I've stumbled upon an issue with CInt and converting a double to an integer.
The issue is the following:
CInt(10.5) 'Result is 10
CInt(10.51) 'Result it 11, but I expected 10...
I got used to C# style conversion where (int) 10.51
is 10.
As pointed out in the question about Integer.Parse vs CInt, the result is just rounded in some fashion.
However, all I need is to get only integer part and throw away the fractional one. How can I achieve such type of conversion in VB.NET? After some research I see that I can use the Fix()
function to do the trick, but is it the best choice?
I think you can try
CInt(Math.Floor(10.51))
hope this helpsFirstly, your assumption that
CInt
is equivalent to(int)
in C# is incorrect.Secondly, the rounding behaviour of
CInt
is not randomly assigned - it actually uses "bankers rounding":The best equivalent to using
(int)
in C# is theFix
function in theVisualBasic
namespace which rounds towards zero (same asMath.Truncate
).This however returns a Double value so you have to do a further conversion to get to your integer using
CInt
.You may use
Int
orFix
functions but return value type of these functions is double so you have to convert it to Integer ifoption strict
ison
.