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I've found such numbers time and time again in verilog and digital logic exercises, but I don't know how I'm supposed to know what number this is or how to write it as 12'b in verilog code.
Can someone explain them to me?
Presumably it's a 12-bit value in hexadecimal notation. Since one hexadecimal digit encodes 4 bits, there are 3 hexadecimal digits:
hexadecimal 0 corresponds to 4 bits 0000
hexadecimal 6 corresponds to 4 bits 0110
hexadecimal f corresponds to 4 bits 1111
So the same value in binary notation is 000001101111.
In Verilog you can write
12'h06f
or
12'b000001101111
Where 12' means "12 bits", h means "hexadecimal" and b means "binary".
The number in the post is known as a literal constant or a number literal or just a literal.
There is a short tutorial document with examples that is a good reference on the subject.
The document can be found here:
https://web.engr.oregonstate.edu/~traylor/ece474/beamer_lectures/verilog_number_literals.pdf
The first page looks like this:
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
I am changing an 8-digit hexadecimal number to a real number Y.
The number is 0xAC0396ED
My question is: Shall I take into consideration the 0x? What is the significance of the 0x?
I researched it and based on wikipedia, I got this:
"use the prefix 0x for numeric constants represented in hex"
What I plan on doing is take the part AC0396ED and change it to binary, then from binary manipulate the 32 bit number byte dividing the number intro 3 parts: sign, exponent, and fraction.
My last question is why do we need hexacdecimal, decimal, octal? Why don't we just stick to binary in all our arithmetic and operations?
Thank you.
My question is: Shall I take into consideration the 0x? What is the significance of the 0x?
No - you should not take that into consideration - its only a notation for hexadecimal base system
What I plan on doing is take the part AC0396ED and change it to binary, then from binary manipulate the 32 bit number byte dividing the number intro 3 parts: sign, exponent, and fraction.
Here is how you calculate it:
AC0396ED =>
10 12 0 3 9 6 14 13 in decimal. Then to binary
1010 1100 0000 0011 1001 0110 1110 1101
here you have all bits i.e 32 bits
1 | 01011000 | 00000111001011011101101
so the first bit is 1 => the number is negative
the second is the exponent
the third is the mantissa
My last question is why do we need hexacdecimal, decimal, octal? Why don't we just stick to binary in all our arithmetic and operations?
Its much more easy to stick to hexadecimal system - compare the 32 bits to the AC0396ED
My instructor stated that "an octal byte consists of 6 bits". I am having difficulty understanding why this is, as an octal digit consists of 3 binary bits. I also do not understand the significance of an octal byte being defined as '6 bits' as opposed to some other number.
Can anyone explain why this is, if it is in fact true, or point me to a useful explanation?
This is all speculation and guesswork, since none of this is in any way standard terminology.
An 8-bit byte can be written as two digits of hexadecimals, because each digit expresses 4 bits. The largest such byte value is 0xFF.
By analogy, two digits of octals can express 2 × 3 = 6 bits. Its largest value is 077. So if you like you can call a pair of two octals an "octal byte", but only if you will also call an 8-bit byte a "hexadecimal byte".
In my personal opinion neither notion is helpful or useful, and you'd be best of just to say how many bits your byte has.
Like #Kerrek SB I'd have to guess the same.
Tell him an octal byte is two octal nibbles, that should sort him out.
Two hexadecimal digits is an 8 bit byte, so each four bits were called a nibble.
as for wiki : The byte is a unit of digital information that most commonly consists of 8 bits. The bit is a contraction of binary digit. The bit represents a logical state with one of two possible values. These values are most commonly represented as either "1" or "0".
so byte is a set of bunch of bits. to be specific of 8 bits.
we see that the definition doesnt say anything about dec oct hex and other notations of integer number.
indeed byte and integer number is not the same. so where
does it start ?
if we type bits of byte like so 01001010 we can find that
we can map 1-to-1 this object to binary notation
of integer numbers 01001010.
(byte)01001010 --> (binary integer) 01001010
the two objects look the same but actually is just mapping.
now we work with integer number instead of abstract object "byte". an integer
number can be designated via different notations like : dec, binary, hex, oct etc.
notation like octal(hex,dec etc) is just a method of designation of integer number. it does not influence the nature of initial byte object.
byte has 8 bits whatever notation is used.
ISO/IEC 2382-1:1993 says:
1.The number of bits in a byte is usually 8.
2.Octet is Byte that consists of eight
Bits
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