如何使用Java解压缩COMP-3数字？

Question

我有一个巨大的大型机文件，在那个文件中有一些打包的数字。我想知道如何使用Java解包以下数字？

public String unpackData(String packedData, int decimalPointLocation) {
        String unpackedData = "";
        char[] characters = packedData.toCharArray();
        final int impliedPositive = 15;
        final int positiveNumber = 12;
        final int negativeNumber = 13;
        for (int currentCharIndex = 0; currentCharIndex < characters.length; currentCharIndex++) {
            byte[] unpackedDigits = unpackByte((byte) characters[currentCharIndex]);
            if(currentCharIndex == (characters.length - 1)) {
                if(unpackedDigits[1] == impliedPositive || unpackedDigits[1] == positiveNumber) {
                    unpackedData += String.valueOf(unpackedDigits[0]);
                } else if(unpackedDigits[1] == negativeNumber) {
                    unpackedData = "-" + unpackedData;
                }
            } else {
                unpackedData += String.valueOf(unpackedDigits[0]) + String.valueOf(unpackedDigits[1]);
            }
        }
        if(decimalPointLocation > 0) {
            unpackedData = unpackedData.substring(0, (decimalPointLocation - 1)) + 
                            "." + 
                            unpackedData.substring(decimalPointLocation);
        }
        return unpackedData;
    }

    private byte[] unpackByte(byte packedData) {
        byte firstDigit = (byte) (packedData >>> 4);
        firstDigit = setBitsToZero(firstDigit, 4, 8);

        //System.out.println(" firstDigit = "+ firstDigit + ", and its bit string after unpacking = " + getBitString(firstDigit, 7));

        byte secondDigit = setBitsToZero(packedData, 4, 8);
        //System.out.println("second digit = " + secondDigit + ", and its bit string of second digit after unpcking = " + getBitString(secondDigit, 7));

        byte[] unpackedData = new byte[2];
        unpackedData[0] = firstDigit;
        unpackedData[1] = secondDigit;
        return unpackedData;
    }

    private byte setBitsToZero(byte number, int startBitPosition, int endBitPosition) {
        for (int i = startBitPosition; i < endBitPosition; i++) {
            number =  (byte) (number & ~(1 << i));
        }
        return number;
    }

有人能告诉我的程序是否正确吗？

Answer 1

COMP-3（或“打包的十进制”）数据如下所示:0x12345s，其中“s”是C表示正，D表示负，或F表示无符号。因此，0x12345C->“12345”，x012345D->“-12345”和0x12345F->“12345”。

您有一个明显的错误：如果符号为负，则忽略包含符号nybble的字节中的nybble（例如，上面的“5”）。此外，你在操纵NYBLE方面太努力了，这是一个简单的按位和或4位移位来隔离NYBLE。

尝试如下所示（未经测试）：

public String unpackData(String packedData, int decimalPointLocation) {
    String unpackedData = "";
    char[] characters = packedData.toCharArray();
    final int negativeSign = 13;
    for (int currentCharIndex = 0; currentCharIndex < characters.length; currentCharIndex++) {
        byte firstDigit = ((byte) characters[currentCharIndex]) >>> 4);
        byte secondDigit = ((byte) characters[currentCharIndex]) & 0x0F;
        unpackedData += String.valueOf(firstDigit);
        if (currentCharIndex == (characters.length - 1)) {
            if (secondDigit == negativeSign) {
                unpackedData = "-" + unpackedData;
            }
        } else {
            unpackedData += String.valueOf(secondDigit);
        }
    }
    if (decimalPointLocation > 0) {
        unpackedData = unpackedData.substring(0, (decimalPointLocation - 1)) + 
                        "." + 
                        unpackedData.substring(decimalPointLocation);
    }
    return unpackedData;
}