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Exception

 
Ranch Hand
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Given:
1 public class Test {
2 public static String output ="";
3
4 public static void foo(int i) {
5 try {
6 if(i==1) {
7 throw new Exception();
8 }
9 output += "1";
10 }
11 catch(Exception e) {
12 output += "2";
13 return;
14 }
15 finally {
16 output += "3";
17 }
18 output += "4";
19 }
20
21 public static void main(String args[]) {
22 foo(0);
23 foo(1);
24
25 }
26 }
What is the value of the variable output at line 23?
 
Greenhorn
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From my point of view it should be
"134234".
After the first invocation,it's "134",because the "finally" block always runs.
And line 23 adds "234" to the string.
Please correct me if any mistake!!
[ December 18, 2005: Message edited by: bibby young ]
 
Greenhorn
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As the last line in the catch block is return. The output would include the last "4"
 
xie li
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young ,are you chinese?
i am
 
Greenhorn
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The output would be "13423". The "4" at last will NOT be included, because of the "return" statement in the catch block.Only finally block will be executed after "return".Please check.

Regards,
Sudhir
 
xie li
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thanks
 
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