I don't understand how integer_decode in num_traits works. For instance: we have
use num_traits::Float;
let num = 2.0f32;
// (8388608, -22, 1)
let (mantissa, exponent, sign) = Float::integer_decode(num);
But how we get those integers?
Binary representation of 2.0f32 has 0 sign bit, 1 bit as leading bit in exponent and mantissa consisting of zeros. How to get integer decode and why we choose this particular decomposition and not 8388608*2 as mantissa and -23 as exponent?
I didn't write the function, so take this answer with a grain of salt, as it's more of a gut feeling than knowledge. The rationale behind it is not explained in the comments in the function implementation, so unless the author of the code speaks up, we can't deliver more than educational guesses.
f32 is based on IEEE-754, which specifies that a 2.0 shall be represented as the following three parts:
the sign bit 0
indicates that 2.0 is positive
the exponent 128
it's one byte that indicates the exponent, with 127 representing 0. 128 means 1
the mantissa 0
the mantissa consists of 23 bits and has an implicit 1. in front of it. So 0 means 1.0.
To get the actual number, you need to do 0 * (-1) + 2 ^ (128 - 127) * 1.0, which is 2 ^ 1 * 1 = 2.
Now this is not the only way to compute that. You could also do:
map the sign bit to 1 and -1
instead of prefixing the mantissa with 1., add a 1 in front of it, making it an integer. (this avoids having to use a float to decode a float, which is nonsense for obvious reasons)
subtract 127 from the exponent, making it signed. Then, remove 23 from it to compensate that our mantissa is now shifted by 23 bits (because the mantissa is 23 bits long and we moved the comma all the way back to make it an integer).
This would, for 2.0 give us:
sign -1
mantissa 0b100000000000000000000000 = 8388608
exponent 128 - 127 - 23 = -22
Now we can do sign * mantissa * 2 ^ exponent, as specified in the documentation to get our value back.
Note how fast calculating those integers was: a binary decision for the sign, a binary or operation for the mantissa and a single u8 subtraction for the exponent (a single one because one can combine - 127 - 23 to - 150 beforehand).
why we choose this particular decomposition and not 8388608*2 as mantissa and -23 as exponent
The short version is that this guarantees that all possible mantissas can be treated the same way. It's 23 bits long and a 1 with the entire mantissa attached to it is always a valid integer. In the case of 0 this is a 1 with 23 0s, 0b100000000000000000000000, which is 8388608.
integer_decode() documentation is quite clear:
Returns the mantissa, base 2 exponent, and sign as integers, respectively. The original number can be recovered by sign * mantissa * 2 ^ exponent.
1 * 8388608 * 2^-22 == 2
Mostly IEEE 754 uses this description of finite floating-point numbers: A floating-point number has the form (−1)s×be×m, where:
s is 0 or 1.
e is any integer emin ≤ e ≤ emax.
m is a number represented by a digit string of the form d0.d1d2…dp−1 where di is an integer digit 0 ≤ di < b.
This form is described in IEEE-754 2019 clause 3.3. b is the base, p is the precision (number of digits in base b), and emin and emax are bounds on the exponent. This form is useful for certain things, such as describing the normalized form as starting with a leading “1.” or “0.” when the base is two. However, the standard also says, in the same clause:
It is also convenient for some purposes to view the significand as an 10 integer; in which case the finite floating-point numbers are described thus:
Signed zero and non-zero floating-point numbers of the form (−1)s×bp×c, where
s is 0 or 1.
q is any integer emin ≤ q+p−1 ≤ emax.
c is a number represented by a digit string of the form d0d1d2…dp−1 where di is an integer digit 0 ≤ di ≤ b (c is therefore an integer with 0 ≤ c < bp).
(My reproduction of the IEEE-754 text changes some of the typography slightly.) Note two things. First, this matches the results Float::integer_decode gives you. In the Float format, p is 24, so m should have 24 bits, not 25, so it can be 8,388,608 (223) and cannot be 16,777,216 (224).
Second, what makes this form useful is m is always an integer and can be any integer in that range—the low digit of m is immediately to the left of the radix point, so the consecutive values representable in this range of the floating-point format are consecutive integers, and we can analyze them and write proofs using number theory.
You could use alternate forms that are mathematically equivalent but let the significand be 224, but then the low digit in such a form would have to be zero (because there is no bit in the format to represent it having any other value), so it is not particularly useful.
Consider this code:
x = 1 # 0001
x << 2 # Shift left 2 bits: 0100
# Result: 4
x | 2 # Bitwise OR: 0011
# Result: 3
x & 1 # Bitwise AND: 0001
# Result: 1
I can understand the arithmetic operators in Python (and other languages), but I never understood 'bitwise' operators quite well. In the above example (from a Python book), I understand the left-shift but not the other two.
