Lambda Expression JAVA-8
I have just started with JAVA 1.8 version and had a question while going through the tutorials regarding lambda expression.
Can we have more than 1 implementation(lambda expression) for the abstract method by creating multiple instances of the Interface WITHIN THE SAME CLASS???
I tried the code and it ran perfectly....
Now my question is that the very concept of interface is that every IMPLEMENTING CLASS WILL HAVE A DEFINITION FOR THE ABSTRACT METHOD. THEN HOW CAN WE HAVE TWO METHOD BODIES(lambda expressions) in the SAME CLASS ???
Consider the below code :
public static void main(String args) {
Interf i = (a, b) -> a + b;
System.out.println("The result is >> " + i.result(10, 20));
Interf i1 = (a, b) -> a - b;
System.out.println("The result is >> " + i1.result(10, 20));
}
Output:
The result is >> 30
The result is >> -10
lambda java-8
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
add a comment |
I have just started with JAVA 1.8 version and had a question while going through the tutorials regarding lambda expression.
Can we have more than 1 implementation(lambda expression) for the abstract method by creating multiple instances of the Interface WITHIN THE SAME CLASS???
I tried the code and it ran perfectly....
Now my question is that the very concept of interface is that every IMPLEMENTING CLASS WILL HAVE A DEFINITION FOR THE ABSTRACT METHOD. THEN HOW CAN WE HAVE TWO METHOD BODIES(lambda expressions) in the SAME CLASS ???
Consider the below code :
public static void main(String args) {
Interf i = (a, b) -> a + b;
System.out.println("The result is >> " + i.result(10, 20));
Interf i1 = (a, b) -> a - b;
System.out.println("The result is >> " + i1.result(10, 20));
}
Output:
The result is >> 30
The result is >> -10
lambda java-8
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
add a comment |
I have just started with JAVA 1.8 version and had a question while going through the tutorials regarding lambda expression.
Can we have more than 1 implementation(lambda expression) for the abstract method by creating multiple instances of the Interface WITHIN THE SAME CLASS???
I tried the code and it ran perfectly....
Now my question is that the very concept of interface is that every IMPLEMENTING CLASS WILL HAVE A DEFINITION FOR THE ABSTRACT METHOD. THEN HOW CAN WE HAVE TWO METHOD BODIES(lambda expressions) in the SAME CLASS ???
Consider the below code :
public static void main(String args) {
Interf i = (a, b) -> a + b;
System.out.println("The result is >> " + i.result(10, 20));
Interf i1 = (a, b) -> a - b;
System.out.println("The result is >> " + i1.result(10, 20));
}
Output:
The result is >> 30
The result is >> -10
lambda java-8
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
I have just started with JAVA 1.8 version and had a question while going through the tutorials regarding lambda expression.
Can we have more than 1 implementation(lambda expression) for the abstract method by creating multiple instances of the Interface WITHIN THE SAME CLASS???
I tried the code and it ran perfectly....
Now my question is that the very concept of interface is that every IMPLEMENTING CLASS WILL HAVE A DEFINITION FOR THE ABSTRACT METHOD. THEN HOW CAN WE HAVE TWO METHOD BODIES(lambda expressions) in the SAME CLASS ???
Consider the below code :
public static void main(String args) {
Interf i = (a, b) -> a + b;
System.out.println("The result is >> " + i.result(10, 20));
Interf i1 = (a, b) -> a - b;
System.out.println("The result is >> " + i1.result(10, 20));
}
Output:
The result is >> 30
The result is >> -10
lambda java-8
lambda java-8
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
asked Nov 15 '18 at 6:30
Lokesh BaranwalLokesh Baranwal
132
132
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Each of your two lambda expressions implements your Interf functional interface separately. Therefore each implementation of that interface has a single implementation of that interface's single abstract method.
Even before Java 8 and lambda expressions you could create two anonymous class instances that implement your Interf interface. Each one of them would have a single implementation of Interf's single method.
The fact that the two lambda expressions are defined in the same class doesn't mean that the two implementations of the functional interface's abstract method belong to the same class.
answered Nov 15 '18 at 6:33
EranEran
287k37467557
add a comment |
Actually they are two different implementation of the same interface. In fact, you can use the existing BinaryOperator to do the same rather than reinventing the wheel like so.
