CODEWITHSUNDEEP
javaprecision
user133466
user133466

Reputation: 3415

Loss of precision

I tried running the following code:

double f = 4.35; 
int n = (int) (100 * f);

n will be 434. Why?

Upvotes: 1

Views: 187

Answers (4)

Óscar López
Óscar López

Reputation: 236140

Try this instead:

double f = 4.35; 
int n = (int) Math.round(100 * f);

Now n will have as value 435, as expected. This kind of problems happen because of rounding errors when dealing with decimal values represented in base 2, take a look at this reference to find out more: What Every Computer Scientist Should Know About Floating-Point Arithmetic.

Upvotes: 1

1''
1''

Reputation: 27125

4.35 cannot be represented exactly as a floating-point number since .35 is not a (negative) power of 2. As a result, there's a bit of roundoff one way or another. There will probably also be roundoff when you multiply by 100. If the resulting number is a bit less than 435, casting to int will take the floor of it, which is 434.

If you'd like to round to the NEAREST integer, you can add 0.5 to the result and then cast to int. This is the most general way to round numbers, although Java has a built-in implementation of rounding.

double f = 4.35; 
int n = (int) (100 * f + 0.5);

Upvotes: 2

lrsjng
lrsjng

Reputation: 2625

this is because floating point numbers are internally represented in base 2 not base 10.

this here will help (have to import Math):

double f = 4.35; 
long n = Math.round(100 * f);

Upvotes: 0

Graham Smith
Graham Smith

Reputation: 25757

You need to watch this. I made first year uni students watch it for homework. http://vimeo.com/7403673 - Jon Skeet and Tony the Pony.

Jon Skeet's presentation at the London Stack Overflow DevDays event: stackoverflow.carsonified.com

At around 6:50 he talks about maths in Java and some of the common problems faced by developers with regards to maths.

Upvotes: 1

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