Question

1. What are the results of the program as written?
2. What results would the program produce if you changed the value of a from 0 to 3?

*************************************************************************************
* Program Summary:   
*   This code includes simple try/catch (exception handling) functionality.
* We put some arithmetic code into a "try" block so that we can "catch" an exception
* that is thrown during the course working through the arithmetic.
* We need to know specifically what kind of exception we intend to catch.
*
* In this example, we suspect that at some point, one of our variables might hold
* a zero. (In a real-world situation, we might change this code so that a user
* specifies the variable values, and a user might type in zero.)
* This is a problem, because we know we are including the division operator. Dividing
* a number by zero is not possible (mathematically, the result is undefined),
* so we need to be able to handle an exception of type ArithmeticException.
* If we had other code in our try block that threw some other kind of exception,
* our catch block would not handle it. We would need to write another catch block
* to handle another kind of exception.
************************************************************************************/
package prg420week5_analyzeassignment;

public class PRG420Week5_AnalyzeAssignment {

public static void main(String args[]){
int a,b,c;       // We define variables of type integer namned a, b, and c.
try {           // We set up the try check block. We will be looking for an invalid value for integer c
a=0;       // Now we assign values to the variables.
b=15;
c=b/a;       // The result of dividing a number by 0 is undefined. Exception!
System.out.println("This line will not be executed when a=0 because when a=0, the previous line throws an exception to the catch block.");
}
catch(ArithmeticException e) { // As soon as an exception is thrown in the "try" block, execution of this code begins.
System.out.println("Catching this exception thrown by the try block: " + e);
}
System.out.println("Execution resumes here after both the try and catch blocks execute.");
}
}

Answer 1

1. What are the results of the program as written?

Ans: as a is zero. so c=b/a will throw an arithmetic exception. So After this statement (c = b/a), control will transfer to catch block. after printing statement in the catch block, it will print "Execution resumes here after both the try and catch blocks execute." this statement on console

Output

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Q2: What results would the program produce if you changed the value of a from 0 to 3?

Ans: c=b/a = 15/3 = 5

there is no exception. so it will execute full try block and then go to last print statement of code.

Output

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