bit

问题 I'm trying to simply convert a byte received from fget into binary. I know the value of the first byte was 49 based on printing the value. I now need to convert this into its binary value. unsigned char byte = 49;// Read from file unsigned char mask = 1; // Bit mask unsigned char bits[8]; // Extract the bits for (int i = 0; i < 8; i++) { // Mask each bit in the byte and store it bits[i] = byte & (mask << i); } // For debug purposes, lets print the received data for (int i = 0; i < 8; i++) {

问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: Cosmic Rays: what is the probability they will affect a program? Is this just a tongue in cheek expression or is this really true, and if so, what precautions should we take in software (or these precautions hardware only)? 回答1: Well, I did dig up this paper, which claims that your RAM will get bit alterations from "Atmospheric Neutrons" (aka: Cosmic Rays) at a rate of about 1.3*10^-12 /bit/hour. An article by

问题 This question already has answers here : Bitshift and integer promotion? (2 answers) Closed 6 years ago . I read somewhere that bitwise shift automatically turns the operand into an int. But I'm not sure if that statement should be qualified with "if the operands are of unequal type." char one = 1, bitsInType = 8; one << (bitsInType - one); Does the default result of the second line result in an int or char? 回答1: The result type is int in normal C implementations. 1 Per C 2011 (N1570) 6.5.7,

问题 Assuming I have a register reg [15:0] my_reg , which contains a 16-bit signed sample: How can I find the place where the first bit change is located? Meaning, that if assuming that my_reg = 16'b0001011011010111 , how can I know that the first change from 0 to 1 is at my_reg [12] ? Same for numbers starting with 1 ,negative numbers, e.g. my_reg = 16'b1111011011010111 would be interested in the position of the first appearing 0 (which is 11 in this case). The ultimate goal (to add a little bit

问题 Let L and B be two machines. L order its bits from LSB (Least Significant Bit) to MSB (Most Significant Bit) while B order from MSB to LSB. Or, in other words, L uses Little Endian while B uses Big Endian bit - not to be confused with byte - ordering. Problem 1 SOLVED : We are writing the following code which we want to be portable: #include <stdio.h> int main() { unsigned char a = 1; a <<= 1; printf("a = %d\n", (int) a); return 0; } on L , it will print 2, but what happens on B ? Will it

问题 Let L and B be two machines. L order its bits from LSB (Least Significant Bit) to MSB (Most Significant Bit) while B order from MSB to LSB. Or, in other words, L uses Little Endian while B uses Big Endian bit - not to be confused with byte - ordering. Problem 1 SOLVED : We are writing the following code which we want to be portable: #include <stdio.h> int main() { unsigned char a = 1; a <<= 1; printf("a = %d\n", (int) a); return 0; } on L , it will print 2, but what happens on B ? Will it

问题 This question directly follows after reading through Bits counting algorithm (Brian Kernighan) in an integer time complexity . The Java code in question is int count_set_bits(int n) { int count = 0; while(n != 0) { n &= (n-1); count++; } } I want to understand what n &= (n-1) is achieving here ? I have seen a similar kind of construct in another nifty algorithm for detecting whether a number is a power of 2 like: if(n & (n-1) == 0) { System.out.println("The number is a power of 2"); } 回答1:

问题 I think understand what they fundamentally do - operate on bits (flip, shift, invert, etc...). My issue is that I don't know when I'd ever need to use them, and I don't think I fully understand bits. I know that there are 8 bits in a byte and I know that bits are either a 0 or 1 . Now here is where I start to get confused... I believe data types define combinations of bits differently. So if I declare an int , 32 bits are set aside for numbers, if I declare a char, 8 bits are set aside and

问题 I need to define a struct which has data members of size 2 bits and 6 bits. Should I use char type for each member?Or ,in order not to waste a memory,can I use something like :2 \ :6 notation? how can I do that? Can I define a typedef for 2 or 6 bits type? 回答1: You can use something like: typedef struct { unsigned char SixBits:6; unsigned char TwoBits:2; } tEightBits; and then use: tEightBits eight; eight.SixBits = 31; eight.TwoBits = 3; But, to be honest, unless you're having to comply with

问题 my_table contains the enabled field which is defined as: enabled BIT NOT NULL DEFAULT 0 . This table has multiple rows with enabled = b'0' , and multiple rows with enabled = b'1' . However, both this: SELECT * from my_table WHERE enabled = b'0'; and this: SELECT * from my_table WHERE enabled = b'1'; show blank in the enabled column: +----+---------+ | id | enabled | +----+---------+ | 1 | | | 2 | | +----+---------+ Why is that? How could I see the value of the enabled field? $ mysql --version