Just wondering on how I would go about converting binary to hexadecimal??
Would I first have to convert the binary to decimal and then to hexadecimal??
For example, 101101001.101110101010011
How would I go about converting a complex binary such as the above to hexadecimal?
Thanks in advance
Each 4 bits of a binary number represents a hexadecimal digit. So the best way to convert from binary to hexadecimal is to pad the binary number with leading zeroes so that the number of bits is divisible by four.
Then you process four bits at a time and convert them to a single hexadecimal digit:
0000 -> 0
0001 -> 1
0010 -> 2
....
1110 -> E
1111 -> F
No, you don't convert to decimal and then to hexadecimal, you convert to a numeric value, and then to hexadecimal.
(Decimal is also a textual representation of a number, just like binary and hexadecimal. Although decimal representation is used by default, a number doesn't have a textual representation in itself.)
As a hexadecimal digit corresponds to four binary digits you don't have to convert the entire string to a number, you can do it four binary digits at a time.
First fill up the binary number so that it has full groups of four digits:
000101101001.1011101010100110
Then you can convert each group to a number, and then to hexadecimal:
0001 0110 1001.1011 1010 1010 0110
169.BAA6
Alternatively, you can split the number into the two parts before and after the period and convert those from binary. The part before the period can be converted stright off, but the part after has to be padded to be correct.
Example in C#:
string binary = "101101001.101110101010011";
string[] parts = binary.Split('.');
while (parts[1].Length % 4 != 0) {
parts[1] += '0';
}
string result =
Convert.ToInt32(parts[0], 2).ToString("X") +
"." +
Convert.ToInt32(parts[1], 2).ToString("X");
You could simply have a small hash table, or other mapping converting each quadruplet of binary digits (as a string, assuming that's your input) into the corresponding hex digit (0 to 9, A to F) for the output string. You'll have to bunch the input bits up by 4, left-padding before the '.' and right-padding after it, with 0 in both cases, as needed.
So...:
locate the '.'
left of the '.', bunch by 4, left-padding the last bunch, going leftwards: in your example, 1001 leftmost, then 0110, finally 0001 (left-padding), that's it;
ditto to the right -- in your example 1011, then 1010, then 1010, finally 0110 (right-padding)
each bunch of 4 binary digits, via a hash or other form of hashing, turns into the hex digit to put in that place in the output string.
Want some pseudo-code for it, e.g., Python?
The simplest approach, especially if you already can convert from binary digits to internal numeric representation and from internal numeric representation to hexadecimal digits, is to go binary->internal->hex. I say internal and not decimal, because even though it may print as decimal, it is actually being stored internally in binary format. That said, it is possible to go straight from one to the other. This does not apply to your specific example, but in many cases when converting from binary to hex, you can go four digits at a time, and simply lookup the corresponding hex values in a table. There are all sorts of ways to convert.
BIN to HEX
Binary and hex are natively compatible. Just group 4 binary digits(bits) and substitute the corresponding HEX-digit.
More reference here:
http://en.wikipedia.org/wiki/Hexadecimal#Binary_conversion
Related
During an EMV transaction, all information exchanged between terminal and card is encoded in byte strings. In order to understand the content of the messages and give meaning to the bits, you should first get familiar with the hexadecimal notation. One byte can be represented by two hexadecimal numbers, or eight binary (0,1) numbers.
What is the binary (i.e., in bits) representation of the byte ‘E3’?
The EMV tags follow Packed BCD format. It means a byte can hold two Hexadecimal values. In your case it becomes [ [ 1110 ] [ 0011 ] ]. The first nibble holds the binary representation for E and the second one holds for 3.
You can simply use Programmer Calculator given in Windows and Linux as well to convert Hexadecimal values into binary as follows:
I see the following code:
binary scan "xyz" H* var
It puzzles me: binary scan is supposed to scan a binary stream and construct string type variables, but here it is "xyz" ...?
I did the following experiment inside tclsh:
% puts $var
78797a <== what is this?
% binary scan $var #1H y <== I mean to get "y"
1
% puts $y <== but I get "3"?
3
I am lost.
Could you explain what is going on?
