bits

问题 Does anyone know the steps for dividing unsigned binary integers using non-restoring division? It's hard to find any good sources online. i.e if A = 101110 and B = 010111 how do we find A divided by B in non-restoring division? What do the registers look like in each step? Thanks! 回答1: (My answer is a little late-reply. But I hope it will be useful for future visitors) Algorithm for Non-restoring division is given in below image : In this problem, Dividend (A) = 101110, ie 46, and Divisor (B)

问题 Does anyone know the steps for dividing unsigned binary integers using non-restoring division? It's hard to find any good sources online. i.e if A = 101110 and B = 010111 how do we find A divided by B in non-restoring division? What do the registers look like in each step? Thanks! 回答1: (My answer is a little late-reply. But I hope it will be useful for future visitors) Algorithm for Non-restoring division is given in below image : In this problem, Dividend (A) = 101110, ie 46, and Divisor (B)

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . If I multiplie two 16-bit numbers, the result will be 32-bit long. But why is this so? What is the clear explanation for this? And for my right understanding: The calculation for this is: n-bit number multiplied with a m-bit number gives a (n+m) bit number? 回答1: (2 n - 1)*(2 m - 1) = 2 n+m - 2 n - 2 m + 1 -(2 n

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . If I multiplie two 16-bit numbers, the result will be 32-bit long. But why is this so? What is the clear explanation for this? And for my right understanding: The calculation for this is: n-bit number multiplied with a m-bit number gives a (n+m) bit number? 回答1: (2 n - 1)*(2 m - 1) = 2 n+m - 2 n - 2 m + 1 -(2 n

问题 I am trying to get the correct int out of an array of bytes. The bytes is read from a RFIDTag via POS for .Net. (Acctually I need 18 bits) In binary the byte array is as follows: 00001110 11011100 00000000 00011011 10000000 What I need to get out of it is: 00 00000000 11101101 (int = 237) From the original bytes that would be the following bits in reverse order: ------10 11011100 00000000 I have been looking at bitArray. Array.Reverse. And several ways of shifting bits. But I just can't wrap

问题 is there a way in python to write less than 1 byte data even when I write the number 0 which represented in 1 bit the file size is 1(8 bits) byte I tried the struct module file.write(struct.pack('b',0)) array module import array data1=array.array('B') x=bin(0)[2:] data1.append(int(0,2)) f2=open('/root/x.txt','wb') data1.tofile(f2) 回答1: No you cannot write less than a byte. A byte is an indivisble amount of memory the computer can handle. The hardware is not equipped to handle units of data <1

问题 I want to find if a bit in a sequense is 1 or 0 (true or false) if i have some bitsequense of 11010011, how can i check if the 4th position is True or False? example: 10010101 (4th bit) -> False 10010101 (3rd bit) -> True 回答1: Without the bit shifting: if bits & 0b1000: ... EDIT: Actually, (1 << 3) is optimized out by the compiler. >>> dis.dis(lambda x: x & (1 << 3)) 1 0 LOAD_FAST 0 (x) 3 LOAD_CONST 3 (8) 6 BINARY_AND 7 RETURN_VALUE >>> dis.dis(lambda x: x & 0b1000) 1 0 LOAD_FAST 0 (x) 3 LOAD

问题 Up to 255, I can understand how the integers are stored in char and unsigned char ; #include<stdio.h> int main() { unsigned char a = 256; printf("%d\n",a); return(0); } In the code above I have an output of 0 for unsigned char as well as char . For 256 I think this is the way the integer stored in the code (this is just a guess): First 256 converted to binary representation which is 100000000 (totally 9 bits). Then they remove the remove the leftmost bit (the bit which is set) because the

问题 This question already has answers here : Implementing Logical Right Shift in C (8 answers) Closed 11 months ago . Right now I am reading the book Computer Systems : Programmer Perspective. One problem in the book says to perform a logical right shift on a signed integer, I can't figure out how to start on this. The following is the actual question from the book: Fill in code for the following C functions. Function srl performs a logical right shift using an arithmetic right shift (given by

问题 What is the fastest way to count the number of set bits (i.e. count the number of 1s) in an UInt32 without the use of a look up table? Is there a way to count in O(1) ? 回答1: Is a duplicate of: how-to-implement-bitcount-using-only-bitwise-operators or best-algorithm-to-count-the-number-of-set-bits-in-a-32-bit-integer And there are many solutions for that problem. The one I use is: int NumberOfSetBits(int i) { i = i - ((i >> 1) & 0x55555555); i = (i & 0x33333333) + ((i >> 2) & 0x33333333);