When I use hexdump on a file with no options, I get rows of hexadecimal bytes:
cf fa ed fe 07 00 00 01 03 00 00 80 02 00 00 00
When I used hexdump -d on the same file, that same data is shown in something called two-byte decimal groupings:
64207 65261 00007 00256 00003 32768 00002 00000
So what I'm trying to figure out here is how to convert between these two encodings. cf and fa in decimal are 207 and 250 respectively. How do those numbers get combined to make 64207?
Bonus question: What is the advantage of using these groupings? The octal display uses 16 groupings of three digits, why not use the same thing with the decimal display?
As commented by #georg.
0xfa * 256 + 0xcf == 0xfacf == 64207
The conversion exactly works like this.
So, if you see man hexdump:
-d, --two-bytes-decimal
Two-byte decimal display. Display the input offset in hexadecimal, followed by eight
space-separated, five-column, zero-filled, two-byte units of input data, in unsigned
decimal, per line.
So, for example:
00000f0 64207 65261 00007 00256 00003 32768 00002 00000
Here, 00000f0 is a hexadecimal offset.
Followed by two-byte units of input data, for eg.: 64207 in decimal (first 16 bits - i.e. two bytes of the file).
The conversion (in your case):
cf fa ----> two-byte unit (the byte ordering depends on your architecture).
fa * 256 + cf = facf ----> appropriately ----> 0xfacf (re-ording)
And dec of oxfacf is 64207.
Bonus question: It is a convention to display octal numbers using three digits (unlike hex and decimal), so it uses a triplet for each byte.
Related
So, I know that when we type characters, each character maps to a number in a character set and then, that number is transformed into a binary format so a computer can understand. They way that number is transformed into a binary format(how many bits gets allocated) depends on character encoding.
So, if I type L, It represents 76. Then 76 gets tranformed into 1 byte binary format because of let's say UTF-8.
Now, I've read the following somewhere:
The Devanagari character क, with code point 2325 (which is 915 in
hexadecimal notation), will be represented by two bytes when using the
UTF-16 encoding (09 15), three bytes with UTF-8 (E0 A4 95), or four
bytes with UTF-32 (00 00 09 15).
So, as you can see it says three bytes with UTF-8 (E0 A4 95). how are E0 A4 95 bytes ? I am asking because i have no idea where E0 A4 95 came from... Why do we need this ? if we know that code point is 2325, all we have to do is in order to use UTF-8, we know that utf-8 will need 3 bytes to transform 2325 into binary... Why do we need E0 A4 95 and what is it ?
E0 A4 95 is the 3-byte UTF-8 encoding of U+0915. In binary:
E 0 A 4 9 5 (hex)
11100000 10100100 10010101 (binary)
1110xxxx 10xxxxxx 10xxxxxx (3-byte UTF-8 encoding pattern)
0000 100100 010101 (data bits)
00001001 00010101 (regrouped data bits to 8-bit bytes)
0 9 1 5 (hex)
U+0915 (Unicode code point)
The first byte's binary pattern 1110xxxx is a lead byte indicating a 3-byte encoding and 4 bits of data. Follow on bytes start with 10xxxxxx and provide 6 more bits of data. There will be two following bytes after a 3-byte leading byte indicator.
For more information read the Wikipedia article on UTF-8 and the standard RFC-3629.
I opened a .exe file and I found a string "Premium" was stored in the following way
50 00 72 00 65 00 6D 00 69 00 75 00 6D 00
I just don't know why "00" is appended to each of characters and what's its usage.
Thanks,
It's probably a UTF-16 encoding of a Unicode string. Here's an example using Python:
>>> u"Premium".encode("utf16")
'\xff\xfeP\x00r\x00e\x00m\x00i\x00u\x00m\x00'
# ^ ^ ^ ^ ^ ^ ^
After the byte marker to indicate endianness, you can see alternating letters and null bytes.
\xff\xfe is the byte-order marker; it indicates that the low-order byte of each 16-bit value comes first. (If the high-order byte came first, the byte marker would be \xfe\xff; there's nothing particularly meaningful about which marker means which.)
Each character is then encoded as a 16-bit value. For many values, the UTF-16 encoding is simply the straightforward unsigned 16-bit representation of its Unicode code point. Specifically, 8-bit ASCII values simply use a null byte as the high-order byte, and its ASCII value as the low-order byte.
I have a .bmp file
I sort of do understand and sort of do not understand. I understand that the first 14 bytes are my Bitmapfileheader. I furthermore do understand that my Bitmapinfoheader contains information about the bitmap as well and is about 40 bytes large (in version 3).
What I do not understand is, how the information is stored in there.
I have this image:
Why is all the colorinformation stored in "FF"? I know that the "00" are "Junk Bytes". What I do not understand is why there is everything in "FF"?!
