Method for returning even numbers

Question

This method will not compile, can't see why. Error is:

Can't find symbol-variable i

The int i is defined in the for-loop. The method should return the even numbers between int a and int b.

The code:

public int partall(int a, int b){
    int partall;
    int største; 
    int minste;
    if(a == b){
        partall = 0;
    }else{
        if(a>b) {
        største = a;
        minste = b; 
        }else if(a<b){
            minste = a;
            største = b;

            for(int i = minste;  i<= største; i++){
                if(i % 2 == 0) {
                     partall = i;
                }
            }

        }

    }
    return i;
}

`i` is only visible in the scope of the loop, because that's where it's declared. Declare `i` inside the method and it will be visible in the `return` statement. — jlordo, Mar 05 '17 at 11:00
We have tried writing int i under int minste and we still got the error. — Akarlsen, Mar 05 '17 at 11:01
Read my solution below. The new error you got after adding i was because you then need to initialize it. — Max Weinzierl, Mar 05 '17 at 11:12

u32i64 · Answer 1 · 2017-03-05T11:30:49.533

2

i is only visible in the scope of for loop, but not outside. When you do return i; it is out of scope. So you need to put the declaration outside of for block.

Solution:

public int partall(int a, int b){
    int partall;
    int største; 
    int minste;
    int i = 0; 
    if(a == b){
        partall = 0;
    }else{
        if(a>b) {
            største = a;
            minste = b; 
        }else if(a<b){
            minste = a;
            største = b;

            for(i = minste;  i<= største; i++){
                if(i % 2 == 0) {
                    partall = i;
                }
            }

        }

    }
    return i;
}

edited Mar 05 '17 at 11:30

answered Mar 05 '17 at 11:01

u32i64

2,384
3
22
36

Unfortunately we still got the same error saying that int i is not defined – Akarlsen Mar 05 '17 at 11:03
@Akarlsen you need to re-compile your code after making the changes. It might be picking the old .class file – Deepak Singh Mar 05 '17 at 11:05
No, this does produce a compilation error due to i potentially not being initialized. It needs to be initialized to a value. – Max Weinzierl Mar 05 '17 at 11:30
@MaxWeinzierl Now I updated the answer so it is always initialized. – u32i64 Mar 05 '17 at 11:31

score 1 · Accepted Answer · answered Mar 05 '17 at 11:02

1

You are returning i which has scope only in for loop not outside that.

From the implementation, I think you might probably want to return partall but not i.

Initialize partall with something, i.e.

int partall = 0;

and in the last line

return partall;

That should work

answered Mar 05 '17 at 11:02

Anand Undavia

3,493
5
19
33

Thank you, this work like a charm – Akarlsen Mar 05 '17 at 11:13

Max Weinzierl · Answer 3 · 2017-03-05T11:12:13.320

0

The variable i is not visible in the scope of the return statement. You need to declare i outside of the for loop and outside of the if-else for it to be visible in the scope of the return. Try declaring int i = 0; (or even int i = minste; and remove the assignment below) at the top of the function where you declare the rest of your variables.

edited Mar 05 '17 at 11:12

answered Mar 05 '17 at 11:04

Max Weinzierl

1,231
10
13

score 0 · Answer 4 · answered Mar 05 '17 at 12:00

you should initialize "partall" and return that variable:

public int partall(int a, int b){
    int partall = 0;
    int største; 
    int minste;
    if(a == b){
        partall = 0;
    }else {
        if(a>b) {
        største = a;
        minste = b; 
    } else if(a<b) {
        minste = a;
        største = b;

    for(int i = minste;  i<= største; i++){
        if(i % 2 == 0) {
            partall = i;
        }
    }

   }

}
return partall;

}

Method for returning even numbers

4 Answers4