[Discuss] IEEE doubles

[Daniel M German]

Hi Alan,

I was curious, so I ran your program in both PowerPC and Intel. In the
PowerPC it generated:

   1.00000e-200    1.00000e+250     0.00000e+00 
    0  
    0  
           -inf 
----------------------------------------------------------------------
In Intel:

   1.00000e-200    1.00000e+250     0.00000e+00 
    1  
    0  
           -inf 
----------------------------------------------------------------------
My gut feeling was that the reason was the optimization done by the
compiler. So I added a new variable (temp) and then rerun it.  See
the code below. It generates the expected output. In a nutshell, the
compiler realizes that the division can be converted into a
multiplication, which is faster to evaluate (at least in an Intel
instruction set). It is a matter of disassembling the code to make
sure this is exactly what is happening. 


[dmg at ti tmp]$ ./rip 
   1.00000e-200    1.00000e+250     0.00000e+00 
1>    1  
2>    0  
    0  
           -inf 
----------------------------------------------------------------------


#include <stdio.h>
#include <math.h>

int main(void) 
{
   double x=1.e-200, xscale=1.e+250;
   double y;
   double temp;
   int greater;
   printf("%15.5e %15.5e %15.5e \n", x, xscale, x/xscale);

   greater = x/xscale > 0.e0;
   printf("1>%5i  \n", greater);
   temp = x/xscale;
   greater = temp > 0.e0;
   printf("2>%5i  \n", greater);


   greater = x/xscale > 1.e-305;
   printf("%5i  \n", greater);
   y = log(x/xscale);
   printf("%15.5e \n", y);

   return(0);
}



-- 
--
Daniel M. German                  
http://turingmachine.org/
http://silvernegative.com/
dmg (at) uvic (dot) ca
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