How compiler find for initialized/ uninitialized variable

[Mansi Agarwal]

Hi, I just read that- Compiler can detect in very simple conditional cases, that a local variable will be initialized, but will complain in complex cases.
e.g.

private void foo() {
   int i;
   System.out.println("i : " + i); //Compiler error
 
   if(true) {
      i = 10;
   } else {
      i = 20;
   }
   System.out.println("i : " + i); //OK. Compiler can figure out i will be definitely initialized.
 
   int j;
   boolean b = true;
   if(b) {
     j = 10;
   } 
   System.out.println("j : " + j); //Compiler error - can't figure out if j will be initialized or not.
}

Can anyone please explain it how to find whether compiler gives error or not!!

[Seetharaman Venkatasamy]

if the condition is compile time constant, then compiler assumes that block executes.

[Stuart A. Burkett]

In your first example either line 6 or 8 will be executed, so there is no path through the code where i would not be initialised.
If you meant to write just

   if(true) {
      i = 10;
   }
   System.out.println("i : " + i); //OK. Compiler can figure out i will be definitely initialized.

and wondered why that compiled and the second example didn't, it's because true is a compile time constant. The compiler knows it will always be true, whereas in your second example b is a variable and even though it is set to true on the previous line, the compiler doesn't take that into account and so complains.
If you made b a final variable, it would again become a compile time constant and the compiler would be happy again.

[Seetharaman Venkatasamy]

Stuart A. Burkett wrote:
and wondered why that compiled and the second example didn't, it's because true is a compile time constant.

even if it is not a compile time constant, compiler wont complaint because one of the block definitely execute.

[Stuart A. Burkett]

Seetharaman Venkatasamy wrote:

Stuart A. Burkett wrote:
and wondered why that compiled and the second example didn't, it's because true is a compile time constant.

even if it is not a compile time constant, compiler wont complaint because one of the block definitely execute.

My comment referred to the modified code in my post. There's only one block there, there's no else.

[Seetharaman Venkatasamy]

Stuart A. Burkett wrote:Look at the code. There's only one block there, there's no else.

Oh, Sorry. I have misinterpreted second example as another example. apologies...

[Stuart A. Burkett]

No problem

[Mansi Agarwal]

Thanks Stuart and Seetharaman!! That clears my doubt!!