Furthermore, I do not understand what type of "encoding" that is? 42 4D equals do "BM". What is that? How can I translate what I see there to colors or letters or numbers?!
What I can read in this case:
Bitmapfileheader:
First 2 bytes. BM if it is a .bmp file: 42 4D = BM (However 42 4D transforms to BM)
Next 4 Bytes: Size of the bitmap. BA 01 00 00. Dont know what size that should be.
Next 4 Bytes: Something reserved.
Next 4 Bytes: Offset (did not quite understand that)
Bitmapinfoheader
Next 4 Bytes: Size of the bitmapinfoheader. 6C 00 00 00 here.
Next 4 Bytes: Width of the .bmp. 0A 00 00 00. I know that that must be 10px since I created that file.
Next 4 Bytes: Height of the .bmp. 0A 00 00 00. I know that that must be 10px since I created that file.
Next 2 Bytes: Something from another file format.
Next two Bytes: Color depth. 18 00 00 00. I thought that can only by 1,2,4,8, 16, 24, 32?
The first 2 bytes of information that you see "42 4D" are what we call the magic numbers. They are the signature of the file, 42 4d is the hex notation of 01000010 01001101 in binary.
Every file has one, .jpg, .gif. You get it.
Here is an image that illustrate a BMP complete header of 54 bytes(24-bit BMP).
BMP Header
The total size of the BMP is calculated by the size of the header + BMP.width x BMP.height * 3 (1 byte for Red, 1 byte for Green, 1 byte for Blue - in the case of 8bits of information per channel) + padding(if it exists).
Junk bytes that you refer, is the padding, they are needed if the size of each scanline(row) is not a multiple of 4.
White in hex notation if ffffff, being the first two red, green and blue.
While in decimal notation each channel will have the value of 255, because 2^8(8 bits) -1 = 255.
Hope this clears a bit(non intended pun) for you.
I am trying to understand SHA256. On the Wikipedia page it says:
append the bit '1' to the message
append k bits '0', where k is the minimum number >= 0 such that the resulting message
length (modulo 512 in bits) is 448.
append length of message (without the '1' bit or padding), in bits, as 64-bit big-endian
integer
(this will make the entire post-processed length a multiple of 512 bits)
So if my message is 01100001 01100010 01100011 I would first add a 1 to get
01100001 01100010 01100011 1
Then you would fill in 0s so that the total length is 448 mod 512:
01100001 01100010 01100011 10000000 0000 ... 0000
(So in this example, one would add 448 - 25 0s)
My question is: What does the last part mean? I would like to see an example.
It means the message length, padded to 64 bits, with the bytes appearing in order of significance. So if the message length is 37113, that's 90 f9 in hex; two bytes. There are two basic(*) ways to represent this as a 64-bit integer,
00 00 00 00 00 00 90 f9 # big endian
and
f9 90 00 00 00 00 00 00 # little endian
The former convention follows the way numbers are usually written out in decimal: one hundred and two is written 102, with the most significant part (the "big end") being written first, the least significant ("little end") last. The reason that this is specified explicitly is that both conventions are used in practice; internet protocols use big endian, Intel-compatible processors use little endian, so if they were decimal machines, they'd write one hundred and two as 201.
(*) Actually there are 8! = 40320 ways to represent a 64-bit integer if 8-bit bytes are the smallest units to be permuted, but two are in actual use.
I have web based system that uses encrypted GET parameters. I need to figure out what encryption is used and create a PHP function to recreate it. Any ideas?
Example URL:
...&watermark=ISpQICAK&width=IypcOysK&height=IypcLykK&...
You haven't provided nearly enough sample data for us to reliably guess even the alphabet used to encode it, much less what structure it might have.
What I can tell, from the three sample values you've provided, is:
There is quite a lot of redundancy in the data — compare e.g. width=IypcOysK and height=IypcLykK (and even watermark=ISpQICAK, though that might be just coincidence). This suggests that the data is neither random nor securely encrypted (which would make it look random).
The alphabet contains a fairly broad range of upper- and lowercase letters, from A to S and from c to y. Assuming that the alphabet consists of contiguous letter ranges, that means a palette of between 42 and 52 possible letters. Of course, we can't tell with any certainty from the samples whether other characters might also be used, so we can't even entirely rule out Base64.
This is not the output of PHP's base_convert function, as I first guessed it might be: that function only handles bases up to 36, and doesn't output uppercase letters.
That, however, is just about all. It would help to see some more data samples, ideally with the plaintext values they correspond to.
Edit: The id parameters you give in the comments are definitely in Base64. Besides the distinctive trailing = signs, they both decode to simple strings of nine printable ASCII characters followed by a line feed (hex 0A):
_Base64___________Hex____________________________ASCII_____
JiJQPjNfT0MtCg== 26 22 50 3e 33 5f 4f 43 2d 0a &"P>3_OC-.
