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
Related
What is the relationship between a DECIMAL(m,n) column specification and the actual representation of that column in a 64-bit MySQL implementation?
I'm defining tables in a context where I know I need an exact value (hence DECIMAL) and I don't know how sensitive I am to truncation errors in the decimal portion. I'd therefore like to choose a column specification that makes reasonable use of the underlying storage (I know it's a 64-bit system).
I haven't yet found an answer in the MySQL documentation despite a reasonable search.
It doesn't matter if you're using a 64-bit MySQL build. DECIMAL supports precision greater than 64 bits.
DECIMAL uses 4 bytes for each 9 digits, plus extra bytes for the leftover digits. For example, DECIMAL(32,0) supports up to 9+9+9+5 digits. It will use 4+4+4 bytes for the first 27 digits, then 3 more bytes for the remaining 5 digits. A total of 12+3, or 15 bytes.
The fractional part of the decimal value (after the decimal point) stores digits similarly. So DECIMAL(32,9) would support up to 9+9+5 digits for the integer portion and another 9 digits for the fractional portion. Thus 4+4+3 bytes for the integer and 4 bytes for the fractional part.
There's a more detailed description with examples down to the byte, in the code comments for the decimal2bin() function here:
https://github.com/mysql/mysql-server/blob/8.0/strings/decimal.cc#L1282-L1343
For example: octal 3 is 011 in the binary digits, and decimal 3 is 0011.
So, is there any way to figure them out or this table must be copied and pasted in my memory?
If I wanted to represent -2455.1152 as 32 bit I know the first bit is 1 (negative sign) but I can get the 2455 to binary as 10010010111 but for the fractional part I'm not too sure. .1152 could have an infinite number of fractional parts. Would that mean that only up to 23 bits are used to represent the fractional part? So since 2445 uses 11 bits, bits 11 to 0 are for the fractional part?
for the binary representation I have 10010010111.00011101001. Exponent is 10. 10+127=137. 137 as binary is 10001001.
full representation would be:
1 10001001 1001001011100011101001
is that right?
It looks like you are trying to devise your own floating-point representation, but you used a fixed-point tag so I will explain how to convert your real number to a traditional fixed-point representation. First, you need to decide how many bits will be used to represent the fractional part of the number. Just for the sake of discussion let's say that 16 bits will be used for the fractional part, 15 bits for the integer part, and one bit reserved for the sign bit. Now, multiply the absolute value of the real number by 2^{16}: 2455.1152 * 65536 = 160898429.747. You can either round to the nearest integer or just truncate. Suppose we just truncate to 160898429. Converting this to hexadecimal we get 0x09971D7D. To make this negative, invert and add a 1 to the LSB, and the final result is 0xF668E283.
To convert back to a real number just reverse the process. Take the absolute value of the fixed-point representation and divide by 2^{16}. In this case we would find that the fixed-point representation is equal to the real number -2455.1151886 . The accuracy can be improved by rounding instead of truncating when converting from real to fixed-point, or by allowing more bits for the fractional part.
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 :-)
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