Re^4: Add 1 to an arbitrary-length binary string

”… characters that are not octets…”

That confuses me a bit. What else would they be made of?

say unpack( "U*", "a");
printf("%04X\n", unpack('W*', decode_utf8("a")));
say join " ", unpack( "U*", "&#128526;");
printf("%04X\n", unpack('W*', decode_utf8("&#128526;")));

__END__

97
0061
240 159 152 142
1F60E
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I read it like this: ”a” - code point 0061 - is one octet and 😎 - code point 1F60E - is four octets long.

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Comment on Re^4: Add 1 to an arbitrary-length binary string Download Code

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Re^5: Add 1 to an arbitrary-length binary string by hv (Prior) on Nov 16, 2023 at 19:26 UTC
Exactly as I said in the immediately following part of that sentence: Unicode characters with codepoints greater than 255. The string `"\x{61}\x{1f60e}"` has a length of two, it consists of two characters. Its internal representation happens to consist of 5 octets, but any time you have to care about the internal representation is an example of the Unicode bug. The string `"\x{61}\x{ff}"` may be stored internally as either 2 or 3 octets; however it also has a length of two, consists of two characters, and will be incremented by my example code to the string `"\x{62}\x{0}"`, regardless of the internal representation.	[reply] [d/l] [select]

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Re^5: Add 1 to an arbitrary-length binary string
by hv (Prior) on Nov 16, 2023 at 19:26 UTC

Exactly as I said in the immediately following part of that sentence: Unicode characters with codepoints greater than 255.

The string "\x{61}\x{1f60e}" has a length of two, it consists of two characters. Its internal representation happens to consist of 5 octets, but any time you have to care about the internal representation is an example of the Unicode bug.

The string "\x{61}\x{ff}" may be stored internally as either 2 or 3 octets; however it also has a length of two, consists of two characters, and will be incremented by my example code to the string "\x{62}\x{0}", regardless of the internal representation.

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