bit

来源： https://stackoverflow.com/questions/64477750/treating-assembly-registers-as-unsigned-vs-signed

来源： https://stackoverflow.com/questions/54054427/different-between-c-struct-bitfields-on-char-and-on-int

来源： https://stackoverflow.com/questions/54054427/different-between-c-struct-bitfields-on-char-and-on-int

来源： https://stackoverflow.com/questions/54054427/different-between-c-struct-bitfields-on-char-and-on-int

来源： https://stackoverflow.com/questions/52007167/how-to-count-1-bits-in-an-integer-in-javascript

问题 Is there a way to show the representation of an int in bits in c#? i.e. 1 = 00001 20 = 10100 etc. I have tried using BitConverter with no luck. This should be simple, but I can't find a solution! 回答1: Convert.ToString(value, base) Converts the value of a 32-bit signed integer to its equivalent string representation in a specified base. Specify 2 for the base. 回答2: Here's a one-liner using linq: var myint = 20; var bytes = Enumerable.Range(0, 32).Select(b => (myint >> b) & 1); // { 0, 0, 1, 0,

问题 Is there a way to show the representation of an int in bits in c#? i.e. 1 = 00001 20 = 10100 etc. I have tried using BitConverter with no luck. This should be simple, but I can't find a solution! 回答1: Convert.ToString(value, base) Converts the value of a 32-bit signed integer to its equivalent string representation in a specified base. Specify 2 for the base. 回答2: Here's a one-liner using linq: var myint = 20; var bytes = Enumerable.Range(0, 32).Select(b => (myint >> b) & 1); // { 0, 0, 1, 0,

问题 I need to check the value of the least significant bit (LSB) and most significant bit (MSB) of an integer in C/C++. How would I do this? 回答1: //int value; int LSB = value & 1; Alternatively (which is not theoretically portable, but practically it is - see Steve's comment) //int value; int LSB = value % 2; Details: The second formula is simpler. The % operator is the remainder operator. A number's LSB is 1 iff it is an odd number and 0 otherwise. So we check the remainder of dividing with 2.

问题 I got some numbers and I need to get how much they should be shifted for their lower bit to be at position 0. ex: 0x40000000 => 30 because 0x40000000 >> 30 = 1 768 = 512+256 => 8 This works if (Math.log2(x) == 31) return 31; if (Math.log2(x) > 31) x = x & 0x7FFFFFFF; return Math.log2(x & -x) Is there any more efficient or elegant way (builtin ?) to do this in javascript ? 回答1: You cannot get that result immediately with a builtin function, but you can avoid using Math.log2 . There is a little

问题 I was reading through some code examples and came across a & on Oracle's website on their Bitwise and Bit Shift Operators page. In my opinion it didn't do too well of a job explaining the bitwise & . I understand that it does a operation directly to the bit, but I am just not sure what kind of operation, and I am wondering what that operation is. Here is a sample program I got off of Oracle's website: http://docs.oracle.com/javase/tutorial/displayCode.html?code=http://docs.oracle.com/javase