About try catch confusion

[Seetharaman Venkatasamy]

Hi, please see the below code

   
 

int method() {
        int i;
        try{
            //int i=5/0;
           return 8;
          //throw new Exception();
        }catch(Exception e){
          i=76;//return 0;
        }finally{
          // return 6;
        }
         return i; 
    }
 
 

i feel bit difficult to understand return statement with try and catch.

also i am geting the value of method=8

i mean System.out.println(method());// it gives 8 and How?


please anyone explain in detail

[Joanne Neal]

Originally posted by seetharaman venkatasamy:
i mean System.out.println(method());// it gives 8 and How?

What did you expect it to print ? Your method has a line that says return 8, no exception is thrown and there is nothing in the finally block, so it returns 8.

[Seetharaman Venkatasamy]

Thanks Neal ... i GOT it ...

one more thing ... in below code...why compilar says that i has not been initialized!

   
int method() {
        int i;
        try{            
           i=30;          
        }catch(Exception e){          
        }
         return i; 
    }
 
 

[ October 15, 2008: Message edited by: seetharaman venkatasamy ]

[Campbell Ritchie]

The values of local variables live on the stack; unless a value is specifically inserted by every path through the method, it is possible that you could return whatever was there on the stack beforehand. So the compiler sees that the "try" might not necessarily set a value to i (even if you don't write an actual Exception) so it insists you set a value to i before the "try."

You may also get "unreachable code" problems with that method.

[Seetharaman Venkatasamy]

Thanks Cambpell ,GOT it

[Campbell Ritchie]

Originally posted by seetharaman venkatasamy:
Thanks Cambpell ,GOT it