Losing precision converting from double to int in Java

Question

I tried to test the following code, and it gave me a 434 on n, the result which I did not anticipate, what's the reason for this loss of precision?

double f = 4.35;
int n =  (int) (100 * f); // n will be 434 - why?
n = (int) Math.round(100*f); // n will be 435

Answer 1

Floating point isn't perfect. It uses approximated values.

In particular, doubles can not represent all numbers. Just like 1/3 can't be precisely represented in decimal using a finite number of digits, so too can't 4.35 be represented in binary with a finite number of bits.

So we get rounded results. Values that are close to 4.35, but not quite equal. In this case, it's a bit smaller, so when you multiply it by 100, you don't get 435, you get almost 435, which is not quite the same.

When you cast to int, you truncate the result, so 434,999... becomes just 434.

In contrast, when you use Math.round, you convert 434,999... to exactly 435 (which can be represented precisely).

If precisely representing 4.35 is necessary, take a look at Java BigDecimal type.

Answer 2

Floating-point cannot represent exactly the floating-point number.

You can try printing it, you'll see that 100 * f = 434.9999999. The int value for that double will be 434 as it is truncating the decimal part.

The round method will round to the closest int number, which is 435.