Hello,
I have a question about the order of the execution of statements in a catch block in Java. when I run the following class Test1 (see below), I expect to have as output first Hi!, then the result of the e.printStackTrace(); statement, and then Bye!. However, I never get this order. Please, look at the outputs, which I have pasted below.
public class Test1 {
public static void calculate() {
try {
int h = 5/0;
} catch (ArithmeticException e) {
System.out.println("Hi!");
e.printStackTrace();
}
System.out.println("Bye!");
}
public static void main(String[] args) {
calculate();
}
}
Output1:
Hi! Bye! java.lang.ArithmeticException: / by zero at Test1.calculate(Test1.java:6) at Test1.main(Test1.java:15)
Output2:
java.lang.ArithmeticException: / by zero at Test1.calculate(Test1.java:6) at Test1.main(Test1.java:15) Hi! Bye!
I have two questions:
1.) The more important question: Why I always have Hi! and Bye! printed always one after the other, even though mye.printStackTrace() in the code is between them?
2.) Why sometimes I have the output of the statement e.printStackTrace() before Hi!, and sometimes after Bye! ? I have run the program many times and I cannot understand under what circumstances I get one or the other print.
Thank you.
I am using Java 6, and Eclipse (Ganymed).