Syntax error Is it possible to have multiple try blocks with only one catch block in java?

Is it possible to have multiple try blocks with only one catch block in java?

An exception is an issue (run time error) occurred during the execution of a program. When an exception occurred the program gets terminated abruptly and, the code past the line that generated the exception never gets executed.

Example

import java.util.Scanner;
public class ExceptionExample {
   public static void main(String args[]) {
      Scanner sc = new Scanner(System.in);
      System.out.println("Enter first number: ");
      int a = sc.nextInt();
      System.out.println("Enter second number: ");
      int b = sc.nextInt();
      int c = a/b;
      System.out.println("The result is: "+c);
   }
}

Output

Enter first number:
100
Enter second number:
0
Exception in thread "main" java.lang.ArithmeticException: / by zero
at ExceptionExample.main(ExceptionExample.java:10)

Multiple try blocks:

You cannot have multiple try blocks with a single catch block. Each try block must be followed by catch or finally. Still if you try to have single catch block for multiple try blocks a compile time error is generated.

Example

The following Java program tries to employ single catch block for multiple try blocks.

class ExceptionExample{
   public static void main(String args[]) {
      int a,b;
      try {
         a=Integer.parseInt(args[0]);
         b=Integer.parseInt(args[1]);
      }
      try {
         int c=a/b;
         System.out.println(c);
      }catch(Exception ex) {
         System.out.println("Please pass the args while running the program");
      }
   }
}

Compile time exception

ExceptionExample.java:4: error: 'try' without 'catch', 'finally' or resource declarations
   try {
   ^
1 error
Updated on: 2020-07-02T14:10:01+05:30

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