[fpc-pascal] fpc 32 / 64 bit / c++/ floating point question

[Michael Müller]

Am 21.11.2010 um 18:54 schrieb Helmut Hartl:

> While stabilizing my bullet physics port i saw some differences in
> floating point behaviour between c++ and fpc32 and fpc 64 bit.
> 
> In question is a simple dot product.
> 
> The function :
> result := a*d+b*e+c*f;
> 
> Gives "different" results( 6.3846106530 vs 6.3846111610 ) depending
> on the usage of intermediate variables for the final result and the fpc version
> 32 bit / 64 bit, while the C++ version does not show this behaviour and always
> gives the same result. (XCode 3.2.1/ GCC on OSX)
> 
> I only show the (extracted) compilable fpc exampe:
> 
> program fpc_floatmul;
> var a,b,c,d,e,f      : single;
>    i,j,k                   : single;
>    result_same_as_c : single;
> begin
>  a:= 1;
>  b:= -5.65984446;
>  c:= -3.4953573;
>  d:= 8.48976051;
>  e:= 1.5;
>  f:= -1.82659933;
>  i := a * d;
>  j := b * e;
>  k := c * f;
>  result_same_as_c := i + j + k;
>  writeln(result_same_as_c:10:10);
>  writeln(a*d+b*e+c*f:10:10); // different result
> end.
> 
> Running the above program gives
> 1) C++ / FPC 64 Bit
> 6.3846106530
> 6.3846106530
> 2) FPC32 Bit
> 6.3846106530
> 6.3846111610
> 
> Is this explainable or wrong behaviour ?

Single has 7 significant digits. So it makes no sense to write more since you can't trust them. If you use in your case 10:6 instead of 10:10 in writeln() you'll get in all cases '6.384611'.
And it makes no sense to store a value with more than these number of digits into a Single type. So perhaps the equation in the last writeln() is done internally in a higher precision than Single. When I use a calculator that can handle all you specified digits than I get 6,384611122290609 which is closer to the last result.
What do you get when you replace Single and float by Double?

Michael