Casting changes value strangely

Question

Why does a1=72 instead of 73 in this (terrible) snippet of C++ code ?

#include <iostream>
#include <string>

using namespace std;

int main (int argc, char* argv[])
{   
    double       a  = 136.73;
    unsigned int a1 = (100*(a-(int)a));

    cout << (a-(int)a)     << endl; // 0.73
    cout << 100*(a-(int)a) << endl; // 73
    cout << a1             << endl; // 72 !
}

You can execute it at http://codepad.org/HhGwTFhw

Answer 1

If you increase the output precision, you'll see that (a - (int) a) prints 0.7299999999999898.

Therefore, the truncation of this value (which you obtain when you cast it to an int) is indeed 72.

(Updated codepad here.)

Answer 2

This is a common precision issue. The literal 136.73 actually stands for the number

136.729999999999989768184605054557323455810546875

and the result of a-(int)a is not 0.73 (even though that is what is displayed), but rather

0.729999999999989768184605054557323455810546875

When you multiply that by 100, you get

72.9999999999989768184605054557323455810546875

And since converting from double to int cuts off everything after the decimal point, you get 72.

Answer 3

0.73 cannot be represented exactly so it is rounded to a number close to it, which in this example is lower then 0.73. when multiplying by 100, you get 72.[something], which is later trimmed to 72.

more info