Having bit type field means that you need to use raw values as a workaround whenever you are inserting/updating that field. That's because PDO by default will bind these values and they will be treated as strings, thus bit will result in 1: DB::table('table')->insert(['bit_field' => 0]); // inserts 1 DB::table('table')->insert(['bit_field'...

If I understand correctly, you're asking how many bits/bytes are used to represent a given number or character. I'll try to cover the common cases: Integer (whole number) values Since most systems use 8-bits per byte, integer numbers are usually represented as a multiple of 8 bits: 8 bits (1...

c,bit,compiler-optimization,bit-shift

That code is very complicated just to check a bit in an integer. Try the standard method: return (t & (1U << n)) != 0; If you have to check n is valid, add an assertion. else masking (n & 7) or modulus (n % 8) (this will be optimized...

c,bit-manipulation,standards,bit

In general, it's not that hard to accommodate unusual platforms for the most cases (if you don't want to simply assume 8-bit char, 2's complement, no padding, no trap, and truncating unsigned-to-signed conversion), the standard mostly gives enough guarantees (a few macros to inspect certain implementation details would be helpful,...

A TINYINT will always have a size of one (1) byte. And accept values between -128 and 127 (if signed). The number you put in the brackets is for display purposes. A BIT(1) on the other hand only take one bit in storage but needs to be aligned to whole...

Variant with array_reduce $s = '0 0 0 1 1 1 1 0 0 1 1 0 0 0 1 0'; $array = explode(' ', $s); $output = ''; // to make count from 1 but 0 array_unshift($array, 0); // keys with value 1 $keys = array_keys($array, 1); $c =...

There is no such helper in redigo. Here is my implementation: func hasBit(n byte, pos uint) bool { val := n & (1 << pos) return (val > 0) } func getBitSet(redisResponse []byte) []bool { bitset := make([]bool, len(redisResponse)*8) for i := range redisResponse { for j:=7; j>=0; j-- {...

Just use : p = p & ~(1u<<3); What happens here ? 1. (1u<<3) 0...01000 2. ~(1u<<3) 1...10111 // Invert the bits 3. p & ~(1u<<3) *****0*** // Here * means the bit representation of p That's how the bit changes to 0. Hope it helps :)...

Simply right shift: (Warning, fails when you want 0 bits, but yours fails for all bits) unsigned dropbits = CHAR_BIT*sizeof(int)-j; //if you want the high bits moved to low bit position, use this: ullong res = (ullong)myInteger >> dropbits; //if you want the high bits in the origonal position, use...

binary to integer integer = parseInt(binary,2); integer to binary binary = integer.toString(2); ...

The best way to do this is: var number = prompt("Enter a number to convert: "); var bitPattern = parseInt(number).toString(2); document.write(bitPattern); ...

else{ value <<= 1;//<-- Value has been changed before it can be used reverseBits( value , c - 1 ); putchar( value & mask ? '1' : '0' ); if( c % 8 == 0 ){ putchar( ' ' ); }//end if }//end else replace with else { reverseBits(value...

The fastest (not sure if that's what you menat by "effective") way of doing this is probably something like void char2bits1(unsigned char c, unsigned char * bits) { int i; for(i=sizeof(unsigned char)*8; i; c>>=1) bits[--i] = c&1; } The function takes the char to convert as the first argument and...

c++,visual-studio-2012,bit,undefined-behavior

Both code-snippets induce undefined behavior if int has 32 bit or less. [expr.shift]/1: The behavior is undefined if the right operand is negative, or greater than or equal to the length in bits of the promoted left operand. Hence an implementation is not in any way obliged to provide consistent...

Just mask and shift. You've already got a comment describing the format: // Message identifiers are constructed: // Unit Type Bit 8-10 // Board ID Bit 5-7 // Unit Specific Message Ids Bit 0-4 So the reversal code is: int messageId = identifier & 0x1f; int boardId = (identifier >>...