BinaryOperator<Integer> sum = (n1, n2) -> n1 + n2;
BinaryOperator<Integer> substract = (n1, n2) -> n1 - n2;
System.out.println(sum.apply(10, 20));
System.out.println(substract.apply(10, 20));
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
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
Each of your two lambda expressions implements your Interf functional interface separately. Therefore each implementation of that interface has a single implementation of that interface's single abstract method.
Even before Java 8 and lambda expressions you could create two anonymous class instances that implement your Interf interface. Each one of them would have a single implementation of Interf's single method.
The fact that the two lambda expressions are defined in the same class doesn't mean that the two implementations of the functional interface's abstract method belong to the same class.
answered Nov 15 '18 at 6:33
EranEran
287k37467557
add a comment |
Each of your two lambda expressions implements your Interf functional interface separately. Therefore each implementation of that interface has a single implementation of that interface's single abstract method.
Even before Java 8 and lambda expressions you could create two anonymous class instances that implement your Interf interface. Each one of them would have a single implementation of Interf's single method.
The fact that the two lambda expressions are defined in the same class doesn't mean that the two implementations of the functional interface's abstract method belong to the same class.
answered Nov 15 '18 at 6:33
EranEran
287k37467557
add a comment |
Each of your two lambda expressions implements your Interf functional interface separately. Therefore each implementation of that interface has a single implementation of that interface's single abstract method.
Even before Java 8 and lambda expressions you could create two anonymous class instances that implement your Interf interface. Each one of them would have a single implementation of Interf's single method.
The fact that the two lambda expressions are defined in the same class doesn't mean that the two implementations of the functional interface's abstract method belong to the same class.
answered Nov 15 '18 at 6:33
EranEran
287k37467557
Each of your two lambda expressions implements your Interf functional interface separately. Therefore each implementation of that interface has a single implementation of that interface's single abstract method.
Even before Java 8 and lambda expressions you could create two anonymous class instances that implement your Interf interface. Each one of them would have a single implementation of Interf's single method.
The fact that the two lambda expressions are defined in the same class doesn't mean that the two implementations of the functional interface's abstract method belong to the same class.
answered Nov 15 '18 at 6:33
EranEran
287k37467557
answered Nov 15 '18 at 6:33
EranEran
287k37467557
answered Nov 15 '18 at 6:33
EranEran
287k37467557
answered Nov 15 '18 at 6:33
EranEran
287k37467557
287k37467557
add a comment |
add a comment |
Actually they are two different implementation of the same interface. In fact, you can use the existing BinaryOperator to do the same rather than reinventing the wheel like so.
BinaryOperator<Integer> sum = (n1, n2) -> n1 + n2;
BinaryOperator<Integer> substract = (n1, n2) -> n1 - n2;
System.out.println(sum.apply(10, 20));
System.out.println(substract.apply(10, 20));
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
add a comment |
Actually they are two different implementation of the same interface. In fact, you can use the existing BinaryOperator to do the same rather than reinventing the wheel like so.
BinaryOperator<Integer> sum = (n1, n2) -> n1 + n2;
BinaryOperator<Integer> substract = (n1, n2) -> n1 - n2;
System.out.println(sum.apply(10, 20));
System.out.println(substract.apply(10, 20));
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
add a comment |
Actually they are two different implementation of the same interface. In fact, you can use the existing BinaryOperator to do the same rather than reinventing the wheel like so.
BinaryOperator<Integer> sum = (n1, n2) -> n1 + n2;
BinaryOperator<Integer> substract = (n1, n2) -> n1 - n2;
System.out.println(sum.apply(10, 20));
System.out.println(substract.apply(10, 20));
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
Actually they are two different implementation of the same interface. In fact, you can use the existing BinaryOperator to do the same rather than reinventing the wheel like so.
BinaryOperator<Integer> sum = (n1, n2) -> n1 + n2;
BinaryOperator<Integer> substract = (n1, n2) -> n1 - n2;
System.out.println(sum.apply(10, 20));
System.out.println(substract.apply(10, 20));
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
answered Nov 15 '18 at 6:35
Ravindra RanwalaRavindra Ranwala
10.7k42040
10.7k42040
add a comment |
add a comment |
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