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UTF-16 can be two bytes character sets or four bytes character sets.
As I understand, the four byte character sets consist of surrogates which are high(16) and low(16).
I know that UTF-8 can be either one, two, three, four, or up to seven bytes.
But we can tell how many bytes are coming from UTF-8 easily by looking at the bits in the first octet.
I wonder how we identify the bytes size coming from UTF-16 plus UTF-32.
I would appreciate your help and welcome opinions on Unicode.
UTF-8
This uses 1, 2, 3 or 4 bytes per codepoint, depending on the codepoint's value (5+ byte variations are not allowed by RFC 3629 in order to maintain compatibility with UTF-16, as they can encode codepoints higher than UTF-16's max codepoint of U+10FFFF). Analyzing the high bits of the first byte will tell you how many total bytes are in the sequence, including that first byte:
if the high bit is 0, the byte's numeric value is the codepoint's value.
if the high 2 bits are 1 and the high 3rd bit is 0, the codepoint value spans 2 bytes. The next byte must have its high bit set to 1 and its high 2nd bit set to 0, or else the sequence is malformed.
if the high 3 bits are 1 and the high 4th bit is 0, the codepoint value spans 3 bytes. The next 2 bytes must have their high bit set to 1 and their high 2nd bit set to 0, or else the sequence is malformed.
if the high 4 bits are 1 and the high 5th bit is 0, the codepoint value spans 4 bytes. The next 3 bytes must have their high bit set to 1 and their high 2nd bit set to 0, or else the sequence is malformed.
if 5 or more high bits are set to 1, the sequence is malformed.
UTF-16
This uses 2 or 4 bytes per codepoint, depending on the codepoint's value. Analyzing the numeric value of the first 2 bytes (taking big/little endian into account) will tell you how many total bytes are in the sequence, including those first 2 bytes.
If the numeric value is in the range of 0x0000..0xD7FF or 0xE000..0xFFFF, it is the codepoint value.
If the numeric value is in the range of 0xD800..0xDBFF, the codepoint value spans 4 bytes. The first 2 bytes are the high surrogate. The numeric value of the next 2 bytes must be in the range of 0xDC00..0xDFFF for the low surrogate, or else the sequence is malformed.
If the numeric value is in the range of 0xDC00..0xDFFF, the sequence is malformed.
UTF-32
This is always 4 bytes per codepoint. The numeric value of the 4 bytes (taking big/little endian into account) is the codepoint value.
It may be a very basic low level architecture questions. I am trying to get my head around it. Please correct if my understanding is wrong, as well.
Word = 64 bit, 32 bit, etc. This is a number of bits computer can read at a time.
Questions:
1.) Would this mean, we can send, 4 numbers (of a 8 bits/byte length each) for 32 bit? Or combination of 8 bit (byte), 32 bit (4 bytes), etc numbers at one time?
2.) If we need to send only 8 bit number, then how does it form a word? Only first byte is filled and rest all bytes are padded with 0s or last byte gets filled while rest of the bytes are padded with 0s? Or I saw somewhere like first byte has information as to how the rest of the bytes are filled. Does that apply here? For example, UTF-8. Here, ASCII is 1 byte, and some other chars take up to 4 bytes. So when we send one char, we send all 4 bytes together, but fill the bytes as required for the char and rest of the bytes 0s?
3.) Now to represent 8 digit number, we would need 27 bits (remember famous question, sorting 1 million 8 digit number with just 1 MB RAM). Can we exactly use 27 bits, which is 32 bits (4 bytes) - 5 bits? and use those 5 digits for something else?
Appreciate your answers!
1- Yes, four 8-bit integers can fit in a 32-bit integer. This can be done using bitwise operations, for example (using C operators):
((a & 255) << 24) | ((b & 255) << 16) | ((c & 255) << 8) | (d & 255)
This example uses C operators, but they are also used for the same purpose in several other languages (see below - a complete, compilable version of this example in C). You may want to look up the bitwise operators AND (&), OR (|), and Left Shift (<<);
2- Unused bits are generally 0. The first byte is sometimes used to represent the type of encoding (Look up "Magic Numbers"), but this is implementation dependent. Sometimes it is a different number of bits.
