division

Write code to determine if a number is divisible by 3. The input to the function is a single bit, 0 or 1, and the output should be 1 if the number received so far is the binary representation of a number divisible by 3, otherwise zero. Examples: input "0": (0) output 1 inputs "1,0,0": (4) output 0 inputs "1,1,0,0": (6) output 1 This is based on an interview question. I ask for a drawing of logic gates but since this is stackoverflow I'll accept any coding language. Bonus points for a hardware implementation (verilog etc). Part a (easy): First input is the MSB. Part b (a little harder): First

This question already has an answer here: What is the difference between '/' and '//' when used for division? 14 answers I saw this in someone's code: y = img_index // num_images where img_index is a running index and num_images is 3. When I mess around with // in IPython, it seems to act just like a division sign (i.e. one forward slash). I was just wondering if there is any reason for having double forward slashes? In Python 3, they made the / operator do a floating-point division, and added the // operator to do integer division (i.e. quotient without remainder); whereas in Python 2, the /

Which of the following techniques is the best option for dividing an integer by 2 and why? Technique 1: x = x >> 1; Technique 2: x = x / 2; Here x is an integer. Use the operation that best describes what you are trying to do. If you are treating the number as a sequence of bits, use bitshift. If you are treating it as a numerical value, use division. Note that they are not exactly equivalent. They can give different results for negative integers. For example: -5 / 2 = -2 -5 >> 1 = -3 (ideone) Cat Plus Plus Does the first one look like dividing? No. If you want to divide, use x / 2 . Compiler

This question already has an answer here: Int division: Why is the result of 1/3 == 0? 15 answers Here's an example: Double d = (1/3); System.out.println(d); This returns 0, not 0.33333... as it should. Does anyone know? Firas Assaad That's because 1 and 3 are treated as integers when you don't specify otherwise, so 1/3 evaluates to the integer 0 which is then cast to the double 0 . To fix it, try (1.0/3) , or maybe 1D/3 to explicitly state that you're dealing with double values. If you have int s that you want to divide using floating-point division, you'll have to cast the int to a double :

I was working on something else, but everything came out as zero, so I made this minimalistic example, and the output is still 0. #include <iostream> int main(int argc, char** argv) { double f=3/5; std::cout << f; return 0; } What am I missing? You are missing the fact that 3 and 5 are integers, so you are getting integer division. To make the compiler perform floating point division, make one of them a real number: double f = 3.0 / 5; It doesn't need to be .0 , you can also do 3./5 or 3/5. or 3e+0 / 5 or 3 / 5e-0 or 0xCp-2 / 5 or... There only needs to be an indicator involved so that the