Does it help to know that the hexadecimal value of the character 'x' is 0x78? Or that binary scan \x78\x79\x7a H* var2 is identical to your example? The examples in the 'binary scan' manual page under the 'H' conversion code explain it pretty well, I think.
In your code:
binary scan "xyz" H* var
The binary string is xyz, which is three bytes that are the ASCII values for x, y and z. We then ask for the var variable to be given a sequence hex digits of the scanned bytes in big endian order (very much the right thing for dealing with strings, BTW!) with twice as many hex digits as there are bytes in the binary string (because *). Let's double check with what the documentation says:
The data is turned into a string of count hexadecimal digits in high-to-low order represented as a sequence of characters in the set “0123456789abcdef”. The data bytes are scanned in first to last order with the hex digits being taken in high-to-low order within each byte. Any extra bits in the last byte are ignored. If count is *, then all of the remaining hex digits in string will be scanned. If count is omitted, then one hex digit will be scanned. For example,
binary scan \x07\xC6\x05\x1f\x34 H3H* var1 var2
will return 2 with 07c stored in var1 and 051f34 stored in var2.
Now, there are three bytes in xyz so there are six digits in 78797a. The first two hex digits, 78 are the the hex for the ASCII version of x (check for yourself), and similarly for 79 and 7a.
When you then do:
binary scan $var #1H y
you move the internal cursor into the string to the byte for the ASCII for 8 (because zero-based indexing), \x38, and because there's no count given to the H, it gets the first hex digit of 38 (i.e., 3) and puts that in the y variable.
To actually retrieve the y, you can just use string index or string range on the original binary string (as all Tcl's string commands work just fine on binary data). Or you use string range to get the hex digits out of var and binary format to convert back:
binary format H* [string range $var 2 3]
It's probably not a good idea to binary scan the results of binary scan. It's totally legal to do so, but the results are going to be unlikely to illuminate.
This is surely a duplicate, but I was not able to find an answer to the following question.
Let's consider the decimal integer 14. We can obtain its binary representation, 1110, using e.g. the divide-by-2 method (% represents the modulus operand):
14 % 2 = 0
7 % 2 = 1
3 % 2 = 1
1 % 2 = 1
but how computers convert decimal to binary integers?
The above method would require the computer to perform arithmetic and, as far as I understand, because arithmetic is performed on binary numbers, it seems we would be back dealing with the same issue.
I suppose that any other algorithmic method would suffer the same problem. How do computers convert decimal to binary integers?
Update: Following a discussion with Code-Apprentice (see comments under his answer), here is a reformulation of the question in two cases of interest:
a) How the conversion to binary is performed when the user types integers on a keyboard?
b) Given a mathematical operation in a programming language, say 12 / 3, how does the conversion from decimal to binary is done when running the program, so that the computer can do the arithmetic?
There is only binary
The computer stores all data as binary. It does not convert from decimal to binary since binary is its native language. When the computer displays a number it will convert from the binary representation to any base, which by default is decimal.
A key concept to understand here is the difference between the computers internal storage and the representation as characters on your monitor. If you want to display a number as binary, you can write an algorithm in code to do the exact steps that you performed by hand. You then print out the characters 1 and 0 as calculated by the algorithm.
Indeed, like you mention in one of you comments, if compiler has a small look-up table to associate decimal integers to binary integers then it can be done with simple binary multiplications and additions.
Look-up table has to contain binary associations for single decimal digits and decimal ten, hundred, thousand, etc.
Decimal 14 can be transformed to binary by multipltying binary 1 by binary 10 and added binary 4.
Decimal 149 would be binary 1 multiplied by binary 100, added to binary 4 multiplied by binary 10 and added binary 9 at the end.
Decimal are misunderstood in a program
let's take an example from c language
int x = 14;
here 14 is not decimal its two characters 1 and 4 which are written together to be 14
we know that characters are just representation for some binary value
1 for 00110001
4 for 00110100
full ascii table for characters can be seen here
so 14 in charcter form actually written as binary 00110001 00110100
00110001 00110100 => this binary is made to look as 14 on computer screen (so we think it as decimal)
we know number 14 evntually should become 14 = 1110
or we can pad it with zero to be
14 = 00001110
for this to happen computer/processor only need to do binary to binary conversion i.e.
