Bitwise operation in C# 
Conditionals in C/C++ effectively compile down to comparisons against zero.
So the expression LdcnGetStat(pico_addr) & MOTOR_MOVING is equivalent
to (LdcnGetStat(pico_addr) & MOTOR_MOVING) != 0. In C# however,
conditionals are done with actual bools, so you can only use the second
expression.
I can explain more if you need me to.

C# Bitwise Operands Vs PHP 
C# uses zerobased arrays but PHP's unpack(..) uses a 1based array.
You could either change
$strHex = unpack("C*",$strReq);
to
$strHex = array_merge(unpack("C*",$strReq));
to make your PHP array zerobased, or simply set $k to an initial value of
1.
After testing either solution the PHP results are the same as the C#
output.

C bitwise operator '&' uses 
Well ... 5 % 5 is 0 (false), so the ?: part goes on to evaluate the
expression to the right of the colon.
That means evaluating 0 bitwiseand:ed with the return value of puts(), so
obviously the function must be called.

SQL Bitwise Masking 
Check whether the result of "privileges" & 3 is actually equal to 3:
SELECT * FROM "accounts" WHERE ("privileges" & 3) == 3;
Otherwise, the query will select records where at least one bit is set.

bitwise OR (on array) 
I think your best bet is to use a BitSet.
That class already has a void or(BitSet bs) method to use.
byte a = new byte[256];
byte b = new byte[256];
byte c = new byte[256];
BitSet bsa = new BitSet();
BitSet bsa = new BitSet();
//fill BitSets with values from your biteArrays
for(int i = 0; i < a.length * 8; i++)
if((a[i/8] & (1 << 7i%8)) != 0)
bsa.set(i);
for(int i = 0; i < a.length * 8; i++)
if((b[i/8] & (1 << 7i%8)) != 0)
bsb.set(i);
//perform OR
bsa.or(bsb);
//write bsa to byteArray c
for(int i = 0, byte h; i < a.length; i++){
h = 0;
for(int j = 7; j >= 0; j++){
if(bsa.get(i*8 + 7  j))
h = h  (1 << j);
}
c[i] = h;
}

C Programming  XOR Bitwise Operation 
It is (A) toggles bits 2 and 5.
The following is the truth table for the XOR operation:
x y x^y
0 0 0
1 0 1
0 1 1
1 1 0
You can see from the table that x XOR 0 = x and x XOR 1 = !x.
XOR is a bitwise operation, so it operates on individual bits. Therefore if
you XOR star with some constant, it will toggle the 1 bits in the constant.
You can find some explanation e.g. here.
Wow, I just found out there is also a fiddle for C  try it yourself!

Bitwise AND function in XACML 
First things first:
In XACML, there is an AND function. The AND function takes two boolean
expressions e.g. AND(isValid, isOpen) or AND(role=="manager",
userGroup==documentGroup). This function is supported in most XACML
implementations  I work for Axiomatics which provides a XACML 3.0 engine
and we definitely support it.
As for custom functions:
XACML is called eXtensible precisely because you can extend it with new
datatypes, functions, combining algorithms, and more. You can therefore
implement your own custom function. The XACML engine you use should be able
to let you load those custom functions. Again, we do.
One group that's been defining several custom datatypes and functions is
GeoXACML. They define geographical datatypes e.g. area. You can check out
what they do here.

How to use Bitwise for security groups 
You AND with the applicable number and check if it's greater than 0, i.e.
10 & 2 > 0, thus you have write permission.

Bitwise operator in js is inconsistent? 
9.9999999999999999 has too many decimals and loses precision in Javascript
representation, becoming 10. You can test this:
9.9999999999999999 === 10 will be true

Optimising bitwise operations in C 
To make an n bit mask:
mask_y = (1U << n)  1;
To start it at bit p:
mask_x = mask_y << p;
Clear the appropriate bits in x:
x &= ~mask_x;
Extract the bits from y:
y &= mask_y;
Upshift them to position p:
y <<= p;
Put it all together:
result = x  y;
Or in a more compact form:
mask = (1U << n)  1;
result = x & ~(mask << p);
result = (y & mask) << p;

What is the point of doing bitwise shifting here? 
The routine is doing a combination of reading in character values and
converting them to an integer value in the same operation (to be fast,
apparently). The shifts are aiding in the summation of the values. The
combined shifts equal x10.

