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SZE SZE XU
9,759 Pointspredict output practice problem
Note : Execution commandline : $ java exception_handling one
class exception_handling {
public static void main(String args[]) {
try {
int a = args.length;
int b = 10 / a;
System.out.print(a);
try {
if (a == 1)
a = a / a - a;
if (a == 2) {
int c = {1};
c[8] = 9;
}
}
catch (ArrayIndexOutOfBoundException e) {
System.out.println("TypeA");
}
catch (ArithmeticException e) {
System.out.println("TypeB");
}
}
}
}
a) TypeA b) TypeB c) 0TypeA d) 0TypeB
Answer: c Explanation: Execution command line is “$ java exception_handling one” hence there is only single string that is in args array, making its length 1, hence “a = a/ a – a” in second try block is executing which throws arithmeticexception which is caught by catch of firts try block as the nested try block does not have a catch block which can detect ArithmeticException. Hence 0TypeA is printed Output: $ javac exception_handling.java $ java exception_handling 0TypeA
I dont get this, if it is passing one to the program, a would be 1, System.out.print(a) will print 1, why it is 0 ?
1 Answer
Seth Kroger
56,416 PointsYou're all wrong. :-P True output (I renamed the class to Test1):
treehouse:~/workspace$ javac Test1.java
Test1.java:5: error: 'try' without 'catch', 'finally' or resource declarations
try {
^
1 error
treehouse:~/workspace$
Seth Kroger
56,416 PointsSeth Kroger
56,416 PointsSeriously, even if you fix the compiler errors the output would be E) 1. There's no ArithmeticException to throw because a = a / a - a evaluates as a = (a/a) - a or a = 1-a