double

问题 Regarding my question I have seen a post on here but did not understand since i am new to C++. I wrote a small script which gets a number from user and script prints out the factorial of entered number. Once I entered bigger numbers like 30, script does not print out all the digits.Output is like 2.652528598 E+32 however What I want is exact number 265252859812191058636308480000000. Could someone explain how to get all digits in long double.Thanks in advance 回答1: You can set the precision of

问题 Regarding my question I have seen a post on here but did not understand since i am new to C++. I wrote a small script which gets a number from user and script prints out the factorial of entered number. Once I entered bigger numbers like 30, script does not print out all the digits.Output is like 2.652528598 E+32 however What I want is exact number 265252859812191058636308480000000. Could someone explain how to get all digits in long double.Thanks in advance 回答1: You can set the precision of

问题 Why Console.WriteLine((1000000f + 0.10f).ToString("N2")); print 1 000 000.00 but no 1 000 000.10? When I use type "double" or type "float" less 1000000 - this problem disappears! 回答1: Use decimal to prevent those accuracy/rounding issues. Console.WriteLine((1000000m + 0.10m).ToString("N2")); Reason: float has only a accuracy of 7 digits (reference) - your number has 8 回答2: According to MSDN, float type has only 7 digits precision. One milion has 7 digits, decimal part is rounded. Double type

问题 I'm just trying to compute a good sigmoid function in C++ (and efficient). So i have to do something like: 1/(1 + exp(-x)) The problem is, when X becomes big (or even small), the result of 1 + e turns to be 0 or 1 For example, 1 + exp(-30) = 1 But this is incorrect... How can we add very small (or big) numbers easily and efficiently ? Datatype I am using : double Here is the code snippet : double Quaternion::sigmoidReal(double v){ return 1.0 / ( 1.0 + exp(-v) ) ; } Thanks ! 回答1: I think you

问题 I'm just trying to compute a good sigmoid function in C++ (and efficient). So i have to do something like: 1/(1 + exp(-x)) The problem is, when X becomes big (or even small), the result of 1 + e turns to be 0 or 1 For example, 1 + exp(-30) = 1 But this is incorrect... How can we add very small (or big) numbers easily and efficiently ? Datatype I am using : double Here is the code snippet : double Quaternion::sigmoidReal(double v){ return 1.0 / ( 1.0 + exp(-v) ) ; } Thanks ! 回答1: I think you

问题 In the most upvoted answer to the question Double precision floating values in Python? it is mentioned that we can use np.float128, if we need precision more than standard float. But I couldn't find a way to do this. Then, in the most upvoted answer to how to declare variable type, C style in python, use of typing is suggested. However, when I tried to use typing to set the datatype as float for a integer type value, Python still used an int data type. n : float=10000 print(type(n)) But still

问题 When i try to convert a number like 1.1 to a double i get thrown a exception: "Input string was not in a correct format." I don't know why this is happening since the input i give is correct and the variable where i am converting to is a double. Below is the code: Controller create method: [ValidateAntiForgeryToken] public ActionResult Create(IFormCollection collection) { try { Riskanalysis riskanalysis = new Riskanalysis() { dateCreation = Convert.ToDateTime(collection["dateCreation"]),

问题 I have an arraybuffer and I want to get double values.For example from [64, -124, 12, 0, 0, 0, 0, 0] I would get 641.5 Any ideas? 回答1: You could adapt the excellent answer of T.J. Crowder and use DataView#setUint8 for the given bytes. var data = [64, -124, 12, 0, 0, 0, 0, 0]; // Create a buffer var buf = new ArrayBuffer(8); // Create a data view of it var view = new DataView(buf); // set bytes data.forEach(function (b, i) { view.setUint8(i, b); }); // Read the bits as a float/native 64-bit

问题 I want to compare two double value as follow: Double doubleValue = Double.valueOf(Long.MAX_VALUE); Double doubleValue2 = Double.valueOf(Long.MAX_VALUE+1d); Apparently doubleValue and doubleValu2 are not equal, 2nd is larger due to the 1d addition. but no matter I use compare() or equals() method, both methods return a equal result for two values. Is there any way I can compare and not losing accuracy here. thanks in advance. 回答1: This is because of the loss of precision while converting long

问题 I want to compare two double value as follow: Double doubleValue = Double.valueOf(Long.MAX_VALUE); Double doubleValue2 = Double.valueOf(Long.MAX_VALUE+1d); Apparently doubleValue and doubleValu2 are not equal, 2nd is larger due to the 1d addition. but no matter I use compare() or equals() method, both methods return a equal result for two values. Is there any way I can compare and not losing accuracy here. thanks in advance. 回答1: This is because of the loss of precision while converting long