I have a Java application which parses a number from somewhere, and checks that it is a valid int (between Integer.MIN_VALUE and Integer.MAX_VALUE) or a valid double (between Double.MIN_VALUE and Doub
This question already has answers here: Closed 11 years ago. Possible Duplicate: Why is floating point arithmetic in C# imprecise?
I have two projects in my eclipse workspace which are having similar tasks. Both projects have a special part where I convert a double to an String.
how can I make this expression not end up being a zero. double dBaseFitness = (double) baseFitness; x.shortfitness = (long)(Math.pow(dbaseFitness, dbaseFitness/8开发者_开发技巧.0)/100.0)
I don\'t know what the correct wording is for what I am trying to achieve so it may already be posted online. Please be kind if it is.
I am currently working on a lab and would like to know how to handle the following problem which I have spent at least two hours on:
I have a very large code that sets up and iteratively solves a system of non-linear partial differential equation, written in fortran.I need all variables to be double precision.In the additional modu
double temp; temp = (double)Convert.ToDouble(\"1234.5678\"); Hey Lads and Lad开发者_JAVA技巧ies, I can\'t for the life of me figure out why the above line isn\'t working. The above line gives me a r
Double (Explicit) Smokepurpp 专辑：Lost Planet 开发者_运维知识库2.0 (Explicit) 语种： 英语
I currently use the following code to print a double: return String.format(\"%.2f\", someDouble); Th开发者_C百科is works well, except that Java uses my Locale\'s decimal separator (a comma) while I
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