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Is it sufficient to convert a double to a BigDecimal just before addition to retain original precision?

We are solving a numeric precision related bug. Our system collects some numbers and spits their sum.
The issue is that the system does not retain the numeric precision, e.g. 300.7 + 400.9 = 701.599..., while expected result would be 701.6. The precision is supposed to adapt to the input values so we cannot just round results to fixed precision.
The problem is obvious, we use double for the values and addition accumulates the error from the binary representation of the decimal value.
The path of the data is following:
XML file, type xsd:decimal
Parse into a java primitive double. Its 15 decimal places should be enough, we expect values no longer than 10 digits total, 5 fraction digits.
Store into DB MySql 5.5, type double
Load via Hibernate into a JPA entity, i.e. still primitive double
Sum bunch of these values
Print the sum into another XML file
Now, I assume the optimal solution would be converting everything to a decimal format. Unsurprisingly, there is a pressure to go with the cheapest solution. It turns out that converting doubles to BigDecimal just before adding a couple of numbers works in case B in following example:
import java.math.BigDecimal;
public class Arithmetic {
public static void main(String[] args) {
double a = 0.3;
double b = -0.2;
// A
System.out.println(a + b);//0.09999999999999998
// B
System.out.println(BigDecimal.valueOf(a).add(BigDecimal.valueOf(b)));//0.1
// C
System.out.println(new BigDecimal(a).add(new BigDecimal(b)));//0.099999999999999977795539507496869191527366638183593750
}
}
More about this:
Why do we need to convert the double into a string, before we can convert it into a BigDecimal?
Unpredictability of the BigDecimal(double) constructor
I am worried that such a workaround would be a ticking bomb.
First, I am not so sure that this arithmetic is bullet proof for all cases.
Second, there is still some risk that someone in the future might implement some changes and change B to C, because this pitfall is far from obvious and even a unit test may fail to reveal the bug.
I would be willing to live with the second point but the question is: Would this workaround provide correct results? Could there be a case where somehow
Double.valueOf("12345.12345").toString().equals("12345.12345")
is false? Given that Double.toString, according to javadoc, prints just the digits needed to uniquely represent underlying double value, so when parsed again, it gives the same double value? Isn't that sufficient for this use case where I only need to add the numbers and print the sum with this magical Double.toString(Double d) method? To be clear, I do prefer what I consider the clean solution, using BigDecimal everywhere, but I am kind of short of arguments to sell it, by which I mean ideally an example where conversion to BigDecimal before addition fails to do the job described above.

If you can't avoid parsing into primitive double or store as double, you should convert to BigDecimal as early as possible.
double can't exactly represent decimal fractions. The value in double x = 7.3; will never be exactly 7.3, but something very very close to it, with a difference visible from the 16th digit or so on to the right (giving 50 decimal places or so). Don't be mislead by the fact that printing might give exactly "7.3", as printing already does some kind of rounding and doesn't show the number exactly.
If you do lots of computations with double numbers, the tiny differences will eventually sum up until they exceed your tolerance. So using doubles in computations where decimal fractions are needed, is indeed a ticking bomb.
[...] we expect values no longer than 10 digits total, 5 fraction digits.
I read that assertion to mean that all numbers you deal with, are to be exact multiples of 0.00001, without any further digits. You can convert doubles to such BigDecimals with
new BigDecimal.valueOf(Math.round(doubleVal * 100000), 5)
This will give you an exact representation of a number with 5 decimal fraction digits, the 5-fraction-digits one that's closest to the input doubleVal. This way you correct for the tiny differences between the doubleVal and the decimal number that you originally meant.
If you'd simply use BigDecimal.valueOf(double val), you'd go through the string representation of the double you're using, which can't guarantee that it's what you want. It depends on a rounding process inside the Double class which tries to represent the double-approximation of 7.3 (being maybe 7.30000000000000123456789123456789125) with the most plausible number of decimal digits. It happens to result in "7.3" (and, kudos to the developers, quite often matches the "expected" string) and not "7.300000000000001" or "7.3000000000000012" which both seem equally plausible to me.
That's why I recommend not to rely on that rounding, but to do the rounding yourself by decimal shifting 5 places, then rounding to the nearest long, and constructing a BigDecimal scaled back by 5 decimal places. This guarantees that you get an exact value with (at most) 5 fractional decimal places.
Then do your computations with the BigDecimals (using the appropriate MathContext for rounding, if necessary).
When you finally have to store the number as a double, use BigDecimal.doubleValue(). The resulting double will be close enough to the decimal that the above-mentioned conversion will surely give you the same BigDecimal that you had before (unless you have really huge numbers like 10 digits before the decimal point - the you're lost with double anyway).
P.S. Be sure to use BigDecimal only if decimal fractions are relevant to you - there were times when the British Shilling currency consisted of twelve Pence. Representing fractional Pounds as BigDecimal would give a disaster much worse than using doubles.