These days, the number of bits per character is not so simple but there are most definitely 8 bits per byte. The number of bits/bytes per character will vary depending on your system and the character set. Original ASCII was 7bits per character. That was extended to 8bits per character...

Integers can directly be used to index arrays; e.g. so you have just a simple array of unsigned 8bit integers containing the set-bit-count for 0x0001, 0x0002, 0x0003... and do a look up by array[number_to_test]. You don't need to implement a hash function to map an 16 bit integer to something...

To test whether all the bits in a mask are on, use: WHERE abilities & b'000000101' = b'000000101' To test whether any of the bits are on, use: WHERE abilities & b'000000101' != 0 To turn on all the bits in the mask: SET abilities = abilities | b'000000101' To...

python,bit-manipulation,bit,pic

Your shifts are wrong. You are shifting by the index of the top-most bit, which isn't right. You must shift by the index of the lowermost (rightmost) bit in each field. So it should be: def fileRegOp(opcode, d, f): return (opcode << 8) | (d << 7) | f This...

sql-server,stored-procedures,where,bit

Try this.. assign @sex with 0,1 or null CREATE TABLE #TEMP(ID INT, SEX BIT) INSERT INTO #TEMP VALUES(1,1) INSERT INTO #TEMP VALUES(2,1) INSERT INTO #TEMP VALUES(3,0) INSERT INTO #TEMP VALUES(4,0) INSERT INTO #TEMP VALUES(5,NULL) DECLARE @sex BIT SET @sex=NULL SELECT * FROM #TEMP WHERE @sex IS NULL or [email protected] DROP...

Looks like a trivial 16 bit rotation, the only complication is that it needs to reverse direction at both ends. Here is what I quickly put together: rcall init loop: rcall next rjmp loop init: ldi r16, 1 ; output ldi r17, 0xc0 ; buffer ldi r18, 0 ; direction...

java,bit-manipulation,bit,bit-shift

Can be achieved using "masking" creating a mask for the position i to j with each bit set to 1 using bitwise OR incrementally blank out the bits in N using bitwise AND and bitwise NOT of the mask select the bits from M using mask with bitwise AND copy...

Read http://www.geeksforgeeks.org/turn-off-the-rightmost-set-bit/ /* unsets the rightmost set bit of n and returns the result */ int fun(unsigned int n) { return n&(n-1); } ...

I solved it with with selvec select ERR <= '0' when "0001", '0' when "0010", '0' when "0100", '0' when "1000", '0' when "0000", '1' when others; Not the neatest code but it does the trick....

Shift the value x bits to the right and then use AND to restrict the number of bits you use. ie: (n >> 8) & 0xff or (n >> 16) & 0xff.

Everything on your computer is represented as a sequence a bits, what you are calling a “bitmap”. So the answer to your question is yes, characters have a binary representation, along with integers, floating-point numbers, machine instructions etc. Different languages use different binary encodings for characters, Objective-C uses Unicode. See...

Basically all you have to do is: shift everything right by n bits using right shift: >> shift the bits you want to rotate all the way to the left: << Combine the shifted right and shifted left bits with or: | See this code for an example implementation using...

python,shell,bit,bitwise-operators

You haven't told Python that you're entering binary numbers; like C, Python will interpret 0010000000000000 as octal (8**13), and 1011110110011001 as decimal. Use the 0b prefix and bin function instead: >>> a = 0b1011110110011001 >>> b = 0b1100001110000101 >>> bin(a|b) '0b1111111110011101' Your 8-bit (Python doesn't know that either, int is...

c++,int,bit-manipulation,bit,bitmask

Here is one way, with a lookup table: static const unsigned short MortonTable256[256] = { 0x0000, 0x0001, 0x0004, 0x0005, 0x0010, 0x0011, 0x0014, 0x0015, 0x0040, 0x0041, 0x0044, 0x0045, 0x0050, 0x0051, 0x0054, 0x0055, 0x0100, 0x0101, 0x0104, 0x0105, 0x0110, 0x0111, 0x0114, 0x0115, 0x0140, 0x0141, 0x0144, 0x0145, 0x0150, 0x0151, 0x0154, 0x0155, 0x0400, 0x0401, 0x0404,...

bgtz doesn't place the next PC address into the return address register, so jr $ra won't return to the instruction after the branching statement. You can either use bgtzal (branch if greater than zero and link), which will give you the behaviour you are looking for, or else you can...