3- Groups of 8-digit numbers can be compressed to use only 27 bits each. This is very similar to the example, except the number of bits and size of the data are different. To do this, you will need 864-bit groups, i.e. 27 32-bit integers to store 32 27-bit numbers. This would be more complex than the example, but it would use the same principles.
Complete, compilable example in C:
#include <stdio.h>
/*Compresses four integers containing one byte of data in the least
*significant byte into a single 32-bit integer*/
__int32 compress(int a, int b, int c, int d){
__int32 compressed = ((a & 255) << 24) | ((b & 255) << 16) |
((c & 255) << 8) | (d & 255);
return compressed;
}
/*Test the compress() function and print the resuts*/
int main(){
printf("%x\n", (unsigned)compress(255, 0, 255, 0));
printf("%x\n", (unsigned)compress(192, 168, 0, 255));
printf("%x\n", (unsigned)compress(84, 94, 255, 2));
return 0;
}
I think that clarification on 2 points is required here :
1. Memory addressing.
2. Word
Memories can be addressed in 2 ways, they are generally either byte addressable or word addressable.
Byte addressable memory means that each byte is given a separate address.
a -> 0th byte
b -> 1st byte
Word addressable memories are those in which each group of bytes that is as wide as the word gets an address. Eg if the Word Length is 32 bits :
a->0th byte
b->4th byte
And so on.
Word
I would say that a word defines the maximum number of bits a processor can handle at a time. For 8086, for eg, it's 16.
It is usually the largest number on which the arithmetic can be performed by the processor. Continuing the example , 8086 can perform operations on 16 bit numbers at a time.
Now i'll try and answer the questions :
1.) Would this mean, we can send, 4 numbers (of a 8 bits/byte length each) for 32 bit? Or combination of 8 bit (byte), 32 bit (4 bytes),
etc numbers at one time?
You can always define your own interpretation for a bunch of bits.
For eg, If it is byte addressable, we can treat every byte individually and thus , we can write code at assemble level that treats each byte as a separate 8 bit number.
If it is not, you can use bit operations to extract individual bytes out.
The point is you can represent 4 8 bit numbers in 32 bits.
2) Mostly, leftover significant bits are stuffed with 0s ( for unsigned numbers)
3.) Now to represent 8 digit number, we would need 27 bits (remember famous question, sorting 1 million 8 digit number with just 1 MB RAM).
Can we exactly use 27 bits, which is 32 bits (4 bytes) - 5 bits? and
use those 5 digits for something else?
Yes, you can do this also. But you know the great space-time tradeoff.
You sure save 5 bits, per number. But you'll need to use bit operations and all the really cool but hard to read stuff. Shooting up time and making code more complex.
But i don't think you'll ever come across a situation where you need such level of saving, unless you are coding for a very constrained system. (embedded etc)
I read some docs about md5, it said that its 128 bits, but why is it 32 characters? I can't compute the characters.
1 byte is 8 bits
if 1 character is 1 byte
then 128 bits is 128/8 = 16 bytes right?
EDIT:
SHA-1 produces 160 bits, so how many characters are there?
32 chars as hexdecimal representation, thats 2 chars per byte.
I wanted summerize some of the answers into one post.
First, don't think of the MD5 hash as a character string but as a hex number. Therefore, each digit is a hex digit (0-15 or 0-F) and represents four bits, not eight.
Taking that further, one byte or eight bits are represented by two hex digits, e.g. b'1111 1111' = 0xFF = 255.
MD5 hashes are 128 bits in length and generally represented by 32 hex digits.
SHA-1 hashes are 160 bits in length and generally represented by 40 hex digits.
For the SHA-2 family, I think the hash length can be one of a pre-determined set. So SHA-512 can be represented by 128 hex digits.
Again, this post is just based on previous answers.
A hex "character" (nibble) is different from a "character"
To be clear on the bits vs byte, vs characters.
1 byte is 8 bits (for our purposes)
8 bits provides 2**8 possible combinations: 256 combinations
When you look at a hex character,
16 combinations of [0-9] + [a-f]: the full range of 0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f
16 is less than 256, so one one hex character does not store a byte.
16 is 2**4: that means one hex character can store 4 bits in a byte (half a byte).
Therefore, two hex characters, can store 8 bits, 2**8 combinations.