Also, what are bitwise operators actually used for? I'd appreciate some examples.
Bitwise operators are operators that work on multi-bit values, but conceptually one bit at a time.
AND is 1 only if both of its inputs are 1, otherwise it's 0.
OR is 1 if one or both of its inputs are 1, otherwise it's 0.
XOR is 1 only if exactly one of its inputs are 1, otherwise it's 0.
NOT is 1 only if its input is 0, otherwise it's 0.
These can often be best shown as truth tables. Input possibilities are on the top and left, the resultant bit is one of the four (two in the case of NOT since it only has one input) values shown at the intersection of the inputs.
AND | 0 1 OR | 0 1 XOR | 0 1 NOT | 0 1
----+----- ---+---- ----+---- ----+----
0 | 0 0 0 | 0 1 0 | 0 1 | 1 0
1 | 0 1 1 | 1 1 1 | 1 0
One example is if you only want the lower 4 bits of an integer, you AND it with 15 (binary 1111) so:
201: 1100 1001
AND 15: 0000 1111
------------------
IS 9 0000 1001
The zero bits in 15 in that case effectively act as a filter, forcing the bits in the result to be zero as well.
In addition, >> and << are often included as bitwise operators, and they "shift" a value respectively right and left by a certain number of bits, throwing away bits that roll of the end you're shifting towards, and feeding in zero bits at the other end.
So, for example:
1001 0101 >> 2 gives 0010 0101
1111 1111 << 4 gives 1111 0000
Note that the left shift in Python is unusual in that it's not using a fixed width where bits are discarded - while many languages use a fixed width based on the data type, Python simply expands the width to cater for extra bits. In order to get the discarding behaviour in Python, you can follow a left shift with a bitwise and such as in an 8-bit value shifting left four bits:
bits8 = (bits8 << 4) & 255
With that in mind, another example of bitwise operators is if you have two 4-bit values that you want to pack into an 8-bit one, you can use all three of your operators (left-shift, and and or):
packed_val = ((val1 & 15) << 4) | (val2 & 15)
The & 15 operation will make sure that both values only have the lower 4 bits.
The << 4 is a 4-bit shift left to move val1 into the top 4 bits of an 8-bit value.
The | simply combines these two together.
If val1 is 7 and val2 is 4:
val1 val2
==== ====
& 15 (and) xxxx-0111 xxxx-0100 & 15
<< 4 (left) 0111-0000 |
| |
+-------+-------+
|
| (or) 0111-0100
One typical usage:
| is used to set a certain bit to 1
& is used to test or clear a certain bit
Set a bit (where n is the bit number, and 0 is the least significant bit):
unsigned char a |= (1 << n);
Clear a bit:
unsigned char b &= ~(1 << n);
Toggle a bit:
unsigned char c ^= (1 << n);
Test a bit:
unsigned char e = d & (1 << n);
Take the case of your list for example:
x | 2 is used to set bit 1 of x to 1
x & 1 is used to test if bit 0 of x is 1 or 0
what are bitwise operators actually used for? I'd appreciate some examples.
One of the most common uses of bitwise operations is for parsing hexadecimal colours.
For example, here's a Python function that accepts a String like #FF09BE and returns a tuple of its Red, Green and Blue values.
def hexToRgb(value):
# Convert string to hexadecimal number (base 16)
num = (int(value.lstrip("#"), 16))
# Shift 16 bits to the right, and then binary AND to obtain 8 bits representing red
r = ((num >> 16) & 0xFF)
# Shift 8 bits to the right, and then binary AND to obtain 8 bits representing green
g = ((num >> 8) & 0xFF)
# Simply binary AND to obtain 8 bits representing blue
b = (num & 0xFF)
return (r, g, b)
I know that there are more efficient ways to acheive this, but I believe that this is a really concise example illustrating both shifts and bitwise boolean operations.
I think that the second part of the question:
Also, what are bitwise operators actually used for? I'd appreciate some examples.
Has been only partially addressed. These are my two cents on that matter.
Bitwise operations in programming languages play a fundamental role when dealing with a lot of applications. Almost all low-level computing must be done using this kind of operations.
In all applications that need to send data between two nodes, such as:
computer networks;
telecommunication applications (cellular phones, satellite communications, etc).
In the lower level layer of communication, the data is usually sent in what is called frames. Frames are just strings of bytes that are sent through a physical channel. This frames usually contain the actual data plus some other fields (coded in bytes) that are part of what is called the header. The header usually contains bytes that encode some information related to the status of the communication (e.g, with flags (bits)), frame counters, correction and error detection codes, etc. To get the transmitted data in a frame, and to build the frames to send data, you will need for sure bitwise operations.
