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:


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.

Have a look at The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!) and the wikipedia page about UTF-8.

A Tutorial on Data Representation, Eight bits is called a byte (why 8-bit unit? However, writing or reading a long sequence of binary bits is cumbersome and error-prone (try to read this To convert 1023(base 4) to base 3: 1023(base 4)/3 => quotient=25D remainder=0 25D/3� Obviously, the best choice of converting a byte sequence into a positive integer is to use the BigInteger(int signum, byte[] magnitude) constructor as: "new BigInteger(1,byteArray)". This avoids getting negative numbers when the first bit of the byte sequence is a negative sign. For converting a positive integer back to a byte sequence, we have

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

Sample code for your convenience:

for n in range(256):

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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 c4in 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".
  • Thank you for this insanely awesome explanation! That really made a lot of pieces add up.
  • Tnx for the reply. I edited the question as I think the answer doesn't fit my question.
  • @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.
  • "é" and "Ĭ" are not defined in ASCII. These numbers are the Unicode code points. For code points below 256, these are the same as Latin 1, which is sometimes referred to as "extended ASCII".