How to get the word which is currently entered/modified from SWT TextBox

Question

I am writing a Spell checker functionality in java. I am facing a trouble, when I enter some text in a Text. From this how can I get the word which is currently entered or currently modified so that I can verify whether that word which is currenlty entered or modified is found in my dictionary or not.

I have achieved highlighting the word which is not found in dictionary by underlining with Red color. But I am achieving this by reading entire text for every modification.

Answer 1

I managed to throw some code together that does not iterate the whole text. Instead it moves left and right from the current cursor position searching for the end of the word. When both ends are found, the word is output:

public static void main(String[] args)
{
    Display display = Display.getDefault();
    final Shell shell = new Shell(display);
    shell.setText("StackOverflow");
    shell.setLayout(new GridLayout(2, false));

    Text text = new Text(shell, SWT.BORDER);
    GridData data = new GridData(SWT.FILL, SWT.BEGINNING, true, true);
    data.horizontalSpan = 2;
    text.setLayoutData(data);
    new Label(shell, SWT.NONE).setText("Current word:");
    final Label label = new Label(shell, SWT.NONE);

    text.addListener(SWT.Verify, new Listener()
    {
        @Override
        public void handleEvent(Event e)
        {
            Text source = (Text) e.widget;

            /* Construct the entered text */
            String oldString = source.getText();
            String textString = oldString.substring(0, e.start) + e.text + oldString.substring(e.end);

            /* Get all the chars */
            char[] text = textString.toCharArray();

            /* Get the cursor position */
            int position = source.getCaretPosition();

            /* Adjust cursor position based on input (necessary for delete operations) */
            if(e.text.equals(""))
                position--;
            else
                position++;

            /* Remember start and end of current word */
            int leftBorder = -1;
            int rightBorder = -1;

            /* Search for left end of the current word */
            for(int i = 1; i < position; i++)
            {
                int left = position - i;

                if(left > 0)
                {
                    if(!Character.isLetter(text[left]))
                    {
                        leftBorder = left + 1;
                        break;
                    }
                }
            }

            /* Search for right end of the current word */
            for(int i = position; i < text.length; i++)
            {
                int right = i;

                if(right < text.length)
                {
                    if(!Character.isLetter(text[right]))
                    {
                        rightBorder = right;
                        break;
                    }
                }
            }

            /* If the word is the first/last, set border accordingly */
            if(leftBorder == -1)
                leftBorder = 0;
            if(rightBorder == -1)
                rightBorder = text.length;

            StringBuilder result = new StringBuilder();
            /* Output the word */
            for(int i = leftBorder; i < rightBorder; i++)
                result.append(text[i]);

            label.setText(result.toString());
            shell.layout(true, true);
        }
    });

    shell.pack();
    shell.setSize(600, 100);
    shell.open();
    while (!shell.isDisposed())
    {
        if (!display.readAndDispatch())
            display.sleep();
    }
    display.dispose();
}

Here are two screenshots highlighting the currently edited word:

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So if the whole text is of length n and the current word has length m, the runtime will be O(m) instead of O(n).