Question

So, I’ve created a HashMap(Resource being my own creation), which each key is the filepath for each resource. Now, let’s say I have a Resource with a filepath “mat\10wdim3.mat” in the HashMap, trying to retrieve it will fail because both hashcodes are not equal.

The hashcode in the Map for this key is: -347056295
The string I’m using to try and retrieve that resource: -2128683668

There is no hidden characters I could find in either strings. Is there any other way that the hashcodes wouldn’t match?

EDIT:

Including some sample code

Initializing in the constructor. Creating and adding the resources(The filepath is trimmed beforehand)

private Map<String, Resource> m_Files;

public ResourceManager() {
    m_Files = new HashMap<String, Resource>();
}

public Resource createResource(String filepath, byte[] contents) {
    if (filepath != null && contents != null && contents.length > 0) {          
        String extension;
        int lastIndex = filepath.lastIndexOf(".");
        Resource res = null;

        if (lastIndex == -1)
            return null;

        extension = filepath.substring(lastIndex).toLowerCase();

        // TODO: Optimize this by putting the most common at the top
        try {
            if (extension.equals(".pup"))
                res = new Puppet(contents);
            else if (extension.equals(".bm"))
                res = new Bitmap(contents);
            else if (extension.equals(".snd"))
                res = new SoundFile(contents);
            else if (extension.equals(".cmp"))
                res = new ColorMap(contents);
            else if (extension.equals(".mat"))
                res = new Mat(contents);
            else if (extension.equals(".sft"))
                res = new Font(contents);
            /*else if (extension.equals(".ai") // Normal AI
                    || extension.equals(".ai0") // Easy AI
                    || extension.equals(".ai2")) // Hard AI
                res = new AIFile(contents);*/

            if (res != null) {
                addResource(filepath, res);

                return res;
            }
        } catch (RuntimeException e) {
            e.printStackTrace();
        }
    }

    return null;
}

public void addResource(String filepath, Resource res) {
    if (filepath != null && res != null && m_Files.put(filepath, res) != null)
        System.out.println("[!] Replacing " + filepath);
}

public Resource getResource(String filepath) {
    return m_Files.get(filepath);
}

To get said resource

ResourceManager rManager = new ResourceManager();

rManager.loadGOB("C:/Documents and Settings/Unrealomega/Desktop/JKDF2/GOB/Resource/Res2.gob");

Resource res = rManager.getResource("mat\10wdim3.mat");

Answer 1

The problem is that you did not escape letter ‘\’ to ‘\\’;

“mat\10wdim3.mat”.hashCode(); //-2128683668

but if you escape letter ‘\’,you will get the follow ouput:

“mat\\10wdim3.mat”.hashCode(); // -347056295

So，just escape the filePath String ,when you fetch the resource from HashMap!