In general, when dealing with that kind of applications, an API is available so you don't have to deal with all those details. For example, all modern programming languages provide libraries for socket connections, so you don't actually need to build the TCP/IP communication frames. But think about the good people that programmed those APIs for you, they had to deal with frame construction for sure; using all kinds of bitwise operations to go back and forth from the low-level to the higher-level communication.
As a concrete example, imagine some one gives you a file that contains raw data that was captured directly by telecommunication hardware. In this case, in order to find the frames, you will need to read the raw bytes in the file and try to find some kind of synchronization words, by scanning the data bit by bit. After identifying the synchronization words, you will need to get the actual frames, and SHIFT them if necessary (and that is just the start of the story) to get the actual data that is being transmitted.
Another very different low level family of application is when you need to control hardware using some (kind of ancient) ports, such as parallel and serial ports. This ports are controlled by setting some bytes, and each bit of that bytes has a specific meaning, in terms of instructions, for that port (see for instance http://en.wikipedia.org/wiki/Parallel_port). If you want to build software that does something with that hardware you will need bitwise operations to translate the instructions you want to execute to the bytes that the port understand.
For example, if you have some physical buttons connected to the parallel port to control some other device, this is a line of code that you can find in the soft application:
read = ((read ^ 0x80) >> 4) & 0x0f;
Hope this contributes.
I didn't see it mentioned above but you will also see some people use left and right shift for arithmetic operations. A left shift by x is equivalent to multiplying by 2^x (as long as it doesn't overflow) and a right shift is equivalent to dividing by 2^x.
Recently I've seen people using x << 1 and x >> 1 for doubling and halving, although I'm not sure if they are just trying to be clever or if there really is a distinct advantage over the normal operators.
I hope this clarifies those two:
x | 2
0001 //x
0010 //2
0011 //result = 3
x & 1
0001 //x
0001 //1
0001 //result = 1
Think of 0 as false and 1 as true. Then bitwise and(&) and or(|) work just like regular and and or except they do all of the bits in the value at once. Typically you will see them used for flags if you have 30 options that can be set (say as draw styles on a window) you don't want to have to pass in 30 separate boolean values to set or unset each one so you use | to combine options into a single value and then you use & to check if each option is set. This style of flag passing is heavily used by OpenGL. Since each bit is a separate flag you get flag values on powers of two(aka numbers that have only one bit set) 1(2^0) 2(2^1) 4(2^2) 8(2^3) the power of two tells you which bit is set if the flag is on.
Also note 2 = 10 so x|2 is 110(6) not 111(7) If none of the bits overlap(which is true in this case) | acts like addition.
Sets
Sets can be combined using mathematical operations.
The union operator | combines two sets to form a new one containing items in either.
The intersection operator & gets items only in both.
The difference operator - gets items in the first set but not in the second.
The symmetric difference operator ^ gets items in either set, but not both.
Try It Yourself:
first = {1, 2, 3, 4, 5, 6}
second = {4, 5, 6, 7, 8, 9}
print(first | second)
print(first & second)
print(first - second)
print(second - first)
print(first ^ second)
Result:
{1, 2, 3, 4, 5, 6, 7, 8, 9}
{4, 5, 6}
{1, 2, 3}
{8, 9, 7}
{1, 2, 3, 7, 8, 9}
This example will show you the operations for all four 2 bit values:
10 | 12
1010 #decimal 10
1100 #decimal 12
1110 #result = 14
10 & 12
1010 #decimal 10
1100 #decimal 12
1000 #result = 8
Here is one example of usage:
x = raw_input('Enter a number:')
print 'x is %s.' % ('even', 'odd')[x&1]
Another common use-case is manipulating/testing file permissions. See the Python stat module: http://docs.python.org/library/stat.html.
For example, to compare a file's permissions to a desired permission set, you could do something like:
import os
import stat
#Get the actual mode of a file
mode = os.stat('file.txt').st_mode
#File should be a regular file, readable and writable by its owner
#Each permission value has a single 'on' bit. Use bitwise or to combine
#them.
desired_mode = stat.S_IFREG|stat.S_IRUSR|stat.S_IWUSR
#check for exact match:
mode == desired_mode
#check for at least one bit matching:
bool(mode & desired_mode)
#check for at least one bit 'on' in one, and not in the other:
bool(mode ^ desired_mode)
#check that all bits from desired_mode are set in mode, but I don't care about
# other bits.
not bool((mode^desired_mode)&desired_mode)
I cast the results as booleans, because I only care about the truth or falsehood, but it would be a worthwhile exercise to print out the bin() values for each one.
Bit representations of integers are often used in scientific computing to represent arrays of true-false information because a bitwise operation is much faster than iterating through an array of booleans. (Higher level languages may use the idea of a bit array.)
