JAVA Bad variable reference? Receiving strange output - possibly memory address?

Question

public class working {

public static String str = "123zASdcvb/;[";

    public static void main(String[] args){
            System.out.println("!!!"+new LetterInventory(str));
    }

public static class LetterInventory {
    public char[] arr;


    public LetterInventory(String str){
        //Strips our string  of all non alphabetic garbage 
        //and assigns it. This regex removes all non a-z
        //without regard for case of alphabet (?!).
        this.arr = str.replaceAll("[^a-z]", "").toLowerCase().toCharArray();
        Arrays.sort(this.arr); //I want to have it sorted to make my life easier.

        System.out.printf("Our freshly sorted LI is : %s\n",this.toString());
    }


    //Simple enough.
    public String toString() {
        return this.arr.toString();
    }




}
}

output:

Our freshly sorted LI is : [C@6d69c9a2
!!![C@6d69c9a2

What is going wrong here? I walked through my debugger and my array does exactly what I want, it is created, cleaned, and sorted just fine. But everytime I try to print my char[].toString() it comes out screwey. Why so?

The array does not Override default implementation of Object to string, sou you end up with hash code in result. — Damian Leszczyński - Vash, Mar 07 '14 at 07:05
Vash, the page you linked contained my answer. I have flagged my post for deletion. — user1695505, Mar 07 '14 at 07:08

score 3 · Answer 1 · edited Jun 20 '20 at 09:12

You're calling toString() on an array - arrays don't override toString(), so you end up with the implementation from Object:

The toString method for class Object returns a string consisting of the name of the class of which the object is an instance, the at-sign character `@', and the unsigned hexadecimal representation of the hash code of the object. In other words, this method returns a string equal to the value of:
getClass().getName() + '@' + Integer.toHexString(hashCode())

As you're dealing with a char[], you should just use the String constructor to create a new string:

return new String(arr);

In other cases you might want to call one of the Arrays.toString overloads, to get a comma-separated representation.

It's not clear why you're doing any of this in a constructor yourself, by the way. Your program would be simpler with just a main method:

// Renamed class to follow Java naming conventions
public class Demo {
    public static void main(String[] args) {
        String input = "123zASdcvb/;[";
        char[] chars = str.replaceAll("[^a-z]", "").toLowerCase().toCharArray();
        Arrays.sort(chars);
        System.out.printf("Our freshly sorted LI is : %s\n", new String(chars));
    }
}

This would work in this case alone as it is a `char` array but in general, `Arrays.toString(arr)` could be used. — Rahul, Mar 07 '14 at 07:10
@Ɍ.Ɉ: Yup, I've edited that in. You almost certainly wouldn't *want* to use `Arrays.toString` in this case though. — Jon Skeet, Mar 07 '14 at 07:11
When you've a `char[]`, its always better to use `new String(char)` but I feel the question title doesn't reflect that and for users seeing this in the future, it'd be better that they knew about `Arrays.toString()` which would ease of their task of writing their own method for printing the array. :) — Rahul, Mar 07 '14 at 07:16

score 0 · Answer 2 · answered Mar 07 '14 at 07:11

0

public String toString() { return String.copyValueOf(arr); }

If you get result [C@6d69c9a2,it means it is a array of chars([ for array,c for char) and its hash code is 0x6d69c9a2

answered Mar 07 '14 at 07:11

tianwei

1,859
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Any reason for preferring a static method call to just calling the constructor here? – Jon Skeet Mar 07 '14 at 07:21
Two ways are similar,just because I post the answer a little bit later... – tianwei Mar 07 '14 at 07:45

JAVA Bad variable reference? Receiving strange output - possibly memory address?

2 Answers2