The >> operator is a "sign-extended" right-shift operator, which means that if the top bit in the original value is set, the bits that are "shifted in" are 1. That keeps the sign of the result the same, basically. You want the >>> operator (the unsigned right shift operator), which...

If I understand correctly, what you are really saying is that the method should return true if and only if each bit in the wordToMask value is set in the corresponding location where the same bit is set in the mask, with the qualification that the mask value should be...

There are several different ways you can count the number of bits in a value stored in the LC3. You can use bit shifting and count the bits that "fall off" at the ends You can use a bit mask to check each of the bits in the value And...

c++,boost,standards,bit,digits

From this std::numeric_limits::digits reference: The value of std::numeric_limits::digits is the number of digits in base-radix that can be represented by the type T without change. For integer types, this is the number of bits not counting the sign bit. And later it states that for char the result is CHAR_BIT...

c++,optimization,bit,bitwise-operators

The quick answer to your question is: there is no way to do this efficiently. The long answer is actually a series of new questions: what do you really intend to achieve with this? What do you even mean exactly by shifting one bit at the beginning of a file?...

I think you need to use dec2bin() Like this example dec2bin(23) Answer : 10111 EDIT » dec2bin(0:7) - '0' ans = 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Also you can check this...

c,floating-point,add,bit,addition

Code is doing signed int shifts and certainly unsigned long long is desired. // unsigned long long x = (1 << (exp2 - exp1 + MANBITS)); unsigned long long x = (1LLU << (exp2 - exp1 + MANBITS)); Notes: Suggest more meaningful variable names like x_mantissa. Rounding not implemented. Rounding...

given the description of the signum (=7) the code should look similar to : void addSet(unsigned char *signalMask, int sigNum) { // turn bit ON *signalMask = *signalMask | (1<<sigNum); } // end function: addSet void selSet(unsigned char *signalMask, int sigNum) { // turn bit OFF *signalMask = *signalMask &...

c,function,bit-manipulation,bit

A bit shift with negative right operand is undefined behavior as per C99 §6.5.7 ¶3. That means that the compiler is free to emit code that may not work reliably in case you have a negative shift (or, as per the usual rules about undefined behavior, it may as well...

For your example method, Modify your BinaryDigit enum to use a constructor and to contain some member variable and get that variable when you loop through your list of BinaryDigits. Check the reference on Java enums here for an example, particularly the Planet enum: Enum Types....

rgb2gray is probably the safest way to convert an M-by-N-by-3 image into M-by-N.

The above code exhibits UB (undefined behaviour), so it should not be trusted to work on any platform. This is because it writes to a member of a union and reads back from a member different from that it last used to write the union with. It also depends heavily...

r=(v<0)?-(unsigned)v : v There are some ways to compute abs without branching but this one is not one of them as < and ?: operators are usually translated to branching instructions. Assuming r is of type unsigned int, the cast to unsigned is here to get a specified behavior...

The second quote is correct, the size of a word varies from computer to computer. The ARM NEON architecture is an example of an architecture with 32-bit words, where 64-bit quantities are referred to as "doublewords" and 128-bit quantities are referred to as "quadwords": A NEON operand can be a...

java,sockets,networking,bit,bitwise-operators

You can cast it to byte : byte toSend = (byte)0xFF; Forget the other part of my answer. I'm a bit sleep deprived. Actually it's -1 that is represented as 11111111. But (byte)0xFF works just fine....

Do you mean something like this: public int getInt(int input){ return (1<<input)-1; } Lets say input=5 then 1<<input (1<<5) is binary 100000 Then it needs to decrease only to 11111 its 31 in decimal...

