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What is the Advantage of Interface over the Inheritance in
OOPS?
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Answer / mms zubeir
1. Provides flexibility in implementing the operations for
the derived classes.
2. Avoid conflicts when more than one interfaces are
derived in a class.
I will club some more later...
| Is This Answer Correct ? | 28 Yes | 9 No |
Answer / iyappan_protech
1)Interface totally keeps away that implementation
knowledge from client.
2)It supports us to change their behaviour dynamically.It
means that, it will act depends on dynamic specilization(or
substitution).
3)This gives very good abstraction about the object to
client.
4)It avoids client broken, even developer made some changes
on implimentation or add new specilization(new
implementation).
5)So it gives open way to extend and implementation.
| Is This Answer Correct ? | 16 Yes | 11 No |
Answer / poorna chandar rao
in inheritence is not support the multiple inhertence but
interface is supported multiple inhertence because imterface
is advatage over the inhertience
| Is This Answer Correct ? | 8 Yes | 10 No |
Answer / rajendra patidar
On OOPS due to multiple inheritance duplicate value may be
received by derived class, so compiler do not understand
what syntax should compile.
Interface Refuse such type of problem in java.
| Is This Answer Correct ? | 0 Yes | 4 No |
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#include <stdio.h> #include <alloc.h> #include <stdlib.h> #include <conio.h> void insert(struct btreenode **, int); void inorder(struct btreenode *); struct btreenode { struct btreenode *leftchild; struct btreenode *rightchild; int data; }; main() { struct btreenode *bt; bt=(struct btreenode *)NULL; int req,i=1,num; clrscr(); printf("Enter number of nodes"); scanf("%d",&req); while(i<=req) { printf("Enter element"); scanf("%d",&num); insert(&bt,num); i++; } inorder(bt); } void insert(struct btreenode **sr, int num) { if(*sr==NULL) { *sr=(struct btreenode *)malloc (sizeof(struct btreenode)); (*sr)->leftchild=(struct btreenode *)NULL; (*sr)->rightchild=(struct btreenode *)NULL; (*sr)->data=num; return; } else { if(num < (*sr)->data) insert(&(*sr)->leftchild,num); else insert(&(*sr)->rightchild,num); } return; } void inorder(struct btreenode *sr) { if(sr!=(struct btreenode *)NULL) { inorder(sr->leftchild); printf("\n %d",sr->data); inorder(sr->rightchild); } else return; } please Modify the given program and add two methods for post order and pre order traversals.
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