How does UTF-8 encoding identify single byte and double byte characters?

Question

Recently I've faced an issue regarding character encoding, while I was digging into character set and character encoding this doubt came to my mind.UTF-8 encoding is most popular because of its backward compatibility with ASCII.Since UTF-8 is variable length encoding format, how it differentiates single byte and double byte characters.For example, "Aݔ" is stored as "410754" (Unicode for A is 41 and Unicode for Arabic character is 0754.How encoding identifies 41 is one character and 0754 is another two-byte character?Why it's not considered as 4107 as one double byte character and 54 as a single byte character?

Answer 1

For example, "Aݔ" is stored as "410754"

That’s not how UTF-8 works.

Characters U+0000 through U+007F (aka ASCII) are stored as single bytes. They are the only characters whose codepoints numerically match their UTF-8 presentation. For example, U+0041 becomes 0x41 which is 01000001 in binary.

All other characters are represented with multiple bytes. U+0080 through U+07FF use two bytes each, U+0800 through U+FFFF use three bytes each, and U+10000 through U+10FFFF use four bytes each.

Computers know where one character ends and the next one starts because UTF-8 was designed so that the single-byte values used for ASCII do not overlap with those used in multi-byte sequences. The bytes 0x00 through 0x7F are only used for ASCII and nothing else; the bytes above 0x7F are only used for multi-byte sequences and nothing else. Furthermore, the bytes that are used at the beginning of the multi-byte sequences also cannot occur in any other position in those sequences.

Because of that the codepoints need to be encoded. Consider the following binary patterns:

• 2 bytes: 110xxxxx 10xxxxxx
• 3 bytes: 1110xxxx 10xxxxxx 10xxxxxx
• 4 bytes: 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx

The amount of ones in the first byte tells you how many of the following bytes still belong to the same character. All bytes that belong to the sequence start with 10 in binary. To encode the character you convert its codepoint to binary and fill in the x’s.

As an example: U+0754 is between U+0080 and U+07FF, so it needs two bytes. 0x0754 in binary is 11101010100, so you replace the x’s with those digits:

11011101 10010100

Answer 2

Short answer:

UTF-8 is designed to be able to unambiguously identify the type of each byte in a text stream:

• 1-byte codes (all and only the ASCII characters) start with a 0
• Leading bytes of 2-byte codes start with two 1s followed by a 0 (i.e. 110)
• Leading bytes of 3-byte codes start with three 1s followed by a 0 (i.e. 1110)
• Leading bytes of 4-byte codes start with four 1s followed by a 0 (i.e. 11110)
• Continuation bytes (of all multi-byte codes) start with a single 1 followed by a 0 (i.e. 10)

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Your example Aݔ, which consists of the Unicode code points U+0041 and U+0754, is encoded in UTF-8 as:

01000001 11011101 10010100

So, when decoding, UTF-8 knows that the first byte must be a 1-byte code, the second byte must be the leading byte of a 2-byte code, the third byte must be a continuation byte, and since the second byte is the leading byte of a 2-byte code, the second and third byte together must form this 2-byte code.

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See here how UTF-8 encodes Unicode code points.

Answer 3

Just to clarify, ASCII mean standard 7-bit ASCII and not extended 8-bit ASCII as commonly used in Europe.

Thus, part of first byte (0x80 to 0xFF) goes to dual byte representation and part of second byte on two bytes (0x0800 to 0xFFFF) takes the full three-byte representation.

Four byte representation uses only the lowest three bytes and only 1.114.111 of the ‭16.777.215‬ available possibilities

You have an xls here

That means that interpreters must 'jump back' a NUL (0) byte when they find those binary patterns.

Hope this helps somebody!