Look at the folowing two snippets, did you find the problem: int test = -2147483647; String converted = Integer.toBinaryString(test); System.out.println(converted); int minusOne = -1; String converted1 = Integer.toBinaryString(test); System.out.println(converted1); You are printing out the same variable test, that's why the output is the same. If you printout "minusOne" it would...

What you are experiencing is integer overflow. The maximum number that a signed integer can represent is 0x7FFFFFFF, since in two's complement 0x80000000 = -2147483648. If the most significant bit is set in a two's complement integer that means the number is negative, with 0x80000000 = -2147483648 and 0xFFFFFFFF =...

python,bytearray,bit,bytestring,bitarray

Use bit shifting and ordinary chr: byte_message = chr(0xaa) + chr((bits6_4 << 4) | bits3_0) ...

To start with, 61 is only equal to 00110110 00110001 if you store it as two ASCII digits. When written in binary, 61 is 111101. Note that the binary representation requires six binary digits, or six "bits" for short. That's what we're taking advantage of in this line: Color =...

This seems to be the best way #include <stdio.h> int bits2Integer(char bitString[]){ int ret = 0; for(int i = 0; i < 32; i++) if(bitString[i] == '1') ret |= 1 << (31-i); return ret; } Bitwise operations rule the world :) Remember that you can't have 33 bits. ...

It is done by bit masking. If you want to check whether or not an i'th bit of a is 1 you will do something like this: if (a & (1 << i)) { // Do something } This way all of the bits of a except the i'th one...

The & operator performs a bitwise AND operation—i.e. it operates independently on each bit, and returns the resulting bits in corresponding locations. So it would, naturally, return 64 if bit 6 was the only location in which both its arguments had a 1. You may be confusing its behaviour with...

c++,bit-manipulation,bit,bitwise-operators

From Wikipedia: The bitwise AND operator is a single ampersand: &. It is just a representation of AND which does its work on the bits of the operands rather than the truth value of the operands. Bitwise binary AND does the logical AND (as shown in the table above) of...

With this you an convert directly from AData to array of boolean: for i := 0 to 39 do bits[i] := (AData[i div 8] and (1 shl (i mod 8))) <> 0; I'm not sure if the order of the booleans is exactly the same, I lost count with all...

python,binary,bit,hamming-distance

Implement it: def hamming2(s1, s2): """Calculate the Hamming distance between two bit strings""" assert len(s1) == len(s2) return sum(c1 != c2 for c1, c2 in zip(s1, s2)) And test it: assert hamming2("1010", "1111") == 2 assert hamming2("1111", "0000") == 4 assert hamming2("1111", "1111") == 0 ...

Here is a proposal: void My_Function () { Bit = (Flag == 0)?0:1; switch(State) { case COUNT: if( GetTime() - time) <= 10 ) { Transmitte(Bit); break; } else { State = BITDETECTION; /* fall through to next case */ } case BITDETECTION: if(Bit != 0) { State = COUNT;...

The constructor for BitArray(bool[]) accepts the values in index order - and then CopyTo uses them in the traditional significance (so bitArray[0] is the least significant bit) - so your true, true, false, false ends up meaning 0011 in binary, not 1100. It's not ignoring the last two bits -...

1) Ascii has 128 value, but they are enumerated from 0-127 like most computer arrays. 0 means null. 2) Either the ascii is fit into 8 bit, or you are using the extended 8 bit version. 3) Define your own program that writes to a bytestream, then you can check...

Why is the key-length 26? C#-strings is in unicode, so you can write all characters out there eg. Chinese, Japanese, etc. etc. That takes two bytes. 13*2=26. Is your wep-key 128 bits You've got a key for 128 bit wep-protocoll which is using 104 bit keys. (fun times) Wiki on...

bit-manipulation,bit,computer-architecture

Think of it this way - a bit can only have 2 possible values i.e. 0 or 1. If you write down the binary representations, you will end up with something like the below: abcd efgh & 0011 1111 ----------- = 0010 0111 Going by the definition of bitwise AND,...

fitsBits(0x80000000, 0x20); This function returns 1, because the first argument of your function is int, which is (in practice these days) a 32 bit signed integer. The largest value that signed 32 bit integer can represent is 0x7FFFFFFF, which is less than the value you are passing in. Because of...