It depends on the Database you are using. If you are using SQL Server you can use data type as numeric(12, 8) where 12 represent numeric value and 8 represents precision. similarly, for my SQL DECIMAL(5,2) you can use.
You won't lose any precision value if you use the above-mentioned datatype.
Java Hibernate Class :
You can define
private double latitude;
Database:

How do you truncate extra decimals from BigDecimal

I'm building a service that uses a currency converter and forwards the BigDecimal amount to another service. Sometimes, the conversion rate makes it so that the converted amount has close to 34 decimal places, which the downstream service does not accept.
Is there a way to simply truncate (not round) the BigDecimal. So, for example, if the converted amount is 1.23456789 I want neither 1.24, nor 1.3, nor 1.20, or anything of that sort. I simply want to get rid of the decimals that appear after 4. So what I want is 1.23.
I saw a lot of questions on SO related to this, but they all rounded the BigDecimal in some way.
Thanks!

RoundingMode.DOWN effectively truncates your decimal values:
Javadoc says:
Rounding mode to round towards zero. Never increments the digit prior
to a discarded fraction (i.e., truncates). Note that this rounding
mode never increases the magnitude of the calculated value.
BigDecimal dec = new BigDecimal(10.2384235254634623524);
System.out.println(dec.setScale(2, RoundingMode.DOWN));
Will give:
10.23

BigDecimal provides RoundingMode, which you need here is RoundingMode.FLOOR,
System.out.println(new BigDecimal("1.234567").setScale(2, RoundingMode.FLOOR)); // 1.23
System.out.println(new BigDecimal("1.236567").setScale(2, RoundingMode.FLOOR)); // 1.23

you could try treating it like a string
System.out.println(new DecimalFormat("#0.##").format(new BigDecimal("1.23456789")));

BigDecimal also provides Rounding Modes. Try this
BigDecimal bd = new BigDecimal(2.2964556655);
System.out.println(bd.setScale(2,BigDecimal.ROUND_DOWN));

Formatting decimal places in Java

I'm trying to format a double to just 2 decimal places in Java.
I've read the following post:
How to round a number to n decimal places in Java
For some reason, every one of those methods fails to work on certain numbers I have..
For example:
DecimalFormat df = new DecimalFormat("#.##");
normalizedValue = Double.valueOf(df.format(normalizedValue));
If I print normalizedValue I get result similar to the following:
-78.64000000000001
18.97
59.469999999999985
-63.120000000000005
(Note: some are formatted correctly... some aren't)
So, these methods seem to round, but I need something that will remove all decimals after 2 decimal places... Any suggestions?
Thank you.

The string representation given by DecimalFormat.format(...) does indeed have only 2 digits in your example. But as soon as you reconvert it into a double those accuracy issues occur. A binary base format like double will always show such effects when you try to represent "exakt" decimal fractions. You may change to BigDecimal to get rid of such effects.

