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33. try {
34. // some code here
35. } catch (NullPointerException e1) {
36. System.out.print(”a”);
37. } catch (RuntimeException e2) {
38. System.out.print(”b”);
39. } finally {
40. System.out.print(”c”);
41. }
What is the result if a NullPointerException occurs on line
34?
1 c
2 a
3 ab
4 ac
- 5 Answers
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Answer / manikandan [ gtec,vellore ]
the answer is 4) ac
because if NullPointerException occurs in a try block it
search for a revelant Exception type so "a" gets printed
then finally ll execute at last and prints c.this both ll
print in same line bcas we didnt use next line that so.
use this code for clear understanding.
class test
{
public static void main(String[]asd)
{
try {
test t=null;
t.a(); //exception occurs in this step
} catch (NullPointerException e1) {
System.out.print("a");
} catch (RuntimeException e2) {
System.out.print("b");
} finally {
System.out.print("c");
}
}
void a()
{
}
}
output:ac
note: if v use RunTimeException b4 the NullPointException it
ll throw the compile time Exception bcas RUnTImeException is
super class for NullPointerException so it ll handel All
it's subclass Exception.
| Is This Answer Correct ? | 14 Yes | 0 No |
Answer / kumar
Once the first catch block executed then rest of the block
skipped.The control transfer to the finally block
the answer is "ac"
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / guest
1)
try {
throw new NullPointerException();
} catch (NullPointerException e1) {
System.out.print("a");
} catch (RuntimeException e2) {
System.out.print("b");
} finally {
System.out.print("c");
}
Ans: ac
2) try {
new NullPointerException();
} catch (NullPointerException e1) {
System.out.print("a");
} catch (RuntimeException e2) {
System.out.print("b");
} finally {
System.out.print("c");
}
Ans: c
| Is This Answer Correct ? | 5 Yes | 2 No |
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