十六进制和十进制转换

发布于 2024-10-07 09:59:51 字数 885 浏览 2 评论 0原文

我有一个二进制文件，其内容的定义如下:(存储所有数据以小端字节序（即，最低有效字节在前））。下面的示例数字是十六进制的，

11 63 39 46             --- Time, UTC in seconds since 1 Jan 1970.
01 00                   --- 0001 = No Fix, 0002 = SPS
97 85 ff e0 7b db 4c 40 --- Latitude, as double
a1 d5 ce 56 8d 26 28 40 --- Longitude, as double
f0 37 e1 42             --- Height in meters, as float
fe 2b f0 3a             --- Speed in km/h, as float
00 00 00 00             --- Heading (degrees ?), as float
01 00                   --- RCR, log reason. 0001=Time, 0004=Distance
59 20 6a f3 4a 26 e3 3f --- Distance in meters, as double,
2a                      --- ? Don't know
a8                      --- Checksum, xor of all bytes above not including 0x2a

来自二进制文件“十六进制”的数据如下，

"F25D39460200269652F5032445401F4228D79BCC54C09A3A2743B4ADE73F2A83"

如果您能支持我根据之前的指令翻译此数据行，我将不胜感激。

原文

I have a binary file , the definition of its content is as below : ( all data is stored
in little endian (ie. least significant byte first)) . The example numbers below are HEX

11 63 39 46             --- Time, UTC in seconds since 1 Jan 1970.
01 00                   --- 0001 = No Fix, 0002 = SPS
97 85 ff e0 7b db 4c 40 --- Latitude, as double
a1 d5 ce 56 8d 26 28 40 --- Longitude, as double
f0 37 e1 42             --- Height in meters, as float
fe 2b f0 3a             --- Speed in km/h, as float
00 00 00 00             --- Heading (degrees ?), as float
01 00                   --- RCR, log reason. 0001=Time, 0004=Distance
59 20 6a f3 4a 26 e3 3f --- Distance in meters, as double,
2a                      --- ? Don't know
a8                      --- Checksum, xor of all bytes above not including 0x2a

the data from the Binary file "in HEX" is as below

"F25D39460200269652F5032445401F4228D79BCC54C09A3A2743B4ADE73F2A83"

I appreciate if you can support me to translate this data line based on the instruction before.

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时光沙漏 2024-10-14 09:59:51

可能是错误的，但这里有一个使用 Ruby 的尝试：

hex   = "F25D39460200269652F5032445401F4228D79BCC54C09A3A2743B4ADE73F2A83"
ints  = hex.scan(/../).map{ |s| s.to_i(16) }
raw   = ints.pack('C*')
fields = raw.unpack( 'VvEEVVVvE')

p fields
#=> [1178164722, 2, 42.2813707974677, -83.1970117467067, 1126644378, 1072147892, nil, 33578, nil]

p Time.at( fields.first )
#=> 2007-05-02 21:58:42 -0600

如果有人精通 #pack 和 < code>#unpack 将向我展示一种更好的方法来完成前三行。

Probably wrong, but here's a shot at it using Ruby:

hex   = "F25D39460200269652F5032445401F4228D79BCC54C09A3A2743B4ADE73F2A83"
ints  = hex.scan(/../).map{ |s| s.to_i(16) }
raw   = ints.pack('C*')
fields = raw.unpack( 'VvEEVVVvE')

p fields
#=> [1178164722, 2, 42.2813707974677, -83.1970117467067, 1126644378, 1072147892, nil, 33578, nil]

p Time.at( fields.first )
#=> 2007-05-02 21:58:42 -0600

I'd appreciate it if someone well-versed in #pack and #unpack would show me a better way to accomplish the first three lines.

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