我有一些需要挖掘数据的旧文件。这些文件是由Lotus123 Release 4 for DOS创建的。我试图通过解析字节而不是使用Lotus打开文件来更快地读取文件。
Dim fileBytes() As Byte = My.Computer.FileSystem.ReadAllBytes(fiPath)
'I loop through all the data getting first/second bytes for each value
do ...
Dim FirstByte As Int16 = Convert.ToInt16(fileBytes(Index))
Dim SecondByte As Int16 = Convert.ToInt16(fileBytes(Index + 1))
loop ...
我可以得到这样的整数值：
Dim value As Int16 = BitConverter.ToInt16(fileBytes, Index + 8) / 2
但是浮点数更加复杂。只有较小的数字存储有两个字节。较大的值占用10个字节，但这是另一个问题。在这里，我们只有两个字节的较小值。以下是一些示例值。我将字节值输入Excel，然后使用= DEC2BIN（）转换为二进制，并根据需要在左侧添加零以获得8位。
First Second
Byte Byte Value First Byte 2nd Byte
7 241 = -1.2 0000 0111 1111 0001
254 255 = -1 1111 1110 1111 1111
9 156 = -0.8 0000 1001 1001 1100
9 181 = -0.6 0000 1001 1011 0101
9 206 = -0.4 0000 1001 1100 1110
9 231 = -0.2 0000 1001 1110 0111
13 0 = 0 0000 1101 0000 0000
137 12 = 0.1 1000 1001 0000 1100
9 25 = 0.2 0000 1001 0001 1001
137 37 = 0.3 1000 1001 0010 0101
9 50 = 0.4 0000 1001 0011 0010
15 2 = 0.5 0000 1111 0000 0010
9 75 = 0.6 0000 1001 0100 1011
137 87 = 0.7 1000 1001 0101 0111
9 100 = 0.8 0000 1001 0110 0100
137 112 = 0.9 1000 1001 0111 0000
2 0 = 1 0000 0010 0000 0000
199 13 = 1.1 1100 0111 0000 1101
7 15 = 1.2 0000 0111 0000 1111
71 16 = 1.3 0100 0111 0001 0000
135 17 = 1.4 1000 0111 0001 0001
15 6 = 1.5 0000 1111 0000 0110
7 20 = 1.6 0000 0111 0001 0100
71 21 = 1.7 0100 0111 0001 0101
135 22 = 1.8 1000 0111 0001 0110
199 23 = 1.9 1100 0111 0001 0111
4 0 = 2 0000 0100 0000 0000
我希望有一种简单的转换方法。也许会更复杂。
我看过BCD：“ BCD在许多早期十进制计算机中使用，并且在诸如IBM System / 360系列之类的机器指令集中实现”和Intel BCD opcode
我不知道这是BCD还是什么。如何将两位转换为浮点数？
答案 0 :(得分：3)
我使用了安德鲁·莫顿（Andrew Morton）在评论中指出的website中的信息。基本上，存储的16位数量由15位二进制补码整数（当lsb为0时）或12位二进制补码整数加上指示要应用于该整数的比例因子的处理代码（当lsb时）组成是1）。我对vb.net不熟悉，因此在这里提供ISO-C代码。下面的程序成功解码了问题中提供的所有数据。
注意：我在下面的代码中转换为8字节的double，而问题是，原来的转换可能是10字节的long double格式（扩展了80位-8087数学协处理器的精度格式）。尝试更多的测试数据以完全覆盖八个定标代码似乎是一个好主意：大整数，如1,000,000和1,000,000,000；小数部分，例如0.0003、0.000005和0.00000007；和二进制分数，如0.125（1/8）和0.046875（3/64）。
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
typedef struct {
uint8_t byte1;
uint8_t byte2;
} num;
num data[] =
{
{ 7, 241}, {254, 255}, { 9, 156}, { 9, 181}, { 9, 206}, { 9, 231},
{ 13, 0}, {137, 12}, { 9, 25}, {137, 37}, { 9, 50}, { 15, 2},
{ 9, 75}, {137, 87}, { 9, 100}, {137, 112}, { 2, 0}, {199, 13},
{ 7, 15}, { 71, 16}, {135, 17}, { 15, 6}, { 7, 20}, { 71, 21},
{135, 22}, {199, 23}, { 4, 0}
};
int data_count = sizeof (data) / sizeof (data[0]);
/* define operators that may look more familiar to vb.net programmers */
#define XOR ^
#define MOD %
int main (void)
{
int i;
uint8_t b1, b2;
uint16_t h, code;
int32_t n;
double r;
for (i = 0; i < data_count; i++) {
b1 = data[i].byte1;
b2 = data[i].byte2;
/* data word */
h = ((uint16_t)b2 * 256) + b1;
/* h<0>=1 indicates stored integer needs to be scaled */
if ((h MOD 2) == 1) {
/* extract scaling code in h<3:1> */
code = (h / 2) MOD 8;
