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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: 02-Jul-2020
7K+ Views
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