ArithmeticException thrown during BigDecimal.divide

I thought java.math.BigDecimal is supposed to be The Answer™ to the need of performing infinite precision arithmetic with decimal numbers.
Consider the following snippet:
import java.math.BigDecimal;
//...
final BigDecimal one = BigDecimal.ONE;
final BigDecimal three = BigDecimal.valueOf(3);
final BigDecimal third = one.divide(three);
assert third.multiply(three).equals(one); // this should pass, right?
I expect the assert to pass, but in fact the execution doesn't even get there: one.divide(three) causes ArithmeticException to be thrown!
Exception in thread "main" java.lang.ArithmeticException:
Non-terminating decimal expansion; no exact representable decimal result.
at java.math.BigDecimal.divide
It turns out that this behavior is explicitly documented in the API:
In the case of divide, the exact quotient could have an infinitely long decimal expansion; for example, 1 divided by 3. If the quotient has a non-terminating decimal expansion and the operation is specified to return an exact result, an ArithmeticException is thrown. Otherwise, the exact result of the division is returned, as done for other operations.
Browsing around the API further, one finds that in fact there are various overloads of divide that performs inexact division, i.e.:
final BigDecimal third = one.divide(three, 33, RoundingMode.DOWN);
System.out.println(three.multiply(third));
// prints "0.999999999999999999999999999999999"
Of course, the obvious question now is "What's the point???". I thought BigDecimal is the solution when we need exact arithmetic, e.g. for financial calculations. If we can't even divide exactly, then how useful can this be? Does it actually serve a general purpose, or is it only useful in a very niche application where you fortunately just don't need to divide at all?
If this is not the right answer, what CAN we use for exact division in financial calculation? (I mean, I don't have a finance major, but they still use division, right???).

If this is not the right answer, what CAN we use for exact division in financial calculation? (I mean, I don't have a finance major, but they still use division, right???).
Then I was in primary school1, they taught me that when you divide by 1 by 3 you get a 0.33333... i.e. a recurring decimal. Division of numbers represented in decimal form is NOT exact. In fact for any fixed base there will be fractions (the result of dividing one integer by another) that cannot be represented exactly as a finite precision floating point number in that base. (The number will have a recurring part ...)
When you do financial calculations involving division, you have to consider the what to do with a recurring fraction. You can round it up, or down, or to the nearest whole number, or something else, but basically you cannot just forget about the issue.
The BigDecimal javadoc says this:
The BigDecimal class gives its user complete control over rounding behavior. If no rounding mode is specified and the exact result cannot be represented, an exception is thrown; otherwise, calculations can be carried out to a chosen precision and rounding mode by supplying an appropriate MathContext object to the operation.
In other words, it is your responsibility to tell BigDecimal what to do about rounding.
EDIT - in response to these followups from the OP.
How does BigDecimal detect infinite recurring decimal?
It does not explicitly detect the recurring decimal. It simply detects that the result of some operation cannot be represented exactly using the specified precision; e.g. too many digits are required after the decimal point for an exact representation.
It must keep track of and detect a cycle in the dividend. It COULD HAVE chosen to handle this another way, by marking where the recurring portion is, etc.
I suppose that BigDecimal could have been specified to represent a recurring decimal exactly; i.e. as a BigRational class. However, this would make the implementation more complicated and more expensive to use2. And since most people expect numbers to be displayed in decimal, and the problem of recurring decimal recurs at that point.
The bottom line is that this extra complexity and runtime cost would be inappropriate for typical use-cases for BigDecimal. This includes financial calculations, where accounting conventions do not allow you to use recurring decimals.
1 - It was an excellent primary school. You may have been taught this in high school.
2 - Either you try to remove common factors of the divisor and dividend (computationally expensive), or allow them to grow without bounds (expensive in space usage and computationally expensive for subsequent operations).

The class is BigDecimal not BigFractional. From some of your comments it sounds like you just want to complain that someone didn't build in all possible number handling algorithms into this class. Financial apps do not need infinite decimal precision; just perfectly accurate values to the precision required (typically 0, 2, 4, or 5 decimal digits).
Actually I have dealt with many financial applications that use double. I don't like it but that was the way they are written (not in Java either). When there are exchange rates and unit conversions then there are both the potential of rounding and bruising problems. BigDecimal eliminates the later but there is still the former for division.