How Does The Bitwise & (AND) Work In Java? 
It's a binary AND operator. It performs an AND operation that is a part of
Boolean Logic which is commonly used on binary numbers in computing.
For example:
0 & 0 = 0
0 & 1 = 0
1 & 0 = 0
1 & 1 = 1
You can also perform this on multiplebit numbers:
01 & 00 = 00
11 & 00 = 00
11 & 01 = 01
1111 & 0101 = 0101
11111111 & 01101101 = 01101101
...

bitwise division by multiples of 2 
Use the operator / for integer division as much as you can.
For instance, when you want to divide 100 by 6 or 10 you should write 100/6
or 100/10.
When you mention bit wise division do you (1) mean an implementation of
operator / or (2) you are referring to the division by a power of two
number.
For (1) a processor should have an integer division unit. If not the
compiler should provide a good implementation.
For (2) you can use 100>>2 instead of 100/4. If the numerator is
known at compile time then a good compiler should automatically use the
shift instruction.

How to use bitwise operators in Python? 
You are looking for the bitwise operators,
>>> 2 & 3
2
>>> 2  3
3
By just doing 2 and 3 you are evaluating 2, which is True, then 3 (also
True) and Python returns that second number. So you get 3.
With 2 or 3, it shortcircuits and just returns 2 since 2 is True.

How to do a bitwise operation on some data in C? 
If you want to print the bytes of raw binary data:
void print_bytes(const void *data, size_t len)
{
const unsigned char *p = data;
for (size_t i = 0; i < len; i++)
printf("%d ", p[i]);
}
You can call it with the address of any object, like this:
unsigned long long u = 1234567890;
print_bytes(&u, sizeof u); // beware of endianness!
const char *s = "hello world";
print_bytes(s, strlen(s));

Confusing PHP bitwise NOT behavior 
Your output is defaulting to a signed int  wrap it in decbin to get a
binary representation.
Consider:
$number = 9;
var_dump( bindec(decbin(~ $number)) );
With two's compliment, the MSB of a signed binary number becomes 0MSB, but
every other bit retains its respective positive values.
So for argument's sake (an 8bit example),
Binary 9: 0000 1001
Inverse: 1111 0110
This results in (128) + 64 + 32 + 16 + 4 + 2 = 10, so PHP is calculating
correctly, its just applying two's compliment to the MSB.

Bitwise wise operations 
You are using ObjectOutputStream, which is intended for portable
serialization of Java objects. If you want to write the single bytes you
should be using a FileOutputStream instead.

Bitwise operations between 128bit integers 
Yes, SSE2 has a 128 bit bitwise AND  you can use it via intrinsics in C or
C++, e.g.
#include "emmintrin.h" // SSE2 intrinsics
__m128i v0, v1, v2; // 128 bit variables
v2 = _mm_and_si128(v0, v1); // bitwise AND
or you can use it directly in assembler  the instruction is PAND.
You can even do a 256 bit AND on Haswell and later CPUs which have AVX2:
#include "immintrin.h" // AVX2 intrinsics
__m256i v0, v1, v2; // 256 bit variables
v2 = _mm256_and_si256(v0, v1); // bitwise AND
The corresponding instruction in this case is VPAND.

How do I perform a bitwise AND of two BIGNUMs? 
It looks like there is no function to do this directly, hence you'll have
to come up with something based on the functionality that is there.
Something like:
BIGNUM *a, *b, *result;
unsigned current = 0;
//Creation of a, b, result
while(!BN_zero(a) && !BN_zero(b)) {
if(BN_is_bit_set(a, current) && BN_is_bit_set(b, current)) {
BN_set_bit(result, current);
} else {
BN_clear_bit(result, current);
}
++current;
BN_rshift1(a, a);
BN_rshift1(b, b);
}
Note that this may need to manually set the higher order bits to 0 if the
bitlength of a is greater than b or viceversa. It should be enough to get
your started, however.

Bitwise operators android 
The event that comes back from event.getAction() is actually an integer.
It contains more than just information about the type of action apparently.
It probably contains flags that describe the action in more detail. And
by passing all that detail in through an int, it saves the use of a class
object to represent the action itself.
The mask is bitwise for the first byte in the integer. Only the first two
bytes out of the eight. And for comparison purposes, you need to remove
the upper parts of the integer before comparing it against the actions,
which are themselves only using the bottom byte of an integer.
The upper parts of the event integer are still useful for other purposes
when checking against them. So the mask is 0x000000ff representing that
you only want information relate

Bitwise & operator returns 255 
You should not expect a bitwise and to be 0 or 1.
Here's how something could be 255:
a = 255 & 255 // then a = 255;
Here's a couple examples
Example 1 11111111 & 11111111 = 11111111 or 255
Example 2: 01010101 & 00000111 = 101 or 5.
http://en.wikipedia.org/wiki/Bitwise_operations_in_C

Return type of bitwise operation in C 
The C99 language specification says that "usual arithmetic conversions
apply" when doing a bitwise AND, val is already an integer and the hex
constant should be an integer too (see page 56) because it's the first type
which can represent it. Both casting and assigning should truncate the int
into a short, but you shouldn't lose any data in this case.

