double

问题 I know that when I would like to check if double == double I should write: bool AreSame(double a, double b) { return fabs(a - b) < EPSILON; } But what when I would like to check if a > b or b > a ? 回答1: There is no general solution for comparing floating-point numbers that contain errors from previous operations. The code that must be used is application-specific. So, to get a proper answer, you must describe your situation more specifically. For example, if you are sorting numbers in a list

问题 I would like to know how to parse several double numbers from a string, but string can be mixed, for instance: String s = "text 3.454 sometext5.567568more_text" . The standard method ( Double.parseDouble ) is unsuitable. I've tried to parse it using the isDigit method, but how to parse other characters and . ? thanks. 回答1: After parsing your doubles with the suitable regular expressions like in this code or in other posts, iterate to add the matching ones to a list. Here you have myDoubles

问题 System.out.println(2.14656); 2.14656 System.out.println(2.14656%2); 0.14656000000000002 WTF? 回答1: The do give the expected results. Your expectations are incorrect. When you type the double-precision literal 2.14656 , what you actually get is the closest double-precision value, which is: 2.14656000000000002359001882723532617092132568359375 the println happens to round this when it prints it out (to 17 significant digits), so you see the nice value that you expect. After the modulus operation

问题 I have an arraylist which consists of latLong object as shown below: ArrayList < LatLng > latLngList = new ArrayList< LatLng >(); Note: LatLng is an object which has latitude and longitude values in double format i.e., Latitude is in double format and Longitude is in double format The below are the values which are stored in the above latLngList object: Lat Long List:[lat/lng: (34.072516,-118.403609), lat/lng: (34.074227,-118.399248), lat/lng: (34.07304,-118.3995), lat/lng: (34.07304,-118

问题 This code I have written to convert double into int getting an exception. Exception in thread "main" java.lang.Error: Unresolved compilation problem: Cannot cast from Double to int This is my code Double d = 10.9; int i = (int)(d); 回答1: Double is a wrapper class on top of the primitive double . It can be cast to double , but it cannot be cast to int directly. If you use double instead of Double , it will compile: double d = 10.9; int i = (int)(d); You can also add a cast to double in the

问题 I have a variable of type double , I need to print it in upto 3 decimals of precision but it shouldn't have any trailing zeros... eg. I need 2.5 // not 2.500 2 // not 2.000 1.375 // exactly till 3 decimals 2.12 // not 2.120 I tried using DecimalFormatter , Am i doing it wrong? DecimalFormat myFormatter = new DecimalFormat("0.000"); myFormatter.setDecimalSeparatorAlwaysShown(false); Thanks. :) 回答1: Try the pattern "0.###" instead of "0.000" : import java.text.DecimalFormat; public class Main {

问题 The following outputs 0.23 . How do I get it to simply output .23 ? printf( "%8.2f" , .23 ); 回答1: The C standard says that for the f and F floating point format specifiers: If a decimal-point character appears, at least one digit appears before it. I think that if you don't want a zero to appear before the decimal point, you'll probably have to do something like use snprintf() to format the number into a string, and remove the 0 if the formatted string starts with "0." (and similarly for "-0.

问题 AFAIK, C supports just a few data types: int, float, double, char, void enum. I need to store a number that could reach into the high 10 digits. Since I'm getting a low 10 digit # from INT_MAX , I suppose I need a double. <limits.h> doesn't have a DOUBLE_MAX. I found a DBL_MAX on the internet that said this is LEGACY and also appears to be C++. Is double what I need? Why is there no DOUBLE_MAX? 回答1: DBL_MAX is defined in <float.h>. Its availability in <limits.h> on unix is what is marked as "

问题 The title is quite self-explanatory, input is given double value, and I want to add/substract the smallest amount possible. 回答1: If your compiler implements C99's math functions/C++11, you can use the nextafter : #include <cfloat> // DBL_MAX #include <cmath> // std::nextafter double x = 0.1; // next representable number after x in the direction of DBL_MAX double xPlusSmallest = std::nextafter(x, DBL_MAX); Even if your compiler doesn't support it, it probably has an instrinsic for it. (MSVC

问题 The title is quite self-explanatory, input is given double value, and I want to add/substract the smallest amount possible. 回答1: If your compiler implements C99's math functions/C++11, you can use the nextafter : #include <cfloat> // DBL_MAX #include <cmath> // std::nextafter double x = 0.1; // next representable number after x in the direction of DBL_MAX double xPlusSmallest = std::nextafter(x, DBL_MAX); Even if your compiler doesn't support it, it probably has an instrinsic for it. (MSVC