A byte represented as a hex character is [0-9a-f][0-9a-f] and that represents both halfs of a byte (we call a half-byte a nibble).
When you look at a regular single-byte character, (we're totally going to skip multi-byte and wide-characters here)
It can store far more than 16 combinations.
The capabilities of the character are determined by the encoding. For instance, the ISO 8859-1 that stores an entire byte, stores all this stuff
All that stuff takes the entire 2**8 range.
If a hex-character in an md5() could store all that, you'd see all the lowercase letters, all the uppercase letters, all the punctuation and things like ¡°ÀÐàð, whitespace like (newlines, and tabs), and control characters (which you can't even see and many of which aren't in use).
So they're clearly different and I hope that provides the best break down of the differences.
MD5 yields hexadecimal digits (0-15 / 0-F), so they are four bits each. 128 / 4 = 32 characters.
SHA-1 yields hexadecimal digits too (0-15 / 0-F), so 160 / 4 = 40 characters.
(Since they're mathematical operations, most hashing functions' output is commonly represented as hex digits.)
You were probably thinking of ASCII text characters, which are 8 bits.
One hex digit = 1 nibble (four-bits)
Two hex digits = 1 byte (eight-bits)
MD5 = 32 hex digits
32 hex digits = 16 bytes ( 32 / 2)
16 bytes = 128 bits (16 * 8)
The same applies to SHA-1 except it's 40 hex digits long.
I hope this helps.
That's 32 hex characters - 1 hex character is 4 bits.
Those are hexidecimal digits, not characters. One digit = 4 bits.
They're not actually characters, they're hexadecimal digits.
For clear understanding, copy the MD5 calculated 128 bit hash value in the Binary to Hex convertor and see the length of the Hex value. You will get 32 characters Hex characters.
The title of the question says it all. I have been researching SHA-1 and most places I see it being 40 Hex Characters long which to me is 640bit. Could it not be represented just as well with only 10 hex characters 160bit = 20byte. And one hex character can represent 2 byte right? Why is it twice as long as it needs to be? What am I missing in my understanding.
And couldn't an SHA-1 be even just 5 or less characters if using Base32 or Base36 ?
One hex character can only represent 16 different values, i.e. 4 bits. (16 = 24)
40 × 4 = 160.
And no, you need much more than 5 characters in base-36.
There are totally 2160 different SHA-1 hashes.
2160 = 1640, so this is another reason why we need 40 hex digits.
But 2160 = 36160 log362 = 3630.9482..., so you still need 31 characters using base-36.
I think the OP's confusion comes from a string representing a SHA1 hash takes 40 bytes (at least if you are using ASCII), which equals 320 bits (not 640 bits).
The reason is that the hash is in binary and the hex string is just an encoding of that. So if you were to use a more efficient encoding (or no encoding at all), you could take only 160 bits of space (20 bytes), but the problem with that is it won't be binary safe.
You could use base64 though, in which case you'd need about 27-28 bytes (or characters) instead of 40 (see this page).
There are two hex characters per 8-bit-byte, not two bytes per hex character.
If you are working with 8-bit bytes (as in the SHA-1 definition), then a hex character encodes a single high or low 4-bit nibble within a byte. So it takes two such characters for a full byte.
My answer only differs from the previous ones in my theory as to the EXACT origin of the OP's confusion, and in the baby steps I provide for elucidation.
A character takes up different numbers of bytes depending on the encoding used (see here). There are a few contexts these days when we use 2 bytes per character, for example when programming in Java (here's why). Thus 40 Java characters would equal 80 bytes = 640 bits, the OP's calculation, and 10 Java characters would indeed encapsulate the right amount of information for a SHA-1 hash.
Unlike the thousands of possible Java characters, however, there are only 16 different hex characters, namely 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. But these are not the same as Java characters, and take up far less space than the encodings of the Java characters 0 to 9 and A to F. They are symbols signifying all the possible values represented by just 4 bits:
0 0000 4 0100 8 1000 C 1100
1 0001 5 0101 9 1001 D 1101
2 0010 6 0110 A 1010 E 1110
3 0011 7 0111 B 1011 F 1111
Thus each hex character is only half a byte, and 40 hex characters gives us 20 bytes = 160 bits - the length of a SHA-1 hash.