Try this code. And operation is used to extract the bits that are same. I don't know C# so I can't type the exact code. The logic is take a random bit mask and AND the bits of c with that bitmask and inverse of that bitmask. The resulting two...

Here you go. Convert your hex string to an int value using the built in parseInt function, then turn that into a binary string. public String hexToBinary(String hexBits) { int intversion = Integer.parseInt(hexBits, 16); String binaryVers = Integer.toBinaryString(intversion); return binaryVers; } Note that this is not padded. If you want...

You can check the entire byte to be zero by applying a bit mask (i.e. a number containing ones in positions that you wish to check and zeros in all other positions*) to it. In your case, you need a bit mask with eight consecutive ones in zero-based positions 7...

python,numpy,bit-manipulation,bit,bitstring

I'm not sure if you want to count the number of "1" bits or to check wether a specific bit is set. To check, I guess the easier way is: bool(n&(1<<b)), where n is the number being tested and b is the bit (starting from 0). To count the number...

performance,comparison,bit,low-level,machine-code

Hardware is fundamentally different from the dominant programming paradigms in that all logic gates (or circuits in general) always do their work independently, in parallel, at all times. There is no such thing as "do this, then do that", only "do this here, feed the result into the circuit over...

I'd do the following: [Replace randInt(n+1) with whatever function returns a random integer in the interval [0,n] inclusive. Define a variable mask with a bit count of exactly n: unsigned int mask = (1<<n)-1;. This is O(1) Perform a slightly modified version of the Fisher-Yates Shuffle; Loop int i =...

Can be done with the bitwise XOR operator, which is ^ in Python. Example: a = 0xF0101010 b = 0xFFFFFFFF print(bin(a)) print(bin(b)) print(bin(a ^ b)) 0b11110000000100000001000000010000 0b11111111111111111111111111111111 0b1111111011111110111111101111 ...

string,go,bit,bitwise-operators

You could create the resulting bitwise OR string by doing character comparisons, or you can perform arbitrary large numeric operations using math/big. Here is an example of such an operation: package main import "fmt" import "math/big" func main() { num1 := "11100" num2 := "00011" var bigNum1 big.Int var bigNum2...

arrays,swift,type-conversion,byte,bit

A possible solution is to enumerate all bits in the array and for all "One" bits set the corresponding bit in the UInt8 array: func bitsToBytes(bits: [Bit]) -> [UInt8] { let numBits = bits.count let numBytes = (numBits + 7)/8 var bytes = [UInt8](count : numBytes, repeatedValue : 0) for...

1) Yes. 2) The computer can't "see" a negative number. It can see a chunk of binary data. Your application is what possesses the intelligence to say "this chunk of binary data is an integer stored in 2's compl". However, almost every CPU in the world has support for 2's...

There are many ways, depending on what operations you can use. It appears you don't have much to choose from. But this should work, using just integer division and multiplication, and a test for equality. (psuedocode): x = 9348 (binary 0010000100100100, bit 0 = 0, bit 1 = 0, bit...

Use the ~ bitwise complement operator. byte flipped = (byte) ~original; ...

There are three basic approaches for extracting encoded information from a bitfield. The first two are related and differ only in the manner the bitfield struct is initialized. The first and shortest is to simply create a bitfield struct defining the bits associated with each member of the struct. The...

java,integer,boolean,long-integer,bit

Use bitwise operators : public static void setUnlocked(int id) { unlocked |= 1L<<id; // set the id'th bit } public static boolean isUnlocked(int id) { return (unlocked & (1L<<id)) != 0; // get the id'th bit } 1<<id is the fastest way to compute 2^id. bitwise OR (|) lets you...

c,byte,bit-manipulation,bit,nibble

Silly restrictions. Who broke your compiler? Something like this should do it: bool sig_nib(int x, int y) { const int top4 = (x >> 28) & 0xf; return !(top4 ^ y); } It uses Boolean inversion (!) to re-interpret the integer result from the bitwise-xor....

