/* ====================================================================
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Licensed to the Apache Software Foundation (ASF) Under one or more
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contributor license agreements. See the NOTICE file distributed with
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this work for Additional information regarding copyright ownership.
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The ASF licenses this file to You Under the Apache License, Version 2.0
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(the "License"); you may not use this file except in compliance with
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the License. You may obtain a copy of the License at
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http://www.apache.org/licenses/LICENSE-2.0
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Unless required by applicable law or agreed to in writing, software
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distributed Under the License is distributed on an "AS IS" BASIS,
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WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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See the License for the specific language governing permissions and
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limitations Under the License.
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==================================================================== */
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using System;
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namespace HH.WMS.Utils.NPOI.HSSF.Util
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{
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/**
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* Utility class for helping convert RK numbers.
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*
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* @author Andrew C. Oliver (acoliver at apache dot org)
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* @author Glen Stampoultzis (glens at apache.org)
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* @author Rolf-Jürgen Moll
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*
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* @see org.apache.poi.hssf.record.MulRKRecord
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* @see org.apache.poi.hssf.record.RKRecord
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*/
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public class RKUtil
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{
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/**
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* Do the dirty work of decoding; made a private static method to
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* facilitate testing the algorithm
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*/
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public static double DecodeNumber(int number)
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{
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long raw_number = number;
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// mask off the two low-order bits, 'cause they're not part of
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// the number
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raw_number = raw_number >> 2;
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double rvalue = 0;
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if ((number & 0x02) == 0x02)
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{
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// ok, it's just a plain ol' int; we can handle this
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// trivially by casting
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rvalue = (double)(raw_number);
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}
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else
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{
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// also trivial, but not as obvious ... left Shift the
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// bits high and use that clever static method in double
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// to convert the resulting bit image to a double
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rvalue = BitConverter.Int64BitsToDouble(raw_number << 34);
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}
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if ((number & 0x01) == 0x01)
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{
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// low-order bit says divide by 100, and so we do. Why?
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// 'cause that's what the algorithm says. Can't fight city
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// hall, especially if it's the city of Redmond
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rvalue /= 100;
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}
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return rvalue;
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}
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}
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}
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