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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.

Categories : C#

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

Categories : C#

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 bitwise-and:ed with the return value of puts(), so obviously the function must be called.

Categories : C

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.

Categories : SQL

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 bite-Arrays for(int i = 0; i < a.length * 8; i++) if((a[i/8] & (1 << 7-i%8)) != 0) bsa.set(i); for(int i = 0; i < a.length * 8; i++) if((b[i/8] & (1 << 7-i%8)) != 0) bsb.set(i); //perform OR bsa.or(bsb); //write bsa to byte-Array 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; }

Categories : Java

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!

Categories : C

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.

Categories : Wso2

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.

Categories : C#

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

Categories : Javascript

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;

Categories : C

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.

Categories : C++

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 multiple-bit numbers: 01 & 00 = 00 11 & 00 = 00 11 & 01 = 01 1111 & 0101 = 0101 11111111 & 01101101 = 01101101 ...

Categories : Java

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.

Categories : C

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 short-circuits and just returns 2 since 2 is True.

Categories : Python

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));

Categories : C

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 0-MSB, but every other bit retains its respective positive values. So for argument's sake (an 8-bit 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.

Categories : PHP

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.

Categories : Java

Bitwise operations between 128-bit 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.

Categories : C

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 vice-versa. It should be enough to get your started, however.

Categories : C++

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

Categories : Android

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

Categories : 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.

Categories : C

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

Categories : C++

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.

Categories : C

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. VHDL-2008 also allows this syntax on std_logic_vectors: anded <= and v; ored <= or v;

Categories : Function

Check bitwise flags that cannot co-exist 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 non-zero and that it's less than or equal to C.

Categories : C#

Scala bitwise-like 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 Set-like (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

Categories : Scala

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

Categories : C

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

Categories : C#

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 }; });

Categories : Angularjs

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 32-bit integers. Ruby on the other hand will produce unsigned 32-bit 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

Categories : Javascript

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.

Categories : C#

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

Categories : C++

Why does this bitwise OR return null on a nullable int?
From http://msdn.microsoft.com/en-us/library/2cf62fcy(v=vs.80).aspx: The predefined unary and binary operators and any user-defined 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.

Categories : C#

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

Categories : C++

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

Categories : C

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/bitwise-flag-code-for-python 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 :)

Categories : Python

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.

Categories : C

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 &.

Categories : Java

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.

Categories : C#



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