From: John Beckett 

"Geo"  wrote in message
news::
> Are you sure it's not just stored as an int? I mean if it's integer math
> then how would it handle this
>
> int n = (rand()/32767)*9999999999999
>
> That shouldn't generate a cpu error, the error would come when it goes to
> store a multibyte value in a single byte location (or maybe at compile
> time?)

I don't follow your meaning. If you intend that the
"9999999999999" is too large for an int, the compiler would
assume it is a long int. If the value is too large for a long int,

In Visual C++, int and long int are the same (32 bits), and one more step
is added to the above. The compiler next tries to fit the value into a
64-bit integer (__int64). If not successful, the compiler gives a warning
(or maybe an error in some situations?).

A C/C++ compiler won't ever take an integer value (9999999999999) and
represent it as a float or double (unless you ask it to do that).

By contrast, VB *will* freely convert stuff to whatever it feels like.

At any rate, rand() is an int, and the other values in your expression were
also ints (or longer). Therefore, int operations (or longer integer
operations) will be performed.

Consider:
double x = sqrt(2);
int a = x;                      // compile error
int b = (int)x;

You can't literally put x in a (x uses maybe a 64-bit floating-point code,
and a uses maybe a 32-bit binary code).

However, if you use the cast, as in '(int)x', you ask the compiler to
generate code that will convert x to an int. The result will discard the
fractional digits, keeping only the integer part. In this case, b == 1.

John

--- BBBS/NT v4.01 Flag-5