2 hex characters mak up a range from 0-255, i.e. 0x00 == 0 and 0xFF == 255. So 2 hex characters are 8 bit, which makes 160 bit for your SHA digest.
SHA-1 is 160 bits
That translates to 20 bytes = 40 hex characters (2 hex characters per byte)
UTF-8 requires 4 bytes to represent characters outside the BMP. That's not bad; it's no worse than UTF-16 or UTF-32. But it's not optimal (in terms of storage space).
There are 13 bytes (C0-C1 and F5-FF) that are never used. And multi-byte sequences that are not used such as the ones corresponding to "overlong" encodings. If these had been available to encode characters, then more of them could have been represented by 2-byte or 3-byte sequences (of course, at the expense of making the implementation more complex).
Would it be possible to represent all 1,114,112 Unicode code points by a UTF-8-like encoding with at most 3 bytes per character? If not, what is the maximum number of characters such an encoding could represent?
By "UTF-8-like", I mean, at minimum:
The bytes 0x00-0x7F are reserved for ASCII characters.
Byte-oriented find / index functions work correctly. You can't find a false positive by starting in the middle of a character like you can in Shift-JIS.
Update -- My first attempt to answer the question
Suppose you have a UTF-8-style classification of leading/trailing bytes. Let:
A = the number of single-byte characters
B = the number of values used for leading bytes of 2-byte characters
C = the number of values used for leading bytes of 3-byte characters
T = 256 - (A + B + C) = the number of values used for trailing bytes
Then the number of characters that can be supported is N = A + BT + CT².
Given A = 128, the optimum is at B = 0 and C = 43. This allows 310,803 characters, or about 28% of the Unicode code space.
Is there a different approach that could encode more characters?
It would take a little over 20 bits to record all the Unicode code points (assuming your number is correct), leaving over 3 bits out of 24 for encoding which byte is which. That should be adequate.
I fail to see what you would gain by this, compared to what you would lose by not going with an established standard.
Edit: Reading the spec again, you want the values 0x00 through 0x7f reserved for the first 128 code points. That means you only have 21 bits in 3 bytes to encode the remaining 1,113,984 code points. 21 bits is barely enough, but it doesn't really give you enough extra to do the encoding unambiguously. Or at least I haven't figured out a way, so I'm changing my answer.
As to your motivations, there's certainly nothing wrong with being curious and engaging in a little thought exercise. But the point of a thought exercise is to do it yourself, not try to get the entire internet to do it for you! At least be up front about it when asking your question.
I did the math, and it's not possible (if wanting to stay strictly "UTF-8-like").
To start off, the four-byte range of UTF-8 covers U+010000 to U+10FFFF, which is a huge slice of the available characters. This is what we're trying to replace using only 3 bytes.
By special-casing each of the 13 unused prefix bytes you mention, you could gain 65,536 characters each, which brings us to a total of 13 * 0x10000, or 0xD0000.
This would bring the total 3-byte character range to U+010000 to U+0DFFFF, which is almost all, but not quite enough.
Sure it's possible. Proof:
224 = 16,777,216
So there is enough of a bit-space for 1,114,112 characters but the more crowded the bit-space the more bits are used per character. The whole point of UTF-8 is that it makes the assumption that the lower code points are far more likely in a character stream so the entire thing will be quite efficient even though some characters may use 4 bytes.
Assume 0-127 remains one byte. That leaves 8.4M spaces for 1.1M characters. You can then solve this is an equation. Choose an encoding scheme where the first byte determines how many bytes are used. So there are 128 values. Each of these will represent either 256 characters (2 bytes total) or 65,536 characters (3 bytes total). So:
256x + 65536(128-x) = 1114112 - 128
Solving this you need 111 values of the first byte as 2 byte characters and the remaining 17 as 3 byte. To check:
128 + 111 * 256 + 17 * 65536 = 1,114,256
To put it another way:
128 code points require 1 byte;
28,416 code points require 2 bytes; and
1,114,112 code points require 3 bytes.
Of course, this doesn't allow for the inevitable expansion of Unicode, which UTF-8 does. You can adjust this to the first byte meaning:
0-127 (128) = 1 byte;
128-191 (64) = 2 bytes;
192-255 (64) = 3 bytes.
This would be better because it's simple bitwise AND tests to determine length and gives an address space of 4,210,816 code points.