s[i] = s[j];
            s[j] = t;
        }

        i = j = 0;
    }
    public void update(byte[] src, int soff, int slen, byte[] dst, int doff)
    {
        int slim;

        slim = soff + slen;
        while (soff < slim) {
            i = (i + 1) & 0xff;
            j = (j + s[i]) & 0xff;
            byte t = s[i];
            s[i] = s[j];
            s[j] = t;

        int start = di, ch;
        int strlen = str.length();

        for( int i = 0; di < dlim && i < strlen; i++ ) {
            ch = str.charAt( i );
            if ((ch >= 0x0001) && (ch <= 0x007F)) {
                dst[di++] = (byte)ch;
            } else if (ch > 0x07FF) {
                if((dlim - di) < 3 ) {
                    break;

        int start = di, ch;
        int strlen = str.length();

        for ( int i = 0; di < dlim && i < strlen; i++ ) {
            ch = str.charAt(i);
            if ( ( ch >= 0x0001 ) && ( ch <= 0x007F ) ) {
                dst[ di++ ] = (byte) ch;
            }
            else if ( ch > 0x07FF ) {
                if ( ( dlim - di ) < 3 ) {
                    break;

    void tryClose() {
        synchronized( LOCK ) {

            /* Yes, there is the potential to drop packets
             * because we might close the socket during a
             * request. However the chances are slim and the
             * retry code should ensure the overall request
             * is serviced. The alternative complicates things
             * more than I think is worth it.
             */

        synchronized ( this.LOCK ) {

            /*
             * Yes, there is the potential to drop packets
             * because we might close the socket during a
             * request. However the chances are slim and the
             * retry code should ensure the overall request
             * is serviced. The alternative complicates things
             * more than I think is worth it.
             */