arithmeticexception

Points calculated using this elliptic curve point multiplication do not lie on the curve and this class brings Arithmetic exception

孤街浪徒 提交于 2019-12-03 09:04:17
I get stack on my error of point multiplication using standard projective coordinates. I don't know what i missed but the multiplied points do not lie on the curve and some times it outputs something like Arithmetic Exception: integer is not invertible. public class ECPointArthimetic { EllipticCurve ec; private BigInteger x; private BigInteger y; private BigInteger z; private BigInteger zinv; private BigInteger one = BigInteger.ONE; private BigInteger zero = BigInteger.ZERO; private boolean infinity; public ECPointArthimetic(EllipticCurve ec, BigInteger x, BigInteger y, BigInteger z) { this.ec

ASP.NET Overflow or underflow in the arithmetic operation when returning large file bigger 1 GB

妖精的绣舞 提交于 2019-12-03 05:17:17
I went across some sort of limitation in ASP.NET. I reduced the problem into a sample project in ASP.NET MVC Project (created with Visual Studio 2010 and .NET 4) and the problem still occurs: In a MVC Controller I have a method which provides a file download: public ActionResult DownloadBigFile() { string file = @"C:\Temp\File.txt"; var readStream = new FileStream(file, FileMode.Open, FileAccess.Read); return File(readStream, "text/plain", "FILE"); } When the file is below 1 GB the download works fine, above 1 GB an exception is thrown: "Overflow or underflow in the arithmetic operation" with

Project Euler #3 out of integer range java [duplicate]

≡放荡痞女 提交于 2019-12-02 07:36:45
This question already has an answer here: The literal xyz of type int is out of range 5 answers The code is supposed to give back the biggest prime number. More about the task here: https://projecteuler.net/problem=3 int checkFactors(double na) { long n = (long) na; int biggestPrimeFactor = 0; for (int i = 1; i < n; i++) if (n % i == 0 && isPrimFaktor(i) && i > biggestPrimeFactor) biggestPrimeFactor = i; return biggestPrimeFactor; } boolean isPrimeFactor(int n) { int length= 0; for (int i = n; i > 0; i--) if (n % i == 0) length++; if (length== 2) return true; return false; } I decided to make

How to prevent arithmetic overflow error when using SUM on INT column?

可紊 提交于 2019-11-30 07:59:08
I am using SQL Server 2008 R2 and I have an INT column where the data inserted never surpasses the max INT , but I have a query which uses the SUM function which when executed surpasses the max INT limit and throws the error mentioned in the title. I want to be able to execute this query without changing the column type from INT to BIGINT . Here is my query: SELECT UserId, SUM( PokemonExp ) AS TotalExp, MAX( PokemonLevel ) AS MaxPokeLevel FROM mytable GROUP BY UserId ORDER BY TotalExp DESC Note: The PokemonExp column is of type INT . Michał Powaga Type of expression in SUM determines return

Java division by zero doesnt throw an ArithmeticException - why?

流过昼夜 提交于 2019-11-27 08:49:18
Why this code doesn't throw an ArithmeticException ? Take a look: public class NewClass { public static void main(String[] args) { // TODO code application logic here double tab[] = {1.2, 3.4, 0.0, 5.6}; try { for (int i = 0; i < tab.length; i++) { tab[i] = 1.0 / tab[i]; } } catch (ArithmeticException ae) { System.out.println("ArithmeticException occured!"); } } } I have no idea! Why can't you just check it yourself and throw an exception if that is what you want. try { for (int i = 0; i < tab.length; i++) { tab[i] = 1.0 / tab[i]; if (tab[i] == Double.POSITIVE_INFINITY || tab[i] == Double

Java division by zero doesnt throw an ArithmeticException - why?

拟墨画扇 提交于 2019-11-26 09:49:30
问题 Why this code doesn\'t throw an ArithmeticException ? Take a look: public class NewClass { public static void main(String[] args) { // TODO code application logic here double tab[] = {1.2, 3.4, 0.0, 5.6}; try { for (int i = 0; i < tab.length; i++) { tab[i] = 1.0 / tab[i]; } } catch (ArithmeticException ae) { System.out.println(\"ArithmeticException occured!\"); } } } I have no idea! 回答1: Why can't you just check it yourself and throw an exception if that is what you want. try { for (int i = 0

Why does division by zero with floating point (or double precision) numbers not throw java.lang.ArithmeticException: / by zero in Java

让人想犯罪 __ 提交于 2019-11-26 03:55:36
问题 The following statement throws java.lang.ArithmeticException: / by zero as obvious. System.out.println(0/0); because the literal 0 is considered to be an int literal and divide by zero is not allowed in integer arithmetic. The following case however doesn\'t throw any exception like java.lang.ArithmeticException: / by zero . int a = 0; double b = 6.199; System.out.println((b/a)); It displays Infinity . The following statement produces NaN (Not a Number) with no exception. System.out.println

ArithmeticException: “Non-terminating decimal expansion; no exact representable decimal result”

亡梦爱人 提交于 2019-11-26 00:39:40
问题 Why does the following code raise the exception shown below? BigDecimal a = new BigDecimal(\"1.6\"); BigDecimal b = new BigDecimal(\"9.2\"); a.divide(b) // results in the following exception. -- java.lang.ArithmeticException: Non-terminating decimal expansion; no exact representable decimal result. 回答1: From the Java 8 docs: When a MathContext object is supplied with a precision setting of 0 (for example, MathContext.UNLIMITED), arithmetic operations are exact, as are the arithmetic methods