bitwise-and | 易学教程

Meaning of bitwise and(&) of a positive and negative number?

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问题 Can anyone help what n&-n means?? And what is the significance of it. 回答1: I believe it is a trick to figure out if n is a power of 2. (n == (n & -n)) IFF n is a power of 2 (1,2,4,8). 回答2: It's an old trick that gives a number with a single bit in it, the bottom bit that was set in n . At least in two's complement arithmetic, which is just about universal these days. The reason it works: the negative of a number is produced by inverting the number, then adding 1 (that's the definition of two

Why does if( -8 & 7) return false [duplicate]

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问题 This question already has answers here : Difference between & and && in C? (3 answers) Closed 25 days ago . When i try to run the following code it prints "FALSE" instead of "TRUE" Can somebody explain why the code returns false? #include <stdio.h> int main(void) { if(-8 & 7) { printf("TRUE"); } else { printf("FALSE"); } return 0; } 回答1: -8 can be represented the following way ( I will use a byte representation 8 = 00001000 ~8 = 11110111 -8 = 11111000 (~8 + 1) That is -8 in the two-complement

What does the bitwise code “$n & ($n - 1)” do?

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问题 What does this code mean and what are other ways accomplish the same without using bit shifting? if ($n & ($n - 1)) 回答1: That formula checks to see whether a number is a power of 2 (if your condition as written is true, then the number is not a power of two). Stated another way, your test checks to see whether there is more than one "1" bit set in the binary representation of $n . If there is zero or only one bit set, then your test will be false. It is by far the most efficient way to

bitwise-ANDing with 0xff is important?

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问题 Doesn't bitwise-ANDing with 0xff essentially mean getting the same value back, for that matter, in this code? byte[] packet = reader.readPacket(); short sh; sh = packet[1]; sh &= 0xFF; System.out.print(sh+" "); Weirdly, I get a -1 if that ANDing is not included but a 255 when included Could someone explain the reason? As I see it 0xff is just 1111 1111. Isn't it? 回答1: Yes, 0xff is just 1111 1111 . But this is attempting to display the unsigned byte value, even though in Java byte s are signed

bitwise and logical AND/OR in terms of hex result

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问题 so if I have x = 0x01 and y = 0xff and I do x & y , I would get 0x01 . If I do x && y do I still get 0x01 , but the computer just says its true if its anything than 0x00 ? My question is are the bits the same after the operation regardless of bit-wise or logical AND/OR, but the computer just interprets the final result differently? In other words are the hex values of the result of & and && (likewise | and ||) operations the same? edit: talking about C here 回答1: The answer depends on the

XOR from only OR and AND

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问题 How do you do the XOR bitwise operation if you only have available the AND and the OR operations? 回答1: Creating my own scripting language - ChrisScript - you just need something like: #!/bin/chrish bit XOR (bit A, bit B) { bit notA; bit notB; IF (A == 0) notA = 1 ELSE notA = 0; IF (B == 0) notB = 1 ELSE notB = 0; F = ((A && notB) || (notA && B)); RETURN F; } Even without NOT, it can be emulated like this. But this is the best solution you're going to get without having some form of inverter.

tracking multiple objects by color OpenCV 2.x

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Currently i am trying to track multiple objects by color. I've based on documentation code. import cv2 import numpy as np cap = cv2.VideoCapture(0) while(1): # Take each frame _, frame = cap.read() # Convert BGR to HSV hsv = cv2.cvtColor(frame, cv2.COLOR_BGR2HSV) # define range of blue color in HSV lower_blue = np.array([110,50,50]) upper_blue = np.array([130,255,255]) # Threshold the HSV image to get only blue colors mask = cv2.inRange(hsv, lower_blue, upper_blue) # Bitwise-AND mask and original image res = cv2.bitwise_and(frame,frame, mask= mask) cv2.imshow('frame',frame) cv2.imshow('mask'

explain arguments meaning in res = cv2.bitwise_and(img,img,mask = mask)

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I am trying to extract blue colour of an input image. For that I create a blue HSV colour boundary and threshold HSV image by using the command mask_img = cv2.inRange(hsv, lower_blue, upper_blue) After that I used a bitwise_and on the input image and the threshold image by using res = cv2.bitwise_and(img,img,mask = mask_img) Where 'img' is the input image. This code I got from the opencv. But I didn't understand why three arguments used in bitwise_and and what is actually each arguments mean? Why the same image is used at src1 and src2 ? And also what is the use of mask keyword here? Please

What does the bitwise AND operator & do?

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Please help to solve this problem and explain the logic. I don't know how the & operator is working here. void main() { int a = -1; static int count; while (a) { count++; a &= a - 1; } printf("%d", count); } If you are referring to a&=a-1; then it is a bitwise and operation of a and a-1 copied into a afterwards. Edit: As copied from Tadeusz A. Kadłubowski in the comment: a = a & (a-1); The expression a&=a-1; clears the least significant bit (rightmost 1) of a . The code counts the number of bits in a (-1 in this case). Starting from a = -1 ; // 11111111 11111111 11111111 11111111 32bits signed

What does the bitwise code “$n & ($n - 1)” do?

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What does this code mean and what are other ways accomplish the same without using bit shifting? if ($n & ($n - 1)) Greg Hewgill That formula checks to see whether a number is a power of 2 (if your condition as written is true, then the number is not a power of two). Stated another way, your test checks to see whether there is more than one "1" bit set in the binary representation of $n . If there is zero or only one bit set, then your test will be false. It is by far the most efficient way to determine that property. First, this code is valid PHP, so your title is poor. Second, the binary