## How to calculate integer based on byte sequence

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So, I'm trying to understand the math involved when trying to translate hexadecimal escape sequences into integers.

So if I have the string "Ã", when I do `"Ã".encode('utf-8')` I get a byte string like this `"\xc3"`. `ord("Ã")` is 195. The math is 16*12+3 which is 195. Things makes sense.

But if I have the character "é" - then the utf8-encoded hex escape sequence is `"\xc3\xa9` - and `ord("é")` is 233. How is this calculation performed? (a9 on its own is 169 so it's clearly not addition).

Similarly with this `'Ĭ'.encode('utf-8')`. This yields `b'\xc4\xac'`. And `ord('Ĭ')` is 300.

Can anyone explain the math involved here?

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From the doc:

ord(c)

Given a string representing one Unicode character, return an integer representing the Unicode code point of that character. For example, ord('a') returns the integer 97 and ord('€') (Euro sign) returns 8364. This is the inverse of chr().

What `ord` returns is the Unicode code point of the character - roughly, a number letting you identify the character among the large number of characters known in Unicode.

When you encode your character with UTF-8, your represent it by a sequence of bytes, which is not directly related to the Unicode code point. There can be some coincidences, mainly for ASCII characters that get represented with a sequence of one byte, but this will fail for all more 'exotic' characters.

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The ASCII-encoding of `"é"` is `0xe9`, which is equal to 233 in decimal base.

```for n in range(256):
print(n,hex(n),chr(n))
```

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So, I thought I'd just wrap this one up and post the answers to the math issues I didn't comprehend before receiving a tons of wisdom from SO.

The first question regarded "é" which yields `"\xc3\xa9"` when encoded with utf8 and where `ord("é")` returns `233`. Clearly 233 was not the sum of 195 (decimal representation of c3) and 169 (ditto for a9). So what's going on?

"é" is has the corresponding unicode point `U+00E9`. The decimal value for the hex `e9` is 233. So that's what the `ord("é")` is all about.

So how does this end up as `"\xc3\xa9"`?

As Jörg W Mittag explained and demonstrated, in utf8 all non-ASCII are "encoded as a multi-octet sequence".

The binary representation of 233 is `11101001`. As this is non-ASCII this needs to be packed in a two-octet sequence which according to Jörg will follow this pattern:

`110xxxxx 10xxxxxx` (110 and 10 are fixed leaving room for five bits in the first octet, and six bits in the second - 11 in total).

So the 8 bits binary representation of 233 is fitted into this pattern replacing the xx-parts... Since there are 11 bits available and we only need 8 bits we pad the 8 bits with 3 more, `000`, (i.e. `00011101001`).

`^^^00011 ^^101001` (`000` followed by our 8 bits representation of 233)

`11000011 10101001` (binary representation of 233 inserted in a two-octet sequence)

`11000011` equals the hex `c3`, as `10101001` equals `a9`- which in other words matches the original sequence `"\xc3\xa9"`

A similar walkthrough for the character "Ĭ":

`'Ĭ'.encode('utf-8')` yields `b'\xc4\xac'`. And `ord('Ĭ')` is 300.

So again the unicode point for this character is `U+012C` which has the decimal value of 300 ((1*16*16)+(2*16*1)+(12*1)) - so that's the ord-part.

Again the binary representation of 300 is 9 bits, `100101100`. So once more there's a need for a two-octet sequence of the pattern `110xxxxx 10xxxxxx`. And again we pad it with a couple of `0` so reach 11 bits (`00100101100`).

`^^^00100 ^^101100` (`00` followed by our 9 bits representation of 300)

`11000100 10101100` (binary representation of 300 inserted in a two octet-sequence).

`11000100` corresponds to `c4`in hex, `10101100` to `ac` - in other words `b'\xc4\xac'`.

Thank you everyone for helping out on this. I learned a lot.

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• FYI, the byte string of `"Ã".encode('utf-8')` is `b'\xc3\x83'`, not `"\xc3"`.
• @jlaur: Your edited question yields the exact same issue. When you do `ord('Ĭ')`, it is the ASCII encoding of `Ĭ` which is used as input to the `ord` function, not the utf8 encoding.