How to implement bit vectors with bitwise operations? 
It seems how this method works, even though I feel like there are better
ways to set a bit. Is to find the index of the ith bit it essentially
divides by 32 because that is the number of bits per word.
Since the operator used here is  the function is setting the bit to one
not toggling the bit
0x1F is actually 31 and when anded with the i you get the remainder (not
sure why they just didn't use %)
And lastly the shift takes the 1 to the proper location and or's it with
the right slot in the vector.
If you are planning to use this code
you could write it a lot clear without defines and using more obvious
methods of doing it, I doubt it would make a difference in speed.
Also you should probably just use std::bitset
the use of the mask to get the remainder particularly annoyed

C bitwise operations relevant bits 
Mask and combine:
finalByte = (B & C)  (A & ~C);
To break down how it works  the result of B & C is a byte containing
all of the bits of B where bits of C are set (a normal masking operation).
A & ~C yields a byte with all of the bits of A where bits of C are
cleared  hence the ~ complement operation. The  combines the two into
the final byte you're looking for.

vhdl bitwise operation on vector 
There are functions called or_reduce and and_reduce which do what you want.
They also call to_X01 on the inputs, which means that H and Ls will be
converted to 1 and 0.
They can be found in std_logic_misc for std_logic_vectors and are proposed
for the next rev of VHDL for ieee.numeric_std for signed and unsigned
vectors.
VHDL2008 also allows this syntax on std_logic_vectors:
anded <= and v;
ored <= or v;

Check bitwise flags that cannot coexist with each other 
To set the bit at position n, you want to set value 2^n.
So if you want to check that only one of the flags is specified, then you
just want to ask if the number is a power of two.
And here is a question about how to do that: How to check if a number is a
power of 2
As GrahamS says, you could read the question as saying that exactly one bit
must be set (i.e. it can't be zero). So to do that, in addition, check that
it's nonzero and that it's less than or equal to C.

Scala bitwiselike method argument 
Enumeration defines an inner type called ValueSet which gives you at least
some of the functionality you are looking for. The methods on it are still
Setlike (you would add a new mode using +, and check for a mode using
contains), but it may serve your purposes.
EDIT: had some fun fiddling around and came up with this:
import scala.collection.BitSet
object OpMode extends Enumeration {
protected case class Val(name: String, val mask: Int) extends
super.Val(nextId, name)
type OpMode = Val
val Read = Val("Read", 1)
val Write = Val("Write", 2)
val Create = Val("Create", 4)
val Modify = Val("Modify", 8)
val Delete = Val("Delete", 16)
val Whatever = Val("Whatever", 32)
case class FlagSet(bits: BitSet) {
def isSet(mode: OpMode) = bits.contains(mode.mask)
def +(mo

Bitwise AND behaves differently than expected? 
It is a bitwise &. That means the result of the operation is the result
of applying & bit by bit on the two operands.
int a = 50; // 110010
int b = 30; // 011110
a & b == 010010 == 18 == true
If you want all the bits to be equal, that's just ==. Or you would bitwise
& it with 111111

Multiple Bitwise Flag Combos 
I believe it is possible. Your Enum must have the [Flags] attribute set,
and then it is as simple as .Where(p=>p.Flags == FilterFlags)
Read more about Enum flags here

Bitwise Angular Expression not working 
From the angularjs github buglist:
https://github.com/angular/angular.js/issues/2838
http://docs.angularjs.org/guide/expression
"Angular Expressions vs. JS Expressions
It might be tempting to think of Angular view expressions as JavaScript
expressions, but that is not entirely correct, since Angular does not use a
JavaScript eval() to evaluate expressions."
You can use a filter to achieve the effect as such:
angular.module('project', [])
.filter('bitwiseAnd', function () {
return function (firstNumber, secondNumber) {
return ((parseInt(firstNumber, 10) & parseInt(secondNumber,
10)) === parseInt(secondNumber, 10));
// return firstNumber % secondNumber > 0
};
});

