How to understand a range in hexadecimal - utf-8

I have this range:
U+F0000..U+FFFFD
Its for UTF, Private use characters. I understand F0000 to FFFFD means a range, but why the U+ is added in the beggining? what does it means?

The "U+" means it's a Unicode codepoint, just like "0x" means what follows is a hexadecimal number. The "U+" implies hexadecimal, so what follows is in hexadecimal notation, but represents a codepoint in Unicode.
In UTF-8, U+F0000 would be encoded as 0xF3 0xB0 0x80 0x80. U+FFFFD would be encoded as 0xF3 0xBF 0xBF 0xBD.

This is called Unicode code point and the U+ prefix is how you write it.

Related

Are there examples of ISO 8859-1 text files which are valid, but different in UTF-8?

I know that UTF-8 supports way more characters than Latin-1 (even with the extensions). But are there examples of files that are valid in both, but the characters are different? So essentially that the content changes, depending on how you think the file is encoded?
I also know that a big chunk of Latin-1 maps 1:1 to the same part in UTF-8. The question is: which code points could change the value if interpreted differently (not invalid, but different)?
Latin-1 is a single-byte encoding (meaning 1 character = 1 byte), which uses all possible byte values. So any byte maps to something in Latin-1. So literally any file is "valid" in Latin-1. So you can interpret any file as Latin-1 and you'll get… something… as a result.
So yes, interpret any valid UTF-8 file in Latin-1. It's valid both in UTF-8 and Latin-1. The first 128 characters are the same for both encodings and both based on ASCII; but if your UTF-8 file uses any non-ASCII characters, those will be interpreted as gibberish (yet valid) Latin-1.
bytes
encoding
text
e6bc a2e5 ad97
UTF-8
漢字
e6bc a2e5 ad97
Latin-1
æ¼¢å­ 👈 valid but nonsensical
Unicode is - somewhat simplified - a character set, and UTF-8 is one of multiple encodings for the binary representation of Unicode.
ISO-8859-1 is both, a character set and encoding.
At the character set level, ISO-8859-1 is a subset of Unicode, i.e. each ISO-8859-1 character also exists in Unicode, and the ISO-8859-1 code is even equal to the Unicode codepoint.
At the encoding level, ISO-8859-1 and UTF-8 use the same binary representation for the ISO-8859-1 characters up to 127. But for the characters between 128 and 255 they differ as UTF-8 needs 2 bytes to represent them.
Example:
Word
ISO-8859-1
UTF-8
Zürich
5a fc 72 69 63 68
5a c3 bc 72 69 63 68

What's rules to specify byte sequence for BOM?

I'm handling some file encoding stuff. When I learn BOM, it says The UTF-8 representation of the BOM is the byte sequence 0xEF,0xBB,0xBF, then I find the Code page layout which is a table that contains many character encoding information. What I am curious is that if there are some rules for the BOM bytes sequence, I mean, why don't use 0xEE,0xFF,0xBB or any other bytes sequence to represent UTF-8? Thanks in advance.
The BOM is specific to Unicode UTF (Unicode Transformation Format) encodings. It is the Unicode character U+FEFF ZERO WIDTH NO-BREAK SPACE encoded to a specific byte sequence according to the rules defined in the specific UTF that it is encoded in, same as for any other Unicode codepoint. What makes the BOM special is that it is the first encoded codepoint at the front of the encoded text, so you can discover which UTF was used to encode the text, if not specified out-of-band through other means.
The BOM for UTF-8 is EF BB BF, for UTF-16LE is FF FE, for UTF-32LE is FF FE 00 00, etc. They are all just different representations of the same Unicode codepoint U+FEFF.
Other encodings, like Windows-1252, which you link to, do not use a BOM and cannot encode that particular character, so there is no alternative "Windows-1252 encoding" of a BOM.

Can UTF8 encoded data missread as Latin-1 produce ASCII artifacts?

UTF-8 single byte characters map perfectly to Latin-1 (ISO 8859-1) characters (those below the character code of 128); basicly the default ASCII characters.
If I have a UTF-8 encoded string and pass it to a function, that expects a Latin-1 string is there any possibility that the Latin-1 function misinterprets parts of UTF-8 mutlibyte characters as ASCII characters?
I imagine something like this could happen:
(imagniray) UTF-8 multibyte character: 0xA330
(mis-)interpreted by Latin-1 function as two Latin-1 characters: 0xA3 0x30
The first of those characters does not lie within the ASCII set, but the second is the ASCII code for the 0 character. Is it possible that an multibyte UTF-8 character produces an artifact that looks like a single-byte UTF-8 / ASCII character like in the example above?
From my understanding of UTF-8 only single-byte characters contain any bytes with the most significant bit unset, so basicly multibyte characters never contain a byte that could be misinterpreted by a Latin-1 function as a valid ASCII character (because all those characters have the most significant bit unset). But I want to make sure this is true and I don't screw up on this, because this may have security implications when dealing with data sanitization - which I am apparently currently doing.
You are correct in your understanding that only single byte characters contain any bytes with the most significant bit unset. There is a nice table showing this at: http://en.wikipedia.org/wiki/UTF-8#Description

Checking all files are encoded as UTF-8

Does anyone know of a Windows app that can scan through a directory and check which scripts are/aren't encoded as a specified charset (UTF-8 in this case)? I could do it manually, but that could take a while and is quite error prone!
UTF-8 isn't a character set, it's an encoding for Unicode characters. And, since this is not programming related, I'm nudging it over to superuser.
If you do want to write a program for detecting those sequences, it's pretty easy:
Illegal UTF-8 initial sequences
UTF-8 Sequence Reason for Illegality
10xxxxxx illegal as initial byte of character (80..BF)
1100000x illegal, overlong (C0 80..BF)
11100000 100xxxxx illegal, overlong (E0 80..9F)
11110000 1000xxxx illegal, overlong (F0 80..8F)
11111000 10000xxx illegal, overlong (F8 80..87)
11111100 100000xx illegal, overlong (FC 80..83)
1111111x illegal; prohibited by spec
Then, provided the first octet is legal, just remember that the number of octets forming a code point can be obtained by counting the number of 1 bits before the first 0 bit.
For example, 11110xxx is the start of a 4-octet sequence so you should skip ahead 4 octets once you've established its legality.
The other thing to do is ensure that all continuation octets start with 10.
Not sure if this is what you're looking for, but I use a command shell for-loop and dump the first few bytes of each file using my hdump utility, which displays the bytes of the file in hexadecimal form. I then look for the leading 3-byte UTF-8 signature (Byte Order Mark) at the start of each file.
My hdump utility is available at: http://david.tribble.com/programs.html

UTF-8 character change to two character

In my text file, I used a character with value larger than 127 for example 0xDC. Then I loaded that text file in a device. Then I read that text file and that character. Then the character was changed to 0xC3 and 0x9C. How come it change to two character?
Thanks
Because that's the sequence for the character when encoded in UTF-8:
>>> '\xc3\x9c'.decode('utf-8')
u'\xdc'
From wikipedia:
"UTF-8 encodes each character (code point) in 1 to 4 octets (8-bit bytes), with the single octet encoding used only for the 128 US-ASCII characters."

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