/* scaled 12-bit integer in h<15:4>. Extract, sign-extend to 32 bits */
n = (int32_t)((((uint32_t)h / 16) XOR 2048) - 2048);
/* convert integer to floating-point */
r = (double)n;
/* scale based on scaling code */
switch (code) {
case 0x0: r = r * 5000; break;
case 0x1: r = r * 500; break;
case 0x2: r = r / 20; break;
case 0x3: r = r / 200; break;
case 0x4: r = r / 2000; break;
case 0x5: r = r / 20000; break;
case 0x6: r = r / 16; break;
case 0x7: r = r / 64; break;
};
} else {
/* unscaled 15-bit integer in h<15:1>. Extract, sign extend to 32 bits */
n = (int32_t)((((uint32_t)h / 2) XOR 16384) - 16384);
/* convert integer to floating-point */
r = (double)n;
}
printf ("[%3d,%3d] n=%08x r=% 12.8f\n", b1, b2, n, r);
}
return EXIT_SUCCESS;
}
该程序的输出如下：
[ 7,241] n=ffffff10 r= -1.20000000
[254,255] n=ffffffff r= -1.00000000
[ 9,156] n=fffff9c0 r= -0.80000000
[ 9,181] n=fffffb50 r= -0.60000000
[ 9,206] n=fffffce0 r= -0.40000000
[ 9,231] n=fffffe70 r= -0.20000000
[ 13, 0] n=00000000 r= 0.00000000
[137, 12] n=000000c8 r= 0.10000000
[ 9, 25] n=00000190 r= 0.20000000
[137, 37] n=00000258 r= 0.30000000
[ 9, 50] n=00000320 r= 0.40000000
[ 15, 2] n=00000020 r= 0.50000000
[ 9, 75] n=000004b0 r= 0.60000000
[137, 87] n=00000578 r= 0.70000000
[ 9,100] n=00000640 r= 0.80000000
[137,112] n=00000708 r= 0.90000000
[ 2, 0] n=00000001 r= 1.00000000
[199, 13] n=000000dc r= 1.10000000
[ 7, 15] n=000000f0 r= 1.20000000
[ 71, 16] n=00000104 r= 1.30000000
[135, 17] n=00000118 r= 1.40000000
[ 15, 6] n=00000060 r= 1.50000000
[ 7, 20] n=00000140 r= 1.60000000
[ 71, 21] n=00000154 r= 1.70000000
[135, 22] n=00000168 r= 1.80000000
[199, 23] n=0000017c r= 1.90000000
[ 4, 0] n=00000002 r= 2.00000000
答案 1 :(得分：2)
只需VB.Net代码posted by njuffa的C转换。
原始的structure已替换为Byte数组，并且数值数据类型适用于.Net类型。就这样。
Dim data As Byte(,) = New Byte(,) {
{7, 241}, {254, 255}, {9, 156}, {9, 181}, {9, 206}, {9, 231}, {13, 0}, {137, 12}, {9, 25},
{137, 37}, {9, 50}, {15, 2}, {9, 75}, {137, 87}, {9, 100}, {137, 112}, {2, 0}, {199, 13},
{7, 15}, {71, 16}, {135, 17}, {15, 6}, {7, 20}, {71, 21}, {135, 22}, {199, 23}, {4, 0}
}
Dim byte1, byte2 As Byte
Dim word, code As UShort
Dim nValue As Integer
Dim result As Double
For i As Integer = 0 To (data.Length \ 2 - 1)
byte1 = data(i, 0)
byte2 = data(i, 1)
word = (byte2 * 256US) + byte1
If (word Mod 2) = 1 Then
code = (word \ 2US) Mod 8US
nValue = ((word \ 16) Xor 2048) - 2048
Select Case code
Case 0 : result = nValue * 5000
Case 1 : result = nValue * 500
Case 2 : result = nValue / 20
Case 3 : result = nValue / 200
Case 4 : result = nValue / 2000
Case 5 : result = nValue / 20000
Case 6 : result = nValue / 16
Case 7 : result = nValue / 64
End Select
Else
'unscaled 15-bit integer in h<15:1>. Extract, sign extend to 32 bits
nValue = ((word \ 2) Xor 16384) - 16384
result = nValue
End If
Console.WriteLine($"[{byte1,3:D}, {byte2,3:D}] number = {nValue:X8} result ={result,12:F8}")
Next