Conversion from C to VBA

[phudgens]

I am attempting to convert the following C code (being run on Solaris) to VBA for use in Excel for the purposes of client presentation:

theta = r / (q * q * q)
if(fabs((double)theta > 1.)
theta = (theta > 0.) ? 1. : -1.;
theta = acos( theta ) / 3. ;

My VBA conversion:

Theta = R / (Q * Q * Q)
If Abs(Theta) > 1 Then
If Theta > 0 Then Theta = 1
If Theta <= 0 Then Theta = -1
End If
Theta = Application.WorksheetFunction.Acos(Theta) / 3

I am wondering first of all, why there is a (double) definition in the C code since theta has been defined previously as double in that code. I am assuming secondly that the VBA Abs is the same as the C code fabs. Lastly, I'm wondering if there is any reason to believe that a calculation of Acos on a Sun machine would be any different from that calculated on a windows PC. I ask because the answers I'm getting from my code are significantly different from the original C code. The section of code here is the only part about which I have doubts.

Thanks for any help
Paul Hudgens
Denver

[helios]

What's with that stray End If?

[phudgens]

The End If is necessary in VBA when there are 2 or more statements following the original if - then.

[helios]

Oh. I didn't see the first if.

I don't understand your first question:

I am wondering first of all, why there is a (double) definition in the C code since theta has been defined previously as double in that code.

is any reason to believe that a calculation of Acos on a Sun machine would be any different from that calculated on a windows PC.

I think different IEEE standards define different results for floating point operations, but I'm not sure.

[phudgens]

Why is the redefinition of theta ("(double)") being used in the following line of code when theta has been defined as double earlier in the C program?:

if(fabs((double)theta > 1.)


Thanks,
Paul H.

[helios]

That's not a redefinition (there's no such thing as a legal redefinition), it's a cast. A redundant cast, but a cast nonetheless.

By the way, that C if is not valid. Is that an error in the copy you posted, or is that how the code actually looks?

[z05DSL3A]

If you call if(fabs(theta > 1.))^[*], you would probably get a error about an ambiguous call to an overloaded function. casting the result of theta > 1. to double removes the ambiguity.


[*] Your missing a closing parenthesis in you post.

[phudgens]

The full if statement is as follows:

if(fabs((double)theta > 1.)
theta = (theta > 0.) ? 1. : -1.;


I assumed that the "(double)" was redundant and that my VBA code would therefore work:

If Abs(Theta) > 1 Then
If Theta > 0 Then Theta = 1
If Theta <= 0 Then Theta = -1
End If

As I say though, my results do not match with those coming from our Sun based software, if not because of these lines than because of some others. I may need to find a C expert to help with this - can anyone recommend someone in the Denver area? I'm only converting about 50 lines of C code.

Thanks,
Paul H.

[z05DSL3A]

I can see the if statment from your fist post

theta = r / (q * q * q)
if(fabs((double)theta > 1.)
    theta = (theta > 0.) ? 1. : -1.;
theta = acos( theta ) / 3. ;

but the if stament that you have posted has 3 (s but only 2 )s, so one is missing somewhere.

My take on the VBA would be somthing like:

If Abs(theta) > 1 Then
    If theta > 0 Then
        theta = 1
    Else
        theta = -1
    End If
End If

[phudgens]

You're right - there is an extra ) after theta in the original C code (of which I only have a hardcopy). C code should be:

if(fabs((double)theta) > 1.)
theta = (theta > 0.) ? 1. : -1.;
theta = acos( theta ) / 3. ;

My VBA code would have worked, but your's is technically better. Thanks.