Handling Exception in Java

Sesha-Savanth — Sun, 24 Oct 2021 19:18:07 +0000

What is exception handling?

An exception is an abnormal condition that disturbs the normal flow of the program. Exception handling is used in handling the runtime errors and maintains the continuity of the program.

Java exceptions

Checked
Unchecked
Errors

Try Block

The statements which throw an exception are to be placed under try block

It is suggested not to write unnecessary statements in the try block because the program terminates at a particular statement where an exception occurs.

Catch Block

It is used to handle the exception by declaring the type of exception within the parameter.

Single or multiple catch blocks can be used after each single try block.

Syntax of try-catch :'

 try{    
   //code that may throw an exception    
    }catch(Exception_class_Name ref){}

Program using Try-Catch :

public class TryCatch {

public static void main(String[] args) {  
    try  
    {  
    int arr[]= {2,4,6,8};  
    System.out.println(arr[10]);    
    }  

    catch(ArrayIndexOutOfBoundsException e)  
    {  
        System.out.println(e);  
    }  
    System.out.println("Program continues"); 
}
}

Output

java.lang.ArrayIndexOutOfBoundsException: 10 Program continues

Program Using multiple catch blocks :

public class Multiple {

public static void main(String[] args) {  

       try{    
            int a[]=new int[5];    

            System.out.println(a[10]);  
           }    
           catch(ArithmeticException e)  
              {  
               System.out.println("Arithmetic Exception occurs");  
              }    
           catch(ArrayIndexOutOfBoundsException e)  
              {  
               System.out.println("ArrayIndexOutOfBounds Exception occurs");  
              }    
           catch(Exception e)  
              {  
               System.out.println("Parent Exception occurs");  
              }             
           System.out.println("Program Continues");    
}  
}

Output

ArrayIndexOutOfBounds Exception occurs Program Continues

Finally Block

The block of statements that are used to execute the important code. It is always executed whether an exception is handled or not.

Program using finally block :

public class Finally{

public static void main(String args[]){

try{

int data=25/0;

System.out.println(data);

}

catch(ArithmeticException e){
System.out.println(e);
}

finally{
System.out.println("finally block is always executed");
}

System.out.println("Program continues");

}

}

Output

java.lang.ArithmeticException: / by zero finally block is always executed Program continues

Floyd-Warshall Algorithm

Sesha-Savanth — Thu, 06 May 2021 11:11:24 +0000

The Floyd-Warshall algorithm is used for finding all pairs shortest path. This algorithm is used to find the shortest path between every pair of vertices in a given edge graph.

Let G = (V,E) be a directed graph with n vertices. Let cost be a cost adjacency matrix for G such that cost(i,i) = 0, 1<=i<=n.

Cost(i,j) = length or cost of edge (i,j), if(i,j) ∈ E(G) and cost(i,j)= ∞ if (i,j) ∉ E(G)

All-pairs shortest path problems is to determine a matrix A such that A(i,j) is the length of a shortest path from i to j

Let k be the highest intermediate vertex between i to j path, then i to k path is a shortest path in graph G going through no vertex with index greater than k-1. Similarly k to j path is a shortest path in graph G going through no vertex with index greater than k-1.

First we need to find highest intermediate vertex k then we need to find the two shortest paths from i to k and k to j.

Then we use A^k(i,j) to represent the length of a shortest path from i to j going through no vertex of index greater than k, we obtain

A^0(i,j)=cost(i,j)

If it goes through highest intermediate vertex k then

A^k(i,j) = A^k-1(i,k)+A^k-1(k,j)

If it does not then highest intermediate vertex is

A^k(i,j) = A^k-1(i,j)

To get a recurrence for $A^{k}$(i,j) we need to combine both

A^k(i,j) =min{ A^k-1(i,j), A^k-1(i,k)+A^k-1(k,j)}, where k>=1

Note: For selected intermediate vertex the path that belongs to that vertex remains same

By taking the above matrix we can get $A^{1}$ matrix.

A^1(2,3) = min{A^0(2,3),A^0(2,1)+A^0(1,3)}

$A^{1}$(2,3) = min{2,8+∞} = 2

A^1(2,4) = min{A^0(2,4),A^0(2,1)+A^0(1,4)}

A^1(2,4) = min{∞,8+7} = 15

A^1(3,2) = min{A^0(3,2),A^0(3,1)+A^0(1,2)}

A^1(3,2) = min{∞,5+3} = 8

A^1(3,4) = min{A^0(4,3),A^0(3,1)+A^0(1,4)}

A^1(3,4) = min{1,5+7} = 1

A^1(4,2) = min{A^0(4,2),A^0(4,1)+A^0(1,2)}

$A^{1}$(4,2) = min{∞,2+3} = 2

A^1(4,3) = min{A^0(4,3),A^0(4,1)+A^0(1,3)}

A^1(4,3) = min{∞,2+∞} = 2

Similarly;

Algorithm

for(k=1;k<=n;k++)
{
  for(i=1;i<=n;i++)
  {
    for(j=1;j<=n;j++)
    {
      if(a[i][j]>a[i][k]+a[k][j])
      {
        a[i][j] = a[i][k]+a[k][j];
      }
    }
  }
}

Program

#include<stdio.h>
void floyd(int a[4][4], int n)
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(a[i][j]>a[i][k]+a[k][j])
                {
                    a[i][j]=a[i][k]+a[k][j];
                }
            }
        }
    }
    printf("All Pairs Shortest Path is :\n");
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                printf("%d ",a[i][j]);
            }
            printf("\n");
        }
}
int main()
{
    int cost[4][4] = {{0, 3, 999, 4}, {8, 0, 2, 999}, {5, 999, 0, 1}, {2, 999, 999, 0}}; 
    int n = 4;

    floyd(cost,n);
}

Output

enter no of vertices :4
The Cost of Adjacency Matrix is :
0 3 9999 7
8 0 2 9999
5 9999 0 1
2 9999 9999 0
All Pairs Shortest Path is :
0 3 5 6
5 0 2 3
3 6 0 1
2 5 7 0

DEV Community: Sesha-Savanth

Handling Exception in Java

What is exception handling?

Java exceptions

Try Block

Catch Block

Syntax of try-catch :'

Program using Try-Catch :

Output

Program Using multiple catch blocks :

Output

Finally Block

Program using finally block :

Output

Floyd-Warshall Algorithm

Similarly;

Program

Output