If you want to work with decimals, not rational numbers, and you need exact arithmetics before the final rounding (rounding to cents or something), here's a little trick.
You can always manipulate your formulas so that there's only one final division. That way you won't lose precision during calculations and you'll always get the correctly rounded result. For instance
a/b + c
equals
(a + bc) / b.

By the way, I'd really appreciate
insight from people who've worked with
financial software. I often heard
BigDecimal being advocated over double
In financial reports we use alwasy BigDecimal with scale = 2 and ROUND_HALF_UP, since all printed values in a report must be lead to a reproducable result. If someone checks this using a simple calculator.
In switzerland they round to 0.05 since they no longer have 1 or 2 Rappen coins.

You should prefer BigDecimal for finance calculations. Rounding should be specified by the business. E.g. an amount (100,00$) has to be split equally across three accounts. There has to be a business rule which account takes the extra cent.
Double, floats are not approriate for use in financial applications because they can not represent fractions of 1 precisely that are not exponentials of 2. E.g. consider 0.6 = 6/10 = 1*1/2 + 0*1/4 + 0*1/8 + 1*1/16 + ... = 0.1001...b
For mathematic calculations you can use a symbolic number, e.g. storing denominator and numerator or even a whole expression (e.g. this number is sqrt(5)+3/4). As this is not the main use case of the java api you won' find it there.

Is there a need for
a=1/3;
b=a*3;
resulting in
b==1;
in financial systems? I guess not. In financial systems it is defined, which roundmode and scale has to be used, when doing calculations. In some situations, the roundmode and scale is defined in the law. All components can rely on such a defined behaviour. Returning b==1 would be a failure, because it would not fulfill the specified behaviour. This is very important when calculating prices etc.
It is like the IEEE 754 specifications for representing floats in binary digits. A component must not optimize a "better" representation without loss of information, because this will break the contract.

To divide save, you have to set the MATHcontext,
BigDecimal bd = new BigDecimal(12.12, MathContext.DECIMAL32).divide(new BigDecimal(2)).setScale(2, RoundingMode.HALF_UP);

I accept that Java doesn't have great support for representing fractions, but you have to realise that it is impossible to keep things entirely precise when working with computers. At least in this case, the exception is telling you that precision is being lost.
As far as I know, "infinite precision arithmetic with decimal numbers" just isn't going to happen. If you have to work with decimals, what you're doing is probably fine, just catch the exceptions. Otherwise, a quick google search finds some interesting resources for working with fractions in Java:
http://commons.apache.org/math/userguide/fraction.html
http://www.merriampark.com/fractions.htm
Best way to represent a fraction in Java?

Notice we are using a computer... A computer has a lot of ram and precision takes ram. So when you want an infinite precision you need
(infinite * infinite) ^ (infinite * Integer.MAX_VALUE) terrabyte ram...
I know 1 / 3 is 0.333333... and it should be possible to store it in ram like "one divided by three" and then you can multiply it back and you should have 1. But I don't think Java has something like that...
Maybe you have to win the Nobel Price for writing something doing that. ;-)

Use of java.math.MathContext

Recently I tried understanding the use of java.math.MathContext but failed to understand properly. Is it used for rounding in java.math.BigDecimal. If yes why does not it round the decimal digits but even mantissa part.
From API docs, I came to know that it follows the standard specified in ANSI X3.274-1996 and ANSI X3.274-1996/AM 1-2000 specifications but I did not get them to read online.
Please let me know if you have any idea on this.

