bit

问题 Why does this property hold true? say the kth bit from right side is the first set bit in number 'x'. (x-1) will toggle every bit upto kth bit from right side. I can verify this property by writing the bit sequence for the numbers but I don't understand as to why this property is true? Can anyone help me with a simple proof or intuition as to why this works? 回答1: Let's do a hand subtraction xxx100...00 - xxx000...01 ───────────── xxx011...11 x represents bits that we don't care Starting from

问题 Is there any efficient algorithm that allows to insert bit bit to position index when working with uint16_t ? I've tried reading bit-by-bit after index , storing all such bits into array of char , changing bit at index , increasing index , and then looping again, inserting bits from array, but could be there a better way? So I know how to get, set, unset or toggle specific bit, but I suppose there could be better algorithm than processing bit-by-bit. uint16_t bit_insert(uint16_t word, int bit

问题 I have two numbers (going from 0-9) and I want to combine them into 1 byte. Number 1 would take bit 0-3 and Number 2 has bit 4-7. Example : I have number 3 and 4. 3 = 0011 and 4 is 0100. Result should be 0011 0100. How can I make a byte with these binary values? This is what I currently have : public Byte CombinePinDigit(int DigitA, int DigitB) { BitArray Digit1 = new BitArray(Convert.ToByte(DigitA)); BitArray Digit2 = new BitArray(Convert.ToByte(DigitB)); BitArray Combined = new BitArray(8);

问题 How does one go about doing a logical right shift of negative numbers in C? Basically I am looking for the C equivalent of >>> in java i.e. int one = -16711936 ; //int two = -16711936 ; //int three = -1; int r, g, b; System.out.println(Integer.toBinaryString(one)); r = one << 8; r >>>= 24; g = one << 16; g >>>= 24; //this always ends up being -1 in C, instead of 255 b = one << 24; b >>>= 24; 回答1: Cast the value to (unsigned int) before shifting. 回答2: Unlike Java, C has unsigned integer types.

问题 How can I write the actual bits of a vector of digits (for example: 65, 66, 67) into a file and then read them as well? I have previous code as well but It doesn't work properly. Writing above (65, 66, 67) should result in these "100000110000101000011" bits but it outputs "001100010100001000001". Here is the link to previous question containing code. Reading the dynamic bitset written data from file cannot read the correct data How can I build write and read to/from file functions so that the

问题 I'm trying to convert a 5 digit decimal value (ranging from 00001 to 99999) and somehow represent it as a 24-bit value split into 3 bytes but have tried every conversion and bitshift tactic I know, but keep getting stuck :/ Example: decimal value is 12345, and I need to send 3 hex values [aa][bb][cc], which would consist of: [aa] - least significant | [bb] - middle | [cc] - most significant I'm hoping I'm not in over my head and that there is a simple answer, thanks in advance! 回答1: Try

问题 I am trying to eliminate an IF statement whereby if I receive the number 32 I would like a '1', but any other number I would like a '0'. 32 is 0010 0000 so I thought about XOR-ing my input number with 1101 1111. Therefore if I get the number 32 I end up with 1111 1111. Now is there any way of AND-ing the individual bits (1111 1111), because if one of my XOR results is a 0, it means my final AND-ed value is 0, otherwise its a 1? EDIT: Using GCC, not Intel compiler (because I know there are a

问题 This question already has answers here : Bit twiddling: which bit is set? (15 answers) Closed 7 years ago . I am trying to find an optimal code to locate a single bit index in a long integer (64 bit). The long integer has exactly one set bit. (Using C language) Currently, I am just shifting the whole thing by one bit, then checking for zero. I have read about lookup tables, but it won't do for the whole 64 bits. I thought about checking each 8 bits for zero, if not use a lookup, but still I

问题 I am not sure of the precise definition of this term. I know that a bitwise XOR operation is going bit by bit and taking the XOR of corresponding bits position wise. Is this result termed the 'XOR sum'? If not, what is an XOR sum, and how do you use XOR to implement this addition? 回答1: In a bit wise XOR operation: a b a^b ----------- 0 0 0 0 1 1 1 0 1 1 1 0 XOR sum refers to successive XOR operations on integers. Suppose you have numbers from 1 to N and you have to find their XOR sum then for

问题 I am not sure of the precise definition of this term. I know that a bitwise XOR operation is going bit by bit and taking the XOR of corresponding bits position wise. Is this result termed the 'XOR sum'? If not, what is an XOR sum, and how do you use XOR to implement this addition? 回答1: In a bit wise XOR operation: a b a^b ----------- 0 0 0 0 1 1 1 0 1 1 1 0 XOR sum refers to successive XOR operations on integers. Suppose you have numbers from 1 to N and you have to find their XOR sum then for