00110001 00110100 to 00001110
and we are all set
How to convert alphabet to binary? I search on Google and it says that first convert alphabet to its ASCII numeric value and than convert the numeric value to binary. Is there any other way to convert ?
And if that's the only way than is the binary value of "A" and 65 are same?
BECAUSE ASCII vale of 'A'=65 and when converted to binary its 01000001
AND 65 =01000001
That is indeed the way which text is converted to binary.
And to answer your second question, yes it is true that the binary value of A and 65 are the same. If you are wondering how CPU distinguishes between "A" and "65" in that case, you should know that it doesn't. It is up to your operating system and program to distinguish how to treat the data at hand. For instance, say your memory looked like the following starting at 0 on the left and incrementing right:
00000001 00001111 000000001 01100110
This binary data could mean anything, and only has a meaning in the context of whatever program it is in. In a given program, you could have it be read as:
1. An integer, in which case you'll get one number.
2. Character data, in which case you'll output 4 ASCII characters.
In short, binary is read by CPUs, which do not understand the context of anything and simply execute whatever they are given. It is up to your program/OS to specify instructions in order for data to be handled properly.
Thus, converting the alphabet to binary is dependent on the program in which you are doing so, and outside the context of a program/OS converting the alphabet to binary is really the exact same thing as converting a sequence of numbers to binary, as far as a CPU is concerned.
Number 65 in decimal is 0100 0001 in binary and it refers to letter A in binary alphabet table (ASCII) https://www.bin-dec-hex.com/binary-alphabet-the-alphabet-letters-in-binary. The easiest way to convert alphabet to binary is to use some online converter or you can do it manually with binary alphabet table.
Right now I'm preparing for my AP Computer Science exam, and I need some help understanding how to convert between decimal, hexadecimal, and binary values by hand. The book that I'm using (Barron's) includes an example but does not explain it very well.
What are the formulas that one should use for conversion between these number types?
Are you happy that you understand number bases? If not, then you will need to read up on this or you'll just be blindly following some rules.
Plenty of books would spend a whole chapter or more on this...
Binary is base 2, Decimal is base 10, Hexadecimal is base 16.
So Binary uses digits 0 and 1, Decimal uses 0-9, Hexadecimal uses 0-9 and then we run out so we use A-F as well.
So the position of a decimal digit indicates units, tens, hundreds, thousands... these are the "powers of 10"
The position of a binary digit indicates units, 2s, 4s, 8s, 16s, 32s...the powers of 2
The position of hex digits indicates units, 16s, 256s...the powers of 16
For binary to decimal, add up each 1 multiplied by its 'power', so working from right to left:
1001 binary = 1*1 + 0*2 + 0*4 + 1*8 = 9 decimal
For binary to hex, you can either work it out the total number in decimal and then convert to hex, or you can convert each 4-bit sequence into a single hex digit:
1101 binary = 13 decimal = D hex
1111 0001 binary = F1 hex
For hex to binary, reverse the previous example - it's not too bad to do in your head because you just need to work out which of 8,4,2,1 you need to add up to get the desired value.
For decimal to binary, it's more of a long division problem - find the biggest power of 2 smaller than your input, set the corresponding binary bit to 1, and subtract that power of 2 from the original decimal number. Repeat until you have zero left.
E.g. for 87:
the highest power of two there is 1,2,4,8,16,32,64!
64 is 2^6 so we set the relevant bit to 1 in our result: 1000000
87 - 64 = 23
the next highest power of 2 smaller than 23 is 16, so set the bit: 1010000
repeat for 4,2,1
final result 1010111 binary
i.e. 64+16+4+2+1 = 87 in decimal
For hex to decimal, it's like binary to decimal, only you multiply by 1,16,256... instead of 1,2,4,8...
For decimal to hex, it's like decimal to binary, only you are looking for powers of 16, not 2. This is the hardest one to do manually.
This is a very fundamental question, whose detailed answer, on an entry level could very well be a couple of pages. Try to google it :-)