Currency Conundrum

In the spirit of illustrating the inaccuracies of floating point math, this is a great example from the book Effective Java:

Question: How many items can you buy if you have $1 if each item you buy costs 10 cents more than the previous purchase, starting at 10 cents?

The answer is 4 items (one item for .10, .20, .30, and .40, all add up to $1). A pretty straight forward math problem involving only four operations and one decimal place. However if you implement the solution using doubles (the highest primitive precision floating point available) thusly:

double funds = 1.00;
int itemsBought = 0;
for (double price = 0.10; funds >= price; price += 0.10) {
funds -= price;
itemsBought++;
}

You will get the itemsBounght equal to 3. Why? Because of a cumulative rounding error even after just three operations. When the loop test comes around for the forth time price is 0.4, but funds is only 0.3999999999999999. In practice this would be rounded up to 0.4 if you were going to display it, but since it's not here, it fails the comparison test.

The moral of this story is: Be mindful of how you use floating point numbers. If exact answers are required use integral data types, or a custom object capable of holding larger scaled numbers (I.E. BigDecimal).

Since money is an important thing to get right, when I implemented the accounting part of Iris, I wrote the Money class as a wrapper around BigDecimal. All math operations using money use the Money class. There have been several (well, 2) occassions where other parties have come up with different invoice amounts when manually calculating invoices out of Iris -- and Iris has proved to be right in both cases.

Here's the answer to that programming problem using the Money class:

int itemsBought = 0;
Money funds = new Money(1.0);
for (Money price = new Money(0.1); funds.compareTo(price) >= 0;
price = price.add(new Money (0.1))) {

  itemsBought++;
  funds = funds.subtract(price);
}

Which gives the correct answer of 4.