variable cannot be resolved ( variable used in if else statement)

Question

I m just learning coding so I got this problem " k cannot be resolved"

I have tried introducing variables before if the operator by:

double a = 2.1223;
int b = (int) a;

if(a-b > 0.5) { 
   int k = b + 1;
}

else { 
   int k = (int)b ;
}

System.out.println(k); // k cannot be resolved.why??

I expected output to be 2

instead got k cannot be resolved

Its called [scope](https://www.java-made-easy.com/variable-scope.html) — XtremeBaumer, Aug 06 '19 at 14:50
More precisely, k is out of scope at that location, which is why it cannot be resolved. — Avi, Aug 06 '19 at 14:50
Your `k` members are within the `if` and `else` scopes. Pretty handy, huh? — x80486, Aug 06 '19 at 14:50
This would be much more obvious if you formatted your code correctly so you can find your own mistakes too, please never put code on a bracket like that again for all of our sakes. — Nexevis, Aug 06 '19 at 14:55
@Nexevis Agreed. Submitted an edit and still hasn't accepted it. — Gaurav Mall, Aug 06 '19 at 14:56

Gaurav Mall · Accepted Answer · 2019-08-06T15:00:03.133

The problem is with your implementation. As you see you define k in the start. But the interesting part is you define it as nothing:

int k;

The other k you have defined are inside the if and else if function and not accessible to the println() function. That means you are passing something that is undefined to be printed. That would surely result in an error. This is called scope as XtremeBaumer said. Meaning it is the property of your variables that define where they are visible and usable.

The correct way to do it would be this:

double a = 2.1223 ;
int b = (int) a ;

if(a-b > 0.5) { 
    k = b + 1;
}

else { 
    k = b; //No casting needed
}

System.out.println(k);

That way you are not creating a new k variable every time you check with the if, or else if. What you are doing is using that k variable again by just saying k = whatever. Your problem was that you created the k variable again by doing int k.

Also, as Andy Turner pointed out you don't need to cast b to int because it already is an int.

(Except that you don't declare `k` anywhere). (And `b` is already an `int`, so you don't need to cast it to one). — Andy Turner, Aug 06 '19 at 14:57

variable cannot be resolved ( variable used in if else statement)

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