digits

My question may seem simple, but still can not get something that works. I need to customize the Math.round rounding format or something to make it work as follows: If the number is 1.6 he should round to 1, if greater than or equal to 1.7 should round to 2.0 . And so to all other decimal results with # .6 The way I'm doing the 1.6 being rounded to 2 shall be rounded to 1. How can I do that? Thank you! Simply do this: double threshold = 0.7; Math.round(x - threshold + 0.5); You could write a method that takes a double variable as input and returns the integer based on the first digit after the

I have the following code: #include <iostream> #include <limits> int main() { std::cout << std::numeric_limits<unsigned long long>::digits10 << std::endl; return 0; } GCC 4.4 returns 19 MS VS 9.0 returns 18 Can someone please explain Why is there a difference between the two? I would have expected such a constant would be the same regardless of the compiler. If Visual C++ 2008 returns 18 for std::numeric_limits<unsigned long long>::digits10 , it is a bug (I don't have Visual C++ 2008 installed to verify the described behavior). In Visual C++ (at least for 32-bit and 64-bit Windows), unsigned

I have an app that prints a 3x3 cv::Mat on the iPhone screen. I need to reduce the decimals, as the screen is not so big, see: [1.004596557012473, -0.003116992336797859, 5.936915104939593; -0.007241746117066327, 0.9973985665720294, -0.2118670500989478; 1.477734234970711e-05, -1.03363734495053e-05, 1.000089074805124] so I would like to reduce the decimals .4 or .6 or six decimals. Any ideas? Cheers If you were using printf cv::Mat data(3, 3, CV_64FC1); for (int y = 0; y < data.rows; ++y) { for (int x = 0;x < data.cols; ++x) { printf("%.6f ", data.at<double>(y, x)); } } If you were using std:

Is there any fast way to find the largest power of 10 smaller than a given number? I'm using this algorithm, at the moment, but something inside myself dies anytime I see it: 10**( int( math.log10(x) ) ) # python pow( 10, (int) log10(x) ) // C I could implement simple log10 and pow functions for my problems with one loop each, but still I'm wondering if there is some bit magic for decimal numbers. An alternative algorithm is: i = 1; while((i * 10) < x) i *= 10; Log and power are expensive operations. If you want fast, you probably want to look up the IEEE binary exponent in table to get the

Converting non-negative Integer to its list of digits is commonly done like this: import Data.Char digits :: Integer -> [Int] digits = (map digitToInt) . show I was trying to find a more direct way to perform the task, without involving a string conversion, but I'm unable to come up with something faster. Things I've been trying so far: The baseline: digits :: Int -> [Int] digits = (map digitToInt) . show Got this one from another question on StackOverflow: digits2 :: Int -> [Int] digits2 = map (`mod` 10) . reverse . takeWhile (> 0) . iterate (`div` 10) Trying to roll my own: digits3 :: Int ->