What, exactly, is the problem? Computer floating point
arithmetic is not exact. 0.999999999999943 is very close
to 1 (5.7E-12% deviation is negligible for most
applications).

Please note that C++ does it right!

#include <iostream
#include <math.h
using namespace std;
int main()
{
double tempRound;
double tempInt, tempDec;
double number;
number = 64.1;

tempRound =floor(number*10+0.5)/10;
tempInt = (int)(tempRound);
tempDec = 10*(tempRound - tempInt);

cout << "tempRound = " << tempRound << endl;
cout << "tempInt = " << tempInt << endl;
cout << "tempDec = " << tempDec << endl;

system("pause");
return 0;
}

gives the right result:

tempRound = 64.1
tempInt = 64
tempDec = 1

Hmmm. Can one trust VBA or not?
Karl
-----Original Message-----
What, exactly, is the problem? Computer floating point
arithmetic is not exact. 0.999999999999943 is very close
to 1 (5.7E-12% deviation is negligible for most
applications).
.

On Thu, 24 Jul 2003 at 03:18:29, Karl Itschen (Karl Itschen
) wrote:
Please note that C++ does it right!

So does VBA if you understand how to manipulate data types. I think C++
may use more bytes to store

Dim number, tempRound As Double

Dim tempInt As Long
Dim tempDec As Long

number = 64.111
tempRound = number
tempInt = tempRound
tempDec = (10 * (tempRound - tempInt))

Debug.Print "tempRound = " & tempRound
Debug.Print "tempInt = " & tempInt
Debug.Print "tempDec = " & tempDec


tempRound = 64.1
tempInt = 64
tempDec = 1
--
Mike

On Thu, 24 Jul 2003 at 03:18:29, Karl Itschen (Karl Itschen
) wrote:
Please note that C++ does it right!

So does VBA if you know how to manipulate the data types

Dim number, tempRound As Double

Dim tempInt As Long
Dim tempDec As Long

number = 64.111
tempRound = number
tempInt = tempRound
tempDec = (10 * (tempRound - tempInt))

Debug.Print "tempRound = " & tempRound
Debug.Print "tempInt = " & tempInt
Debug.Print "tempDec = " & tempDec

tempRound = 64.111
tempInt = 64
tempDec = 1

I don't know C++, but my guess is that 'cout' has some default value of
number of dp to display
--
Mike

"Karl Itschen" wrote in message
...
Please note that C++ does it right!

#include <iostream
#include <math.h
using namespace std;
int main()
{
double tempRound;
double tempInt, tempDec;
double number;
number = 64.1;

tempRound =floor(number*10+0.5)/10;
tempInt = (int)(tempRound);
tempDec = 10*(tempRound - tempInt);

cout << "tempRound = " << tempRound << endl;
cout << "tempInt = " << tempInt << endl;
cout << "tempDec = " << tempDec << endl;

system("pause");
return 0;
}

gives the right result:

tempRound = 64.1
tempInt = 64
tempDec = 1


Well of course it does

The default precison used for doubles by cout
is only 6 decimal places

Try

cout.precision(15);

if you want a real comparison.

http://msdn.microsoft.com/library/de...recisio n.asp

Keith