What about one-liner? var mask = (Amenities)5722635; var result = Enum.GetValues(typeof(Amenities)) .Cast<Amenities>() .Where(value => mask.HasFlag(value)) .ToList(); You can cache result of Enum.GetValues(typeof(Amenities)).Cast<Amenities>() in order to improve performance....

It depends on what is your 32 bits register meaning. If this is byte address, it can handle 4G bytes address space. If it is index of 512 bytes sectors it can handle 2Tera bytes of storage space. And finaly if it is a bit index only 512M bytes (4G...

The code assumes a 32 bit integer and 8 bit bytes. A 32 bit integer is made up of 4 bytes: Let's say these 4 bytes are laid out in memory like so: +---------------------------------+ |Byte 4 | Byte 3 | Byte 2 | Byte 1| +---------------------------------+ This could relate to...

That is how the two's complement works. To negate a number it is inverting all the bits and adding 1 to the result. Example: negating 11110001 ----------> 00001111...

In your code n |= (n & 0X0000) | 0x6000; is wrong beacuse of is equal to 0xABCDE98 & 0x0000 = 0 and 0x0000 | 0x6000 = 0x6000 and 0xABCDE98 | 0x6000 = 0xABCFDE98 Instead you must write n = (n & 0XFFF0FFF) | 0x6000; ...

bit-manipulation,bit,bitwise-operators,hamming-distance

You can simply do this : int need=__builtin_popcountll(A^B); cout<<need; ...

c#,concatenation,bit,bit-shift,uint16

I already tried to (logically) bit-rightshift r[0] by 8, but then the upper bits get lost because they are stored in the first 8 bits of r[1]. Well they're not "lost" - they're just in r[1]. It may be simplest to break it down step by step: byte val1LowBits...

datetime,sql-server-2012,bit,data-type-conversion,sqldatatypes

you get this error because DF____Person__onvac__59062A42 sql object Depends on onvacation column. You can Find Dependency of Person table by Right Click-->View Dependancy remove that dependent object and try to alter column...

You have n possible problems, and each one can either be included or excluded from the problem set. This means that there are 2 options for every problem, so with n problems there are 2^n options for creating possible problem sets. With the line for(i=0; i<(1<<n); i++) you are iterating...

Just shift each one into place, and OR them together: #include <stdint.h> int main(void) { uint8_t a = 0xAB, b = 0xCD, c = 0xEF; /* * 'a' must be first cast to uint32_t because of the implicit conversion * to int, which is only guaranteed to be at least...

This equation can be solved by making a simple observation about + and | on a single bit value: When both values are 0, both operations produce 0, When the values are 1 and 0 or 0 and 1, both operations produce 1, When both values are 1, the results...

Um... when you initialize an unsigned int object withsigned int value, there's definitely a need to convert unsigned int to signed int. This is exactly what you asked the language to do by mixing these two different types in an initialization. When you initialize an object of type T with...

If you want to check against one bitmask, then if ((value & mask) == mask) will give you an exact match ("all bits in the mask"), and if ((value & mask) != 0) will supply a loose match ("any bit in the mask"). The compiler will further optimize the check...

Ideally you would be encoding this as binary. 11001100 is a single byte 0xCC. Storing as varchar means it takes 8 bytes and declared as nvarchar it takes 16 bytes. You could then also use CLR and bitwise operators. Answering the question you asked though using a CLR function would...

c++11,int,bit-manipulation,bit

There are two reasonable solutions: either set all don't-care bits to zero, and then test the result, or set the don't-care bits to one and test the result: (x & 0xC0) == 0x80 (x | ~0xC0) == ~0x40 As harold pointed out in the comment, the first form is far...