C++ Store double in int variable error

Question

This is a small program that calculates the gaps of 'interval' size that are beetween two numbers 'from' 'to'. Then I calculate the 'size' (number of gaps) and store it in an int variable, and give me a smaller value sometimes.

Here is the code:

double from=0, to=1, interval=0.1;

cout << "WORKING WITH VARIABLES: " << endl;
double operation = (to-from)/interval +1;
cout << "Size: " << operation << endl;

int size = operation;

cout << "Size after storing: " << size << endl << endl;

cout << "WORKING WITHOUT VARIABLES: " << endl;
cout << "Size: " << (to-from)/interval +1 << endl;

size = (to-from)/interval +1;

cout << "Size after storing: " << size << endl << endl;

Problem seems to be in how it's stored interval. If interval=1 everything is good, but if is 0.1, as in the example it give me 10 instead 11 in the "Size after storing" of the second case.

I've found out that it works well with interval=0.25 (2^-2).

EDIT: I haven't found that it fails in the first case, always does in the second.

Answer 1

Floating point numbers are stored with a finite precision, and in binary. 0.25 is easy. That's just 1/4, so 0.01 binary. 0.1 is 1/10, which cannot be represented by a finite binary string. It's 1/16+1/32+ ...

So 1/10 is rounded down, and 10 * 1/10 is slightly less than 1.

As for the different results in the first and second case, that's probably because intermediate values are rounded to more digits than double has.

Answer 2

You are suffering from the inaccuracies inherent in floating point arithmetic. What you probably get in the 0.1 case is 10.999.... instead of 11. Converting double to int truncates instead of rounding so you get 10. Add a small value before converting to int to combat this.

int size=operation+0.0000000001;

Answer 3

If you want to convert a double that is close to an integer (due to rounding errors, as explained by others) to an int, always round the value to the nearest integer, e.g. with round(), before converting the value to int, since double to int conversion truncates the value and will yield an incorrect result if the error is negative.