double-precision

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 ? 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 or other data structure, you should not use any tolerance for comparison. Usually, if your program needs to

I was testing some of my code, in javascript I added .1+.2 and it gives me .30000000000000004 instead of .3 . I don't understand this. But when I added .1+.3 it gives me .4. I googled it and find its something about Double Precision addition. But I don't know what it is. Here's the obligatory link: What Every Computer Scientist Should Know About Floating-Point Arithmetic Basically, there are many base 10 numbers that cannot be exactly represented in the floating point format used by most computers, so you'll get issues like the ones you highlight. If you can't stay awake for What Every

I'm attempting to recode an old C++ program in Fortran to make use of LAPACK (I'm aware that C++ does have LAPACK++, but I'm having a lot of trouble installing it, so I gave up). I originally didn't have any problems with compilation, but that was when I had all variables declared as REAL . When I started coding the section of the program that required LAPACK, I found that all parameters passed to DSYEV need to be DOUBLE PRECISION . So I tried to change everything to double precision (including changing all hard coded numbers to their double precision counterparts, i.e. 0.0 -> 0.0D0) Now when

I have a unit test, testing boundaries: [TestMethod] [ExpectedException(typeof(ArgumentOutOfRangeException))] public void CreateExtent_InvalidTop_ShouldThrowArgumentOutOfRangeException() { var invalidTop = 90.0 + Double.Epsilon; new Extent(invalidTop, 0.0, 0.0, 0.0); } public static readonly double MAX_LAT = 90.0; public Extent(double top, double right, double bottom, double left) { if (top > GeoConstants.MAX_LAT) throw new ArgumentOutOfRangeException("top"); // not hit } I thought I'd just tip the 90.0 over the edge by adding the minimum possible positive double to it, but now the exception

I'm running on a 32-bit machine and I'm able to confirm that long values can tear using the following code snippet which hits very quickly. static void TestTearingLong() { System.Threading.Thread A = new System.Threading.Thread(ThreadA); A.Start(); System.Threading.Thread B = new System.Threading.Thread(ThreadB); B.Start(); } static ulong s_x; static void ThreadA() { int i = 0; while (true) { s_x = (i & 1) == 0 ? 0x0L : 0xaaaabbbbccccddddL; i++; } } static void ThreadB() { while (true) { ulong x = s_x; Debug.Assert(x == 0x0L || x == 0xaaaabbbbccccddddL); } } But when I try something similar

Given a double, I want to round it to a given number of points of precision after the decimal point , similar to PHP's round() function. The closest thing I can find in the Dart docs is double.toStringAsPrecision(), but this is not quite what I need because it includes the digits before the decimal point in the total points of precision. For example, using toStringAsPrecision(3): 0.123456789 rounds to 0.123 9.123456789 rounds to 9.12 98.123456789 rounds to 98.1 987.123456789 rounds to 987 9876.123456789 rounds to 9.88e+3 As the magnitude of the number increases, I correspondingly lose

I am currently writing a calculator program in java. It is my first java program, I am used to c++. I have noticed that doubles in java are not at all like doubles in c++. try this in java and c++ 4.1*3 that/.1 it should be 12.3 then 123, and c++ gives this result but java gives 12.299999999999999 and 122.99999999999999 How can I do math like in c++ with doubles, I understand that anything you would use 12.299999999999999 in a program at all would make no difference compared to 12.3, but when a user is reading the numbers this is very ugly. I have looked into the BigDecimal class but I cannot

I have the following problem: double a = 6.005; double b = 5.995; I want to set precision of doubles 2 digits after point, for example double c = a+b;// I would like to get 11.99 not 12.00 . How can I do this? Precision is one thing; rounded for display is quite another. I think this is wrong headed. You should want all the precision you can get and worry about rounding for display when the results are complete. UPDATE: You should not be representing currency using doubles. Last time I looked, C++ was an object-oriented language. You should create an abstraction for Money that does the right

I would like some advice from people who have more experience working with primitive double equality in Java. Using d1 == d2 for two doubles d1 and d2 is not sufficient due to possible rounding errors. My questions are: Is Java's Double.compare(d1,d2) == 0 handling rounding errors to some degree? As explained in the 1.7 documentation it returns value 0 if d1 is numerically equal to d2 . Is anyone certain what exactly they mean by numerically equal? Using relative error calculation against some delta value, is there a generic (not application specific) value of delta you would recommend? Please