Daniel Hoelbling-Inzko talks about programming
Ok, this is really simple, but it just has cost me 15 minutes because I ignored a compiler warning:
double result = myInt1/myInt2;
The compiler will underline that there is a possible loss in fraction, but I ignored it while not seeing anything wrong with the code at all.
Whatever values I calculated, the result was always an integer. Turns out, dividing two integers will also create an integer instead of a double (so I got weird results like 31/10 = 3 ).
To bypass this, you need to promote at least one of your integers to a double so a double divide will occur:
double result = (double)myInt1 / myInt2;
Hope this helps, took me far too long to figure out.