A nice and fairly simple example of this is the general solution to the game of Nim. Take a look at the Python code on the Wikipedia page. It makes heavy use of bitwise exclusive or, ^.
There may be a better way to find where an array element is between two values, but as this example shows, the & works here, whereas and does not.
import numpy as np
a=np.array([1.2, 2.3, 3.4])
np.where((a>2) and (a<3))
#Result: Value Error
np.where((a>2) & (a<3))
#Result: (array([1]),)
i didnt see it mentioned, This example will show you the (-) decimal operation for 2 bit values: A-B (only if A contains B)
this operation is needed when we hold an verb in our program that represent bits. sometimes we need to add bits (like above) and sometimes we need to remove bits (if the verb contains then)
111 #decimal 7
-
100 #decimal 4
--------------
011 #decimal 3
with python:
7 & ~4 = 3 (remove from 7 the bits that represent 4)
001 #decimal 1
-
100 #decimal 4
--------------
001 #decimal 1
with python:
1 & ~4 = 1 (remove from 1 the bits that represent 4 - in this case 1 is not 'contains' 4)..
Whilst manipulating bits of an integer is useful, often for network protocols, which may be specified down to the bit, one can require manipulation of longer byte sequences (which aren't easily converted into one integer). In this case it is useful to employ the bitstring library which allows for bitwise operations on data - e.g. one can import the string 'ABCDEFGHIJKLMNOPQ' as a string or as hex and bit shift it (or perform other bitwise operations):
>>> import bitstring
>>> bitstring.BitArray(bytes='ABCDEFGHIJKLMNOPQ') << 4
BitArray('0x142434445464748494a4b4c4d4e4f50510')
>>> bitstring.BitArray(hex='0x4142434445464748494a4b4c4d4e4f5051') << 4
BitArray('0x142434445464748494a4b4c4d4e4f50510')
the following bitwise operators: &, |, ^, and ~ return values (based on their input) in the same way logic gates affect signals. You could use them to emulate circuits.
To flip bits (i.e. 1's complement/invert) you can do the following:
Since value ExORed with all 1s results into inversion,
for a given bit width you can use ExOR to invert them.
In Binary
a=1010 --> this is 0xA or decimal 10
then
c = 1111 ^ a = 0101 --> this is 0xF or decimal 15
-----------------
In Python
a=10
b=15
c = a ^ b --> 0101
print(bin(c)) # gives '0b101'
You can use bit masking to convert binary to decimal;
int a = 1 << 7;
int c = 55;
for(int i = 0; i < 8; i++){
System.out.print((a & c) >> 7);
c = c << 1;
}
this is for 8 digits you can also do for further more.
I am trying to find the formula to calculate the maximum bit-width required to contain a sum of M n-bit unsigned binary numbers. Thanks!
The maximum bit-width needed should be ceil(log_2(M * (2^n - 1))).
Edit: Thanks to #MBurnham I realize now that it should be floor(log_2(M * (2^n - 1))) + 1 instead.
Assuming positive integers, you need floor(log2(x)) + 1 bits to store x. and the largest value the sum of m n-bit numbers can produce would be m * 2^n.
So I believe the formula should be
floor(log2(m * 2^n)) + 1
bits.
If I add 2 numbers the I need 1 bit more than the wider of the 2 numbers to store the result. So, if I add 2 n-bit numbers, I need n+1 bits to store the result.
if I add another n-bit number, I need (n+1)+1 bits to store the result (that's 3 n-bit numbers added so far)
if I add another n-bit number, I need ((n+1)+1)+1 bits to store the result (that's 4 n-bit numbers added so far)
if I add another n-bit number, I need (((n+1)+1)+1)+1 bits to store the result (that's 5 n-bit numbers added so far)
So, I think your formula is
n + M - 1
I am doing a divider circuit in verilog and using the non-restoring division algorithm.
I am having trouble representing the remainder as a fractional binary number.
For example if I do 0111/0011 (7/3) I get the quotient as 0010 and remainder as 0001 which is correct but I want to represent it as 0010.0101.
Can Someone help ??
Suppose, as in your example, you are dividing 4 bit numbers, but want an extra 4 bits of fractional precision in the result.
One approach is to simply multiply the numerator by 2^4 before doing the division.
i.e.
instead of
0111/0011 = 0010 (+remainder)
do
01110000/0011 = 00100101 (+remainder)
hi just do mathematics !!!
you have already got the Q(quotient) and R(remainder) , now with the remainder you multiply that with 10(decimal) in binary 1010 that for example
7/3 gives 2 as Q and 1 as remainder than just multiply this 1 with 10 then again apply your logic which gives 10/3 gives 3 as Q so your answer will be
3(Q(first_division)).3(second division-Q)
try it it is working . and very easy to implement in verilog .