Convert double to int in C++ without round down errors

Question

I have the following codes to cast a double into an int:

double dblValue = 7.30;
int intValue = (int)(dblValue*100);  //I want intValue to store exactly 730;
std::cout << intValue;

Output: 729

I know that the compiler is reading dblValue as 7.2999999 before casting it to int.

My question is: Is it possible to cast it as 730 by preventing the round down errors?

It would be best if your solution avoid using C++11 or other predefined functions. The only pre-processor directive I am using here is <iostream>.

Answer 1

You cannot prevent rounding errors when converting a number that is not an integer (in the mathematical sense) to an integer, the only thing you can do is try to achieve proper rounding.

The easiest way to achieve a sensible (although not perfect) rounding is the following:

int intValue = (int)(dblValue < 0 ? dblValue - 0.5 : dblValue + 0.5);

And of course, since your question is tagged both c++ and casting I cannot resist replacing your c-style cast with a c++ style cast:

int intValue = static_cast<int>(dblValue < 0 ? dblValue - 0.5 : dblValue + 0.5);

Answer 2

You can define your own integer truncation function that increases the value by the smallest possible amount, to make sure the rounded result is over the integer threshold.

#include <limits>

int int_cast(double x)
{
    return (int)(x * (1.0 + std::numeric_limits<double>::epsilon()));
}

If you don't want to depend on <limits> you can use DBL_EPSILON from <float.h> or substitute your own very small number. See also this question.

Answer 3

C++11 added lround that helps to avoid precision loss on implicit double to long truncating cast:

int intValue = (int) lround(dblValue*100)

long to int cast won't loose precision too.

There is no iround(), unfortunately.

Update

I guess there is no iround() because any 32-bit integer will allways fit completely in 52 precision bits of 64-bit double. So there is no chance of precision loss on direct double to int truncation:

int intValue = (int) round(dblValue*100)

Answer 4

It's not any type of error, it's the way computer stores floating-point values. You can do something like this:

int intValue = (int)(dblValue*100 + 0.00001);