression is to implement Functional Interface. It is the interface with only one method. Here is awesone article about predefined and legacy functional interfaces.
Anyway, if you want to implement your own functional interface, make it. Just for simple example:
public interface MyFunctionalInterface {
String makeIt(String s);
}
So let's make a class, where we will create a method, which accepts the type of MyFunctionalInterface :
public class Main {
static void printIt(String s, MyFunctionalInterface f) {
System.out.println(f.makeIt(s));
}
public static void main(String[] args) {
}
}
The last thing you should do is to pass the implementation of the MyFunctionalInterface to the method we've defined:
public class Main {
static void printIt(String s, MyFunctionalInterface f) {
System.out.println(f.makeIt(s));
}
public static void main(String[] args) {
printIt("Java", s -> s + " is Awesome");
}
}
That's it!
Do the following ..
You have declared method(lambda l)
All you want to do is create a Interface with the name lambda and declare one abstract method
public int add(int a,int b);
method name does not matter here..
So when u call MyClass.method( (a,b)->a+b)
This implementation (a,b)->a+b will be injected to your interface add method .So whenever you call l.add it is going to take this implementation and perform addition of a and b and return l.add(2,3) will return 5.
- Basically this is what lambda does..
For functions that do not have more than 2 parameters, you can pass them without defining your own interface. For example,
class Klass {
static List<String> foo(Integer a, String b) { ... }
}
class MyClass{
static List<String> method(BiFunction<Integer, String, List<String>> fn){
return fn.apply(5, "FooBar");
}
}
List<String> lStr = MyClass.method((a, b) -> Klass.foo((Integer) a, (String) b));
In BiFunction<Integer, String, List<String>>, Integer and String are its parameters, and List<String> is its return type.
For a function with only one parameter, you can use Function<T, R>, where T is its parameter type, and R is its return value type. Refer to this page for all the interfaces that are already made available by Java.
Here's roughly how C# handles this problem (but expressed as Java code). Something like this could handle almost all your needs:
import static org.util.function.Functions.*;
public class Test {
public static void main(String[] args)
{
Test.invoke((a, b) -> a + b);
}
public static void invoke(Func2<Integer, Integer, Integer> func)
{
System.out.println(func.apply(5, 6));
}
}
package org.util.function;
public interface Functions {
//Actions:
public interface Action {
public void apply();
}
public interface Action1<T1> {
public void apply(T1 arg1);
}
public interface Action2<T1, T2> {
public void apply(T1 arg1, T2 arg2);
}
public interface Action3<T1, T2, T3> {
public void apply(T1 arg1, T2 arg2, T3 arg3);
}
public interface Action4<T1, T2, T3, T4> {
public void apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4);
}
public interface Action5<T1, T2, T3, T4, T5> {
public void apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5);
}
public interface Action6<T1, T2, T3, T4, T5, T6> {
public void apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6);
}
public interface Action7<T1, T2, T3, T4, T5, T6, T7> {
public void apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6, T7 arg7);
}
public interface Action8<T1, T2, T3, T4, T5, T6, T7, T8> {
public void apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6, T7 arg7, T8 arg8);
}
//Functions:
public interface Func<TResult> {
public TResult apply();
}
public interface Func1<T1, TResult> {
public TResult apply(T1 arg1);
}
public interface Func2<T1, T2, TResult> {
public TResult apply(T1 arg1, T2 arg2);
}
public interface Func3<T1, T2, T3, TResult> {
public TResult apply(T1 arg1, T2 arg2, T3 arg3);
}
public interface Func4<T1, T2, T3, T4, TResult> {
public TResult apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4);
}
public interface Func5<T1, T2, T3, T4, T5, TResult> {
public TResult apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5);
}
public interface Func6<T1, T2, T3, T4, T5, T6, TResult> {
public TResult apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6);
}
public interface Func7<T1, T2, T3, T4, T5, T6, T7, TResult> {
public TResult apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6, T7 arg7);
}
public interface Func8<T1, T2, T3, T4, T5, T6, T7, T8, TResult> {
public TResult apply(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6, T7 arg7, T8 arg8);
}
}
For anyone who is googling this, a good method would be to use java.util.function.BiConsumer.
ex:
Import java.util.function.Consumer
public Class Main {
public static void runLambda(BiConsumer<Integer, Integer> lambda) {
lambda.accept(102, 54)
}
public static void main(String[] args) {
runLambda((int1, int2) -> System.out.println(int1 + " + " + int2 + " = " + (int1 + int2)));
}
The outprint would be: 166
Lambda expression can be passed as a argument.To pass a lambda expression as an argument the type of the parameter (which receives the lambda expression as an argument) must be of functional interface type.
If there is a functional interface -
interface IMyFunc {
boolean test(int num);
}
And there is a filter method which adds the int in the list only if it is greater than 5. Note here that filter method has funtional interface IMyFunc as one of the parameter. In that case lambda expression can be passed as an argument for the method parameter.
public class LambdaDemo {
public static List<Integer> filter(IMyFunc testNum, List<Integer> listItems) {
List<Integer> result = new ArrayList<Integer>();
for(Integer item: listItems) {
if(testNum.test(item)) {
result.add(item);
}
}
return result;
}
public static void main(String[] args) {
List<Integer> myList = new ArrayList<Integer>();
myList.add(1);
myList.add(4);
myList.add(6);
myList.add(7);
// calling filter method with a lambda expression
// as one of the param
Collection<Integer> values = filter(n -> n > 5, myList);
System.out.println("Filtered values " + values);
}
}
Source: Stackoverflow.com