I will relinquish the award for shortest example because this adds  
one line. This long example demonstrates that FB handles this  
correctly if multiplied times 1000 or anything other than 100. This  
is a real problem for me because my program has records using long  
integers to store real numbers. I thought FB had binary number issues  
like this handled at one time.

dim x#,y&,y2&
x#=1.15
y&=100*x#
y2&=1000*x#
stop str$(y&)+str$(y2&)

Doug



On Feb 27, 2007, at 3:42 PM, George Beckman wrote:

> OK, first,  I believe Doug should get an award for the shortest  
> example of a code problem.   4 lines.   Not bad.
>
>
> Doug, I tend to be fairly clueless about #, & and ! but this  
> works:  (I think a signed integer (&) must be the deal here when  
> you need a real to receive the result of the math)
>
> dim x#,y!
> x#=1.15
> y!=100.00*x#
> stop str$(y!)
>
>
> Now comes the big question.   Is this close to the shortest code  
> solution?
>
>
>
>
> On Feb 27, 2007, at 10:12 AM, Douglas Stemen wrote:
>
>> dim x#,y&
>> x#=1.15
>> y&=100.00*x#
>> stop str$(y&)
>
> ----
> Best Wishes,
>
> George
> http://www.pggp.com
>
>
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