decimal

I am attempting to tryparse using decimal; however, I keep getting an "Object Required" run-time error. I'm not certain what I'm doing wrong. I'm used to doing a tryparse in C#. This is VBA, so the language translation is not clicking just yet. Any help appreciated. Sub try() Dim val As Variant Dim res As Boolean res = Decimal.TryParse("2.5", val) MsgBox (res & ":" & val) End Sub You can try CInt and check for a specific error using On Error Goto. res = cBool(Val("2.5")) should do the trick here, since any value <> 0 will evaluate as True 来源： https://stackoverflow.com/questions/23655251

This question already has an answer here: C++: How to round a double to an int? [duplicate] 5 answers round() for float in C++ 21 answers I am stuck in problem where the double number is not getting properly converted to integer. In this case-> int x=1000; double cuberoot=pow(x,(1/(double)3)); int a=cuberoot; cout<<"cuberoot="<<cuberoot<<endl; cout<<"a="<<a<<endl; Output: cuberoot=10 a=9 Why here a=9 and not 10? Any solution to this problem?? Also I don't want to round the value..if a=3.67 then it should be converted to 3 only and not 4. Because the cuberoot is very close to 10 but not quite

This question already has an answer here: Is it possible in matlab to explicitly format the output numbers? 7 answers I have modified some code in MatLab so that it will give me the root of the function cos(x) - 3*x. When I run the code and ask for it to return the value of xnew (as xnew should equal the root of the function) it returns xnew to only 4 decimal points. I would like for it to be more than this. Does anyone know why it capping this value? x = 0; N = 100000; Tol = 0.00001; count = 1; while count <= N f = cos(x) - 3*x; Df = -sin(x) - 3; d = (f/Df); xnew = x - (d); if (abs(xnew - x))

Let's say we have 0.00045. I want to find a way to count the number of "significant" zeros after the decimal point (3 in this case). I've been trying to implement strpos or substr , but I'm getting stuck. Other exs.... 3.006405: Should return "2" 0.0000062: Should return "5" 9.0100000008: Should return "1" Any ideas? strspn($num, "0", strpos($num, ".")+1) strspn finds the length of a sequence of zeroes. strpos finds the position of the decimal point, and we start from 1 position past that. However, this doesn't work for 0.0000062 because it gets converted to scientific notation when converted

I would appreciate it if you could tell me how I can convert hexadecimal letters within an NSString , e.g. @"50A6C2" , to decimals using Objective-C. Thanks in advance. The easiest way is to use an NSScanner , specifically the methods scanHexInt: or scanHexLongLong: . Another possibility is to get the C string from the NSString and use C-style functions such as strtol (with base 16). 来源： https://stackoverflow.com/questions/6002196/how-to-convert-hexadecimal-string-to-decimal

This question already has an answer here: Why python decimal.Decimal precision differs with equable args? 2 answers I know this has been asked numerous times and I've come across many blogs and SO answers but this one's making me pull my hair out. I just want to multiply a two decimal number by 100 so I get rid of its decimals: >>> 4321.90 * 100 432189.99999999994 >>> Decimal(4321.90) * Decimal(100) Decimal('432189.9999999999636202119291') I'm scared to use rounding for such seemingly trivial operation. Would it be safe? What if the precision problem plays tricks on me and the result is close

I'm having some trouble with a problem I've run into dealing with floating points. I'm having a hard time moving from floating point representation to decimal values and also from format A of the representation to format B of the representation. The problem: Consider the following two 9-bit floating-point representations based on the IEEE floating-point format. Format A There is one sign bit. There are k = 5 exponent bits. The exponent bias is 15. There are n = 3 fraction bits. Format B There is one sign bit There are k = 4 exponent bits. The exponent bias is 7. There are n = 4 faction bits