For rounding just the fractional part of a BigDecimal, check out the BigDecimal.setScale(int newScale, int roundingMode) method.
E.g. to change a number with three digits after the decimal point to one with two digits, and rounding up:
BigDecimal original = new BigDecimal("1.235");
BigDecimal scaled = original.setScale(2, BigDecimal.ROUND_HALF_UP);
The result of this is a BigDecimal with the value 1.24 (because of the rounding up rule)

#jatan
Thanks for you answer. It makes sense. Can you please explain me MathContext in the context of BigDecimal#round method.
There's nothing special about BigDecimal.round() vs. any other BigDecimal method. In all cases, the MathContext specifies the number of significant digits and the rounding technique. Basically, there are two parts of every MathContext. There's a precision, and there's also a RoundingMode.
The precision again specifies the number of significant digits. So if you specify 123 as a number, and ask for 2 significant digits, you're going to get 120. It might be clearer if you think in terms of scientific notation.
123 would be 1.23e2 in scientific notation. If you only keep 2 significant digits, then you get 1.2e2, or 120. By reducing the number of significant digits, we reduce the precision with which we can specify a number.
The RoundingMode part specifies how we should handle the loss of precision. To reuse the example, if you use 123 as the number, and ask for 2 significant digits, you've reduced your precision. With a RoundingMode of HALF_UP (the default mode), 123 will become 120. With a RoundingMode of CEILING, you'll get 130.
For example:
System.out.println(new BigDecimal("123.4",
new MathContext(4,RoundingMode.HALF_UP)));
System.out.println(new BigDecimal("123.4",
new MathContext(2,RoundingMode.HALF_UP)));
System.out.println(new BigDecimal("123.4",
new MathContext(2,RoundingMode.CEILING)));
System.out.println(new BigDecimal("123.4",
new MathContext(1,RoundingMode.CEILING)));
Outputs:
123.4
1.2E+2
1.3E+2
2E+2
You can see that both the precision and the rounding mode affect the output.

I would add here, a few examples. I haven't found them in previous answers, but I find them useful for those who maybe mislead significant digits with number of decimal places. Let's assume, we have such context:
MathContext MATH_CTX = new MathContext(3, RoundingMode.HALF_UP);
For this code:
BigDecimal d1 = new BigDecimal(1234.4, MATH_CTX);
System.out.println(d1);
it's perfectly clear, that your result is 1.23E+3 as guys said above. First significant digits are 123...
But what in this case:
BigDecimal d2 = new BigDecimal(0.000000454770054, MATH_CTX);
System.out.println(d2);
your number will not be rounded to 3 places after comma - for someone it can be not intuitive and worth to emphasize. Instead it will be rounded to the first 3 significant digits, which in this case are "4 5 4". So above code results in 4.55E-7 and not in 0.000 as someone could expect.
Similar examples:
BigDecimal d3 = new BigDecimal(0.001000045477, MATH_CTX);
System.out.println(d3); // 0.00100
BigDecimal d4 = new BigDecimal(0.200000477, MATH_CTX);
System.out.println(d4); // 0.200
BigDecimal d5 = new BigDecimal(0.000000004, MATH_CTX);
System.out.println(d5); //4.00E-9
I hope this obvious, but relevant example would be helpful...

If I'm understanding you correctly, it sounds like you're expecting the MathContext to control how many digits should be kept after the decimal point. That's not what it's for. It specifies how many digits to keep, total. So if you specify that you want 3 significant digits, that's all you're going to get.
For example, this:
System.out.println(new BigDecimal("1234567890.123456789",
new MathContext(20)));
System.out.println(new BigDecimal("1234567890.123456789",
new MathContext(10)));
System.out.println(new BigDecimal("1234567890.123456789",
new MathContext(5)));
will output:
1234567890.123456789
1234567890
1.2346E+9

It's not for fun. Actually I found some online example, which stated the use of MathContext to round the amounts/numbers stored in BigDecimal.
For example,
If MathContext is configured to have precision = 2 and rounding mode = ROUND_HALF_EVEN
BigDecimal Number = 0.5294, is rounded to 0.53
So I thought it is a newer technique and used it for rounding purpose. However it turned into nightmare because it started rounding even mentissa part of number.
For example,
Number = 1.5294 is rounded to 1.5
Number = 10.5294 is rounded to 10
Number = 101.5294 is rounded to 100
.... and so on
So this is not the behavior I expected for rounding (as precision = 2).
It seems to be having some logic because from patter I can say that it takes first two digits (as precision is 2) of number and then appends 0's till the no. of digits become same as unrounded amount (checkout the example of 101.5294 ...)