Getting the same result from Ruby as Javascript for bitwise XOR 
Those two are the same result, modulo 232. In Ruby you could &
4294967295 to make the result the same as in Javascript.
To cover all the cases, you need to take into account that Javascript
considers binary values to be signed 32bit integers. Ruby on the other
hand will produce unsigned 32bit integers from the & 4294967295
operation.
So, in Javascript simply:
c = a ^ b
To get the same thing in Ruby:
c = (a ^ b) & 4294967295
c = 4294967296 if c > 2147483647

C# AND assignment operator (&=) with bitwise enums 
You want bitwise OR () rather than bitwise AND (&):
var returnVar = PatientRecord.NoRecord;
....
if (condition...)
{
returnVar = PatientRecord.SameEnrollmentDate;
// ^ Bitwise OR assignment
}
AND'ing with zero (NoRecord=0x0) will always result in zero.

Radix sort using bitwise operations 
First of all, you don't need to convert an integer to bits, because it
already is stored as bits. An int is usually 4 bytes, so 32 bits. You can
access the bits using bit operators.
Radix sort is shown here in detail.
https://en.wikipedia.org/wiki/Radix_sort
This example sorts based on base 10 digits.
To sort based on bit, you would change the algorithm slightly to use 2
instead of 10 in all places:
void radixsort(int *a, int n) {
...
while (m / exp > 0) {
int bucket[2] = { 0 };
for (i = 0; i < n; i++) bucket[a[i] / exp % 2]++;
bucket[1] += bucket[0];
for (i = n  1; i >= 0; i) b[bucket[a[i] / exp % 2]] = a[i];
for (i = 0; i < n; i++) a[i] = b[i];
exp *= 2;
...
}
}
But if you needed to use bit wise operators instead, you could r

Why does this bitwise OR return null on a nullable int? 
From http://msdn.microsoft.com/enus/library/2cf62fcy(v=vs.80).aspx:
The predefined unary and binary operators and any userdefined operators
that exist for value types may also be used by nullable types. These
operators produce a null value if the operands are null; otherwise, the
operator uses the contained value to calculate the result.

C++ Operator precedence for Bitwise AND and Logical OR 
Precedence just means that the expression is written as below
( (a++  (b++ & c++)))
Once you do that, short circuiting means that just the first expression is
evaluated.
This is why a = 2 but b and c are unchanged.
codepad

Compare two characters using Bitwise operators 
Equality is easy to check for with xor.
Assuming the case is significant, the upper and lower case characters
differ at bit 6 so masking with 0x20 checks the case so:
int Lcmp(char unsigned first,char unsigned sec)
{
unsigned char diff = first^sec;
if (!diff)
{
// bits the same
return 0;
}
if (diff & 0x20)
{
// case differs
if (first & 0x20)
{
// sec is capital
return 1;
}
// first is capital
return 1;
}
// same case  find highest different bit
char unsigned mask=0x80;
while (!(mask & diff))
{
mas

python library for bitwise storage? 
There is some discussion and some source code for this kind of thing at
http://codereview.stackexchange.com/questions/23187/bitwiseflagcodeforpython
It allows you to set flags like:
# define your flags
class sec(FlagType):
admin = 1
read = 2
write = 4
usage = 8
flags = +sec.read sec.write +sec.usage
flags.read
>>> True
Lots of feedback saying its not Pythonic though :)

Applications of bitwise operators in C and their efficiency? 
Example 1
If you have 10 booleans that "work together" you can do simplify your code
a lot.
int B1 = 0x01;
int B2 = 0x02;
int B10 = 0x0A;
int someValue = get_a_value_from_somewhere();
if (someValue & (B1 + B10)) {
// B1 and B10 are set
}
Example 2
Interfacing with hardware. An address on the hardware may need bit level
access to control the interface. e.g. an overflow bit on a buffer or a
status byte that can tell you the status of 8 different things. Using bit
masking you can get down the the actual bit of info you need.
if (register & 0x80) {
// top bit in the byte is set which may have special meaning.
}
This is really just a specialized case of example 1.

Why are my bitwise operations not working with `int`s in Java? 
The expression inside if must explicitly be a boolean (that is, boolean or
Boolean  see JLS §14.9 for further details):
if ((a & c) != 0) {
Also note that the second set of parenthesis are required here since != has
a higher precedence than &.

Bitwise or on a variable amount of numbers C# 
You can construct a combination of "flag" enums from a set of bools as
follows:
Permissions p = ...
var res = (p.AllowCopy ? PdfWriter.ALLOW_COPY : 0)
 (p.AllowFillIn ? PdfWriter.ALLOW_FILL_IN : 0)
 // ...and so on.
