Floating Point Roundoff Error Java
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Round Off Error
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Floating Point Error
here: Is floating point math broken? 27 answers Floating point multiplication in java [duplicate] 3 answers Ok, so i made a scientific calculator recently and one of my friends was using it when he said that if he types in (2.1 * 3) the answer should be 6.3 but he recieves something like: 6.30000001. Why does this happen? Is this some sort of bug in java? EDIT: Its not just a problem
Machine Epsilon
with my calc, i also made a small java file for multiplying 2 numbers and i face the same probelem even in cmd java rounding multiplication share|improve this question asked Aug 28 '14 at 1:07 namsnath 49210 marked as duplicate by Gunther Fox, Boann, Anirudh Ramanathan, Code-Apprentice, EJPjava Users with the java badge can single-handedly close java questions as duplicates and reopen them as needed. Aug 28 '14 at 1:12 This question was marked as an exact duplicate of an existing question. See this answer to a similar question stackoverflow.com/a/588014/3440545 –AbcAeffchen Aug 28 '14 at 1:11 powerfield-software.com/?p=30 –paxdiablo Aug 28 '14 at 1:12 question about floating point precision has been asked thousands times here. –Baby Aug 28 '14 at 1:13 It's a bug in base 2 floating point representation. –Ted Hopp Aug 28 '14 at 1:18 add a comment| active oldest votes Browse other questions tagged java rounding multiplication or ask your own question. asked 2 years ago viewed 1846 times Linked 1288 Is floating point math broken? 0 Floating point multiplication in java Related 3598Is Java “pass-by-reference” or “pass-by-value”?671How to round a number to n decimal places in Java194Round a double to 2 decimal places3Java round off 0.05 for money0Floating point multiplication in java3Java Roun
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Java Double Precision
Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only java rounding takes a minute: Sign up Java Rounding Off Issue up vote 3 down vote favorite Basically, I don't understand why the code below will output 434 when 4.35 * 100 = 435.0 which is converted to the int of 435, right? http://stackoverflow.com/questions/25539222/round-off-error-java What is the simple explanation for this and is this an issue which crops up often? How would you get around this? public class RoundOffDemo { public static void main(String[] args) { double price = 4.35; int cents = (int) (100 * price); // Should be 435 System.out.println(cents); // Prints 434! } } java floating-point rounding floating-accuracy share|improve this question edited Dec 9 '15 at 10:10 Brian Tompsett - 汤莱恩 3,091132775 asked Dec 28 '11 at 16:44 mino 1,850144162 http://stackoverflow.com/questions/8658719/java-rounding-off-issue possible duplicate of Floating point arithmetic not producing exact results in Java. And at least a hundred other questions. –dan04 Dec 28 '11 at 23:46 add a comment| 6 Answers 6 active oldest votes up vote 7 down vote accepted The problem is that 4.35 cannot be exactly represented in binary. So 4.35 * 100 is not exactly 435.0. (Every fraction in binary is the sum of inverse powers of 2, all of which are terminating decimals. Unfortunately, 0.35 decimal has a non-terminating binary expansion. So 4.35 is more like 4.349999...x (where x is something beyond which everything is zero, courtesy of the machine's finite representation of floating point numbers.) Integer truncation then produces 434. You can use Math.round() to avoid this problem. share|improve this answer answered Dec 28 '11 at 16:48 Ted Hopp 163k28240345 Little addition, I suppose that Math.rint() is more correct to use in such case (casting to int). –4ndrew Dec 28 '11 at 17:11 @4ndrew - Perhaps so. One could also cast the argument to a float so round returns an int. (Although, for the numbers OP is using, using float to start with instead of double might be the best option.) –Ted Hopp Dec 28 '11 at 17:19 2 A good example from the "real world" is 1/3, which cannot be exactly represented in decimal. If computers used decimal instead of binary, 1/3 * 3 would be 0.9999999, not 1.0. –us
a number and its exact mathematical value due to rounding. This is a form of quantization error.[3] One of the goals of numerical https://en.wikipedia.org/wiki/Round-off_error analysis is to estimate errors in calculations, including round-off error, when using approximation equations and/or algorithms, especially when using finitely many digits to represent real numbers (which in theory have infinitely many digits).[4] When a sequence of calculations subject to rounding error is made, errors may accumulate, sometimes dominating the calculation. In ill-conditioned problems, significant floating point error may accumulate.[5] Contents 1 Representation error 2 See also 3 References 4 External links Representation error[edit] The error introduced by attempting to represent a number using a finite string of digits is a form of round-off error called representation error.[6] Here are some examples of representation error in decimal representations: Notation Representation Approximation Error floating point roundoff 1/7 0.142857 0.142857 0.000000142857 ln 2 0.69314718055994530941... 0.693147 0.00000018055994530941... log10 2 0.30102999566398119521... 0.3010 0.00002999566398119521... ∛2 1.25992104989487316476... 1.25992 0.00000104989487316476... √2 1.41421356237309504880... 1.41421 0.00000356237309504880... e 2.71828182845904523536... 2.718281828459045 0.00000000000000023536... π 3.14159265358979323846... 3.141592653589793 0.00000000000000023846... Increasing the number of digits allowed in a representation reduces the magnitude of possible round-off errors, but any representation limited to finitely many digits will still cause some degree of round-off error for uncountably many real numbers. Additional digits used for intermediary steps of a calculation are known as guard digits.[7] Rounding multiple times can cause error to accumulate.[8] For example, if 9.945309 is rounded to two decimal places (9.95), then rounded again to one decimal place (10.0), the total error is 0.054691. Rounding 9.945309 to one decimal place (9.9) in a single step introduces less error (0.045309). This commonly occurs when performing arithmetic operations (See Loss of Significance). See also[edit] Precision (arithmetic) Truncation Rounding Loss of significance Floating point Kahan summation algorithm Machine epsilon Wilkin