return updated value in call function

Question

Hi =) I have a question in my coding and I have no idea on it so I make a similar coding post here for help.

this is my coding:

package test;

public class Test {

public static void main(String[] args) {
    int i=0, j=5;
    while(i<10){
        skip(i,j);
        System.out.println(i);
        i++;
    }
}

public static void skip(int i, int j){
    if(i==j){
        i+=1;
    }
}
}

My problem is how can I skip the integer 5 with the call function skip?

my outcome should be like this: ("5" did't print out)

`if (i == 5) { continue; }` - don't forget to increment `i` also. — August, Nov 11 '14 at 04:42
You can't. Java is strictly [pass-by-value](http://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value), so nothing you do in your `skip()` method is going to affect the loop counter in the calling code. — azurefrog, Nov 11 '14 at 04:43
hey dude, I mean using call function...... this is just my similar coding, i not prefer use continue..... if really use continue also can't cause the output is 0 to 4 oni and 6 to 9 is gone.... — user3832964, Nov 11 '14 at 04:45

Elliott Frisch · Accepted Answer · 2014-11-11T04:54:50.157

2

You can't modify the caller's reference to a primitive value. They have no reference, and the wrapper type Integer is immutable. I think you're looking for something like

public static void main(String[] args) {
    int i = 0;
    while (i < 10) {
        System.out.println(i);
        i = skip(i);
    }
}

public static int skip(int i) {
    return (i == 4) ? i + 2 : i + 1;
}

The above ternary could be expressed as

public static int skip(int i) {
    if (i == 4) return i + 2;
    return i + 1;
}

Edit

Or,

public static int skip(int i) {
    if (i + 1 == 5) return i + 2;
    return i + 1;
}

edited Nov 11 '14 at 04:54

answered Nov 11 '14 at 04:46

Elliott Frisch

198,278
20
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I not understand why have to use i == 4 ? – user3832964 Nov 11 '14 at 04:52
Because you want to skip 5, and 4+1 is 5. – Elliott Frisch Nov 11 '14 at 04:53
But i see there is a return of i+2, but why return i+2?? – user3832964 Nov 11 '14 at 04:58
@user3832964 Because 4+2 is 6 and you want to skip 5. – Elliott Frisch Nov 11 '14 at 04:59
can do it without i = skip(i); can do with just a skip(i)? – user3832964 Nov 11 '14 at 05:06
@user3832964 No, because *you can't modify the caller's reference to a primitive value.* – Elliott Frisch Nov 11 '14 at 05:09
see my question again i have edit something.... my problem is inside my function have other complex stuff so i want to do it skip that value in the call function – user3832964 Nov 11 '14 at 05:16
1

@user3832964 You still can't modify the caller's reference. Please read my answer again. – Elliott Frisch Nov 11 '14 at 05:18
Sorry now i edit again already, see again maybe u will understand wat i mean – user3832964 Nov 11 '14 at 05:21
@user3832964 I do see what you mean. **`skip` cannot be `void`**. – Elliott Frisch Nov 11 '14 at 05:22
i pass 2 value in the function now.... tat mean i dunwan to use i = skip(i).... now the function is skip(i,j) – user3832964 Nov 11 '14 at 05:27
@user3832964 `i = skip(i,j);` but skip must return `i`. – Elliott Frisch Nov 11 '14 at 05:29

Naman Gala · Answer 2 · 2014-11-11T05:09:27.477

1

if (i == 5) {
    i++;
    continue;
}

You need to increment i by one before continue, otherwise it will end up for infinite loop because i value is not changing after 5.

Update: Try to run this code. It is not printing abc, so it is going it infinite loop

public static void main(String[] args) {
    int i=0;
    while(i<10){
        if (i == 5) {
            System.out.println(i);
            //i++;
            continue;
        }
        System.out.println(i);
        i++;
    }
    System.out.println("abc");
}

edited Nov 11 '14 at 05:09

answered Nov 11 '14 at 04:54

Naman Gala

4,670
1
21
55

It won't make a difference.. won't end up in infinite loo.p – SonalPM Nov 11 '14 at 05:01
Can any one please tell me the reason for down vote. So that I can keep in mind for next time. – Naman Gala Nov 11 '14 at 07:50

R A Khan · Answer 3 · 2014-11-11T05:15:35.987

try

package test;

public class Test {

    public static void main(String[] args) {
        int i=0;
        while(i<10){
            i=skip(i);
         }
    }

    public static int skip(int i){
        if(i!=5){
            System.out.println(i);
        }
        return i+1;
    }
}

EDIT

package test;

public class Test {

    public static void main(String[] args) {
        int i=0;
        while(i<10){
            i=skip(i);
            System.out.println(i);
         }
    }

    public static int skip(int i){
        if(i==4){
           i +=1;
        }
        return i+1;
    }
}

the system.print i not prefer put in call function – user3832964 Nov 11 '14 at 04:55 — user3832964, Nov 11 '14 at 04:55

SonalPM · Answer 4 · 2014-11-11T04:59:53.857

0

Try this then :)

if(i==5){
   i=skip(i);               
}

private static int skip(int i) {
    i+=1;
    return i;   
}

edited Nov 11 '14 at 04:59

answered Nov 11 '14 at 04:51

SonalPM

1,317
8
17

score 0 · Answer 5 · answered Nov 11 '14 at 05:32

Look at the below code.

package test;

public class Test {
    int i=0;
    public void skip(int x)
    {
        if(x==i)
        {
            i++;
        }
    }
    public static void main(String args[])
    {
        int j=5;
        Test t=new Test();
        while(t.i<10)
        {
            t.skip(j);
            System.out.println(t.i);
            t.i++;
        }
    }
}

return updated value in call function

5 Answers5