Code isnt even executing. JAVA

Question

For some reason my if statement is not even being executed at all. Cannot figure out what is wrong, I am trying to test a word called "fuor" and do some manipulations in the hash table I have.

            if(table[getHashIndex(c.toString())].contains(c.toString()))

this is the line that is not executing

Tableclass

char[] c = word.toCharArray();
            for(int i=0;i<word.length()-1;i++)
            {
                char tempChar= c[i];
                c[i]=c[i+1];
                c[i+1]=tempChar;

                if(table[getHashIndex(c.toString())].contains(c.toString()))
                {
                    list.add(c.toString());                 
                   System.out.println("GOT IT BABY");
                }
                c = word.toCharArray();

            }



    public int getHashIndex(String word){
    int key = 7;
    //Adding ASCII values of string
    //To determine the index
    for(int i = 0 ; i < word.length(); i++){
        key = key*BASE+(int)word.charAt(i);
        //Accounting for integer overflow
        if(key<0)
        {
            key*=-1;
        }
    }
    key %= sizeOfTable;
    return key;
}



//Bucket class
public boolean contains(String word){

    Node insert = start;

    //Traversing the list to find a match
    while(insert!=null){
        if(word.equalsIgnoreCase(insert.item))
            return true;

        insert = insert.next;
    }
    //did not find a match
    return false;

}

So what exactly do you want us to do? Have you stepped through the code in the debugger? — OldProgrammer
– OldProgrammer, Commented Mar 22, 2014 at 2:36
why don't you break "table[getHashIndex(c.toString())].contains(c.toString())" into several smaller lines of code to see what's happening? I mean, the debugger is the right way to go, but it's no sin to do this when you're learning — Leo
– Leo, Commented Mar 22, 2014 at 2:40
I found that when I call c.toString on an array it returns some garbage looking values, not sure if its an address or not. This is resulting in a wrong computation of the hash index. — LeatherFace
– LeatherFace, Commented Mar 22, 2014 at 2:41
It's an object identity string -- not an address but related to it. It's of little use except to identify that object. — keshlam
– keshlam, Commented Mar 22, 2014 at 2:43

Elliott Frisch · Accepted Answer · 2014-03-22 02:49:26Z

1

The toString on a Java Array returns the default Object toString() (because arrays are not a primitive type e.g. c.toString() looks like this),

char[] t = new char[] { 'a' };
System.out.println(t.toString());

Output (for example)

[C@25f45022

I think you really wanted something like

char[] t = new char[] { 'a' };
System.out.println(java.util.Arrays.toString(t));

Which outputs

[a]

Or, maybe you wanted something like this

char[] t = new char[] { 'a' };
System.out.println(new String(t));

Which outputs

edited Mar 22, 2014 at 2:49

answered Mar 22, 2014 at 2:40

Elliott Frisch

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2 Comments

LeatherFace Over a year ago

Thank you I just realized that, How do I convert a char array back into a string?

keshlam Over a year ago

new String(characterArray)

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