rkutil.java

java 报表 to office文档： 本包由java语言开发

/* ====================================================================
   Copyright 2002-2004   Apache Software Foundation

   Licensed under the Apache License, Version 2.0 (the "License");
   you may not use this file except in compliance with the License.
   You may obtain a copy of the License at

       http://www.apache.org/licenses/LICENSE-2.0

   Unless required by applicable law or agreed to in writing, software
   distributed under the License is distributed on an "AS IS" BASIS,
   WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
   See the License for the specific language governing permissions and
   limitations under the License.
==================================================================== */

package org.apache.poi.hssf.util;

/**
 * Utility class for helping convert RK numbers.
 *
 * @author Andrew C. Oliver (acoliver at apache dot org)
 * @author Glen Stampoultzis (glens at apache.org)
 * @author Rolf-J黵gen Moll
 *
 * @see org.apache.poi.hssf.record.MulRKRecord
 * @see org.apache.poi.hssf.record.RKRecord
 */
public class RKUtil
{
    private RKUtil()
    {
    }

    /**
     * Do the dirty work of decoding; made a private static method to
     * facilitate testing the algorithm
     */

    public static double decodeNumber(int number)
    {
        long raw_number = number;

        // mask off the two low-order bits, 'cause they're not part of
        // the number
        raw_number = raw_number >> 2;
        double rvalue = 0;

        if ((number & 0x02) == 0x02)
        {
            // ok, it's just a plain ol' int; we can handle this
            // trivially by casting
            rvalue = ( double ) (raw_number);
        }
        else
        {

            // also trivial, but not as obvious ... left shift the
            // bits high and use that clever static method in Double
            // to convert the resulting bit image to a double
            rvalue = Double.longBitsToDouble(raw_number << 34);
        }
        if ((number & 0x01) == 0x01)
        {

            // low-order bit says divide by 100, and so we do. Why?
            // 'cause that's what the algorithm says. Can't fight city
            // hall, especially if it's the city of Redmond
            rvalue /= 100;
        }

        return rvalue;
    }

}