How do I fetch tokens on demand not all at once up front?

/*
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 Copyright (c) 2005-2006 Terence Parr
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import org.antlr.runtime.*;
import org.antlr.runtime.BitSet;

import java.util.*;

/** The most common stream of tokens is one where every token is buffered up
 *  and tokens are prefiltered for a certain channel (the parser will only
 *  see these tokens and cannot change the filter channel number during the
 *  parse).
 *
 *  This is a test variation where I try out a dynamic version that only
 *  fetches from lexer when you need it in LT(1) or whatever.
 */
public class DynamicTokenStream extends CommonTokenStream {
	public DynamicTokenStream(TokenSource tokenSource) {
		super(tokenSource);
	}

	public DynamicTokenStream(TokenSource tokenSource, int channel) {
		super(tokenSource, channel);
	}

	/** Load all tokens from the token source and put in tokens.
	 *  This is done upon first LT request because you might want to
	 *  set some token type / channel overrides before filling buffer.
	 */
	protected void fillBuffer() {
		throw new UnsupportedOperationException("should not be called");
	}

	public void fetch() {
		Token t = tokenSource.nextToken();
		// is there a channel override for token type?
		if ( channelOverrideMap!=null ) {
			Integer channelI = (Integer)
				channelOverrideMap.get(new Integer(t.getType()));
			if ( channelI!=null ) {
				t.setChannel(channelI.intValue());
			}
		}
		boolean discard = false;
		if ( discardSet!=null &&
			 discardSet.contains(new Integer(t.getType())) )
		{
			discard = true;
		}
		else if ( discardOffChannelTokens && t.getChannel()!=this.channel ) {
			discard = true;
		}
		if ( !discard )	{
			int index = tokens.size();
			t.setTokenIndex(index);
			tokens.add(t);
		}
	}

	/** Fill between p and n; do nothing if n is p.  If p==-1 and n==0,
	 *  fill in first token.
	 */
	protected void fill(int n) {
		for (int i=p; i<n; i++) {
			fetch();
		}
		if ( p<0 ) {
			p=0; // init
		}
	}

	/** Given a start and stop index, return a List of all tokens in
	 *  the token type BitSet.  Return null if no tokens were found.  This
	 *  method looks at both on and off channel tokens.
	 *  Make sure we have at least stop tokens.
	 */
	public List getTokens(int start, int stop, BitSet types) {
		fill(stop);
		return super.getTokens(start, stop, types);
	}

	/** Get the ith token from the current position 1..n where k=1 is the
	 *  first symbol of lookahead.
	 */
	public Token LT(int k) {
		if ( k==0 ) {
			return null;
		}
		if ( k<0 ) {
			return LB(-k);
		}
		//System.out.println("LT(p="+p+","+k+")");
		fill(p+k);
		//System.out.println(tokens.get(p+k-1));
		int i = p;
		int n = 1;
		// find k good tokens
		while ( n<k ) {
			// skip off-channel tokens
			i = skipOffTokenChannels(i+1); // leave p on valid token
			n++;
		}
		return (Token)tokens.get(i);
	}

	/** Look backwards k tokens on-channel tokens */
	protected Token LB(int k) {
		//System.out.print("LB(p="+p+","+k+") ");
		if ( k==0 ) {
			return null;
		}
		if ( (p-k)<0 ) {
			return null;
		}

		int i = p;
		int n = 1;
		// find k good tokens looking backwards
		while ( n<=k ) {
			// skip off-channel tokens
			i = skipOffTokenChannelsReverse(i-1); // leave p on valid token
			n++;
		}
		if ( i<0 ) {
			return null;
		}
		return (Token)tokens.get(i);
	}

	public String toString(int start, int stop) {
		if ( start<0 || stop<0 ) {
			return null;
		}
		if ( stop>=tokens.size() ) {
			stop = tokens.size()-1;
		}
		 StringBuffer buf = new StringBuffer();
		for (int i = start; i <= stop; i++) {
			Token t = (Token)tokens.get(i);
			buf.append(t.getText());
		}
		return buf.toString();
	}

	public String toString(Token start, Token stop) {
		if ( start!=null && stop!=null ) {
			return toString(start.getTokenIndex(), stop.getTokenIndex());
		}
		return null;
	}
}