double

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: Float to binary in C++ I have a very small double var, and when I print it I get -0. (using C++). Now in order to get better precision I tried using cout.precision(18); \\i think 18 is the max precision i can get. cout.setf(ios::fixed,ios::floatfield); cout<<var;\\var is a double. but it just writes -0.00000000000... I want to see the exact binary representation of the var. In other words I want to see what

问题 I know this has been discussed time and time again, but I can't seem to get even the most simple example of a one-step division of doubles to result in the expected, unrounded outcome in C# - so I'm wondering if perhaps there's i.e. some compiler flag or something else strange I'm not thinking of. Consider this example: double v1 = 0.7; double v2 = 0.025; double result = v1 / v2; When I break after the last line and examine it in the VS debugger, the value of "result" is 27.999999999999996. I

问题 At this moment i am trying to get a double value from textbox like this: String.IsNullOrEmpty(textBox1.Text) ? 0.0 : Double.Parse(textBox1.Text) But there is a problem, i cant get how to parse empty textbox? For example if to try this code with OleDb and Excel with empty textbox, we will get error System.FormatException: Input string was not in a correct format. 回答1: double val; if(!double.TryParse(textBox.Text,out val)) val = 0.0 回答2: Did you try Double.TryParse(String, NumberStyles,

问题 I need to turn double into long preserving its binary structure, not number value. Just change type, but leave binary value as it is. Is there a native way to do it? 回答1: There is Double with to doubleToLongBits and doubleToLongRawBits. Javadoc is your friend. 来源： https://stackoverflow.com/questions/15065869/double-to-long-without-conversion-in-java

问题 I need to turn double into long preserving its binary structure, not number value. Just change type, but leave binary value as it is. Is there a native way to do it? 回答1: There is Double with to doubleToLongBits and doubleToLongRawBits. Javadoc is your friend. 来源： https://stackoverflow.com/questions/15065869/double-to-long-without-conversion-in-java

问题 In this code i got the above error in lines i commented. public double bigzarb(long u, long v) { double n; long x; long y; long w; long z; string[] i = textBox7.Text.Split(','); long[] nums = new long[i.Length]; for (int counter = 0; counter < i.Length; counter++) { nums[counter] = Convert.ToInt32(i[counter]); } u = nums[0]; int firstdigits = Convert.ToInt32(Math.Floor(Math.Log10(u) + 1)); v = nums[1]; int seconddigits = Convert.ToInt32(Math.Floor(Math.Log10(v) + 1)); if (firstdigits >=

问题 This question already has answers here : Convert Double to Binary representation? (5 answers) Closed 5 years ago . I am generating sequence of numbers of type double like 0.9913963644564902 0.0341175990344773 0.13181105852355268 0.45773616980747556 and I have to convert it to binary representation in java. 回答1: This will help you. I have tried out using one of your input double value - 0.9913963644564902 public static void main(String[] args){ double d = 0.9913963644564902; System.out.println

问题 What does it mean when their are square brackets in front of double . For example double[] and is there any special way to use the return function when using them in a method. For Example: public double[] MethodName(){ } What is the type of the return value? 回答1: Square brackets [] would indicate an array. In your case of code: public double[] MethodName(){ .... } You have a public method MethodName which returns an array. Double indicates what type the array is, i.e. stores an array of

问题 I am trying to read a Fortran double-precision number like 1.2345D+02 into python, but I got the following error: >>> float('1.2345D+02') Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: invalid literal for float(): 1.2345D+02 By following the advice on Python scientific notation using D instead of E, I tried numpy but I also get the same error: import numpy >>> numpy.float("1.2345D+02") Traceback (most recent call last): File "<stdin>", line 1, in <module>

问题 How do I convert double value to a char array in C? double a=2.132; char arr[8]; Is there any way to do this in standard C? Can anyone give any solution? 回答1: If you are about to store the double DATA, not the readable representation, then: #include <string.h> double a=2.132; char arr[sizeof(a)]; memcpy(arr,&a,sizeof(a)); 回答2: To make a character string representing the number in human-readable form, use snprintf(), like in code below. To access the bytes of the double, use a union. For