JikwPClUPENICg== 26 29 30 3c 29 54 3c 43 48 0a &)0<)T<CH.
(I've replaced non-printable characters with a . in the ASCII column above.) On the assumption that all the other parameters are Base64 too, let's see what they decode to:
_Base64___Hex________________ASCII_
ISpQICAK 21 2a 50 20 20 0a !*P .
IypcOysK 23 2a 5c 3b 2b 0a #*\;+.
IypcLykK 23 2a 5c 2f 29 0a #*\/).
ISNAICAK 21 23 40 20 20 0a !## .
IyNAPjIK 23 23 40 3e 32 0a ###>2.
IyNAKjAK 23 23 40 2a 30 0a ###*0.
ISggICAK 21 28 20 20 20 0a !( .
IikwICAK 22 29 30 20 20 0a ")0 .
IilAPCAK 22 29 40 3c 20 0a ")#< .
So there's definitely another encoding layer involved, but we can already see some patterns:
All decoded values consist of a constant number of printable ASCII characters followed by a trailing line feed character. This cannot be a coincidence.
Most of the characters are on the low end of the printable ASCII range (hex 20 – 7E). In particular, the lowest printable ASCII character, space = hex 20, is particularly common, especially in the watermark strings.
The strings in each URL resemble each other more than they resemble the corresponding strings from other URLs. (But there are resemblances between URLs too: for example, all the decoded watermark values begin with ! = hex 21.)
In fact, the highest numbered character that occurs in any of the strings is _ = hex 5F, while the lowest (excluding the line feeds) is space = hex 20. Their difference is hex 3F = decimal 63. Coincidence? I think not. I'll guess that the second encoding layer is similar to uuencoding: the data is split into 6-bit groups (as in Base64), and each group is mapped to an ASCII character simply by adding hex 20 to it.
In fact, it looks like the second layer might be uuencoding: the first bytes of each string have the right values to be uuencode length indicators. Let's see what we get if we try to decode them:
_Base64___________UUEnc______Hex________________ASCII___re-UUE____
JiJQPjNfT0MtCg== &"P>3_OC- 0b 07 93 fe f8 cd ...... &"P>3_OC-
JikwPClUPENICg== &)0<)T<CH 25 07 09 d1 c8 e8 %..... &)0<)T<CH
_Base64___UUEnc__Hex_______ASC__re-UUE____
ISpQICAK !*P 2b + !*P``
IypcOysK #*\;+ 2b c6 cb +.. #*\;+
IypcLykK #*\/) 2b c3 c9 +.. #*\/)
ISNAICAK !## 0e . !##``
IyNAPjIK ###>2 0e 07 92 ... ###>2
IyNAKjAK ###*0 0e 02 90 ... ###*0
ISggICAK !( 20 !(```
IikwICAK ")0 25 00 %. ")0``
IilAPCAK ")#< 26 07 &. ")#<`
This is looking good:
Uudecoding and re-encoding the data (using Perl's unpack "u" and pack "u") produces the original string, except that trailing spaces are replaced with ` characters (which falls within acceptable variation between encoders).
The decoded strings are no longer printable ASCII, which suggests that we might be closer to the real data.
The watermark strings are now single characters. In two cases out of three, they're prefixes of the corresponding width and height strings. (In the third case, which looks a bit different, the watermark might perhaps have been added to the other values.)
One more piece of the puzzle — comparing the ID strings and corresponding numeric values you give in the comments, we see that:
The numbers all have six digits. The first two digits of each number are the same.
The uudecoded strings all have six bytes. The first two bytes of each string are the same.
Coincidence? Again, I think not. Let's see what we get if we write the numbers out as ASCII strings, and XOR them with the uudecoded strings:
_Num_____ASCII_hex___________UUDecoded_ID________XOR______________
406747 34 30 36 37 34 37 25 07 09 d1 c8 e8 11 37 3f e6 fc df
405174 34 30 35 31 37 34 25 07 0a d7 cb eb 11 37 3f e6 fc df
405273 34 30 35 32 37 33 25 07 0a d4 cb ec 11 37 3f e6 fc df
What is this 11 37 3f e6 fc df string? I have no idea — it's mostly not printable ASCII — but XORing the uudecoded ID with it yields the corresponding ID number in three cases out of three.
More to think about: you've provided two different ID strings for the value 405174: JiJQPjNfT0MtCg== and JikwPCpVXE9LCg==. These decode to 0b 07 93 fe f8 cd and 25 07 0a d7 cb eb respectively, and their XOR is 2e 00 99 29 33 26. The two URLs from which these ID strings came from have decoded watermarks of 0e and 20 respectively, which accounts for the first byte (and the second byte is the same in both, anyway). Where the differences in the remaining four bytes come from is still a mystery to me.
That's going to be difficult. Even if you find the encryption method and keys, the original data is likely salted and the salt is probably varied with each record.
That's the point of encryption.