GCC beaten by ICC in stupid trig test!

[Andrew Pinski]

On Mar 14, 2004, at 15:39, Scott Robert Ladd wrote:

> Hello,
>
> Consider the following program, compiled and run on a Pentium 4 
> (Northwood) system:
>
>     #include <math.h>
>     #include <stdio.h>
>
>     double doit(double a)
>     {
>         double s = sin(a);
>         double c = cos(a);
>
>         // should always be 1
>         return s * s + c * c;
>     }
>
>     int main(void)
>     {
>         double a = 1.0, r = 0.0;
>
>         for (int i = 0; i < 100000000; ++i)
>             r += doit(a);
>
>         printf("r = %f\n",r);
>         return 0;
>     }
>

The point here if you know that it is 1.0 then just return 1.0 instead 
of trying to
play tricks with trig functions.  Yes GCC should do better for trig 
functions
but in most cases, the developer was just doing something dumb like the 
above example
which by the way is not a good benchmark anyways because you know that 
the trig
functions can be reduced to just a load of a constant (as ICC does this 
transformation
while GCC does not but could).

Actually what is happening here is that the function doit is being 
inlined and the math in the inner loop is not being pulled out of the 
loop as it is constant just like a is.
So doing the following (aka forces GCC not to inline) will at least get 
GCC to be about
the same ball park (but still nowhere near) as ICC.  The reason why 
still is that ICC will just unroll the loop to be "r = 
doit(a)*100000000.0" so that is the reasons why
ICC is better than GCC at doing this stupid trig test (note this is 
transformation
is only valid if fast-math is on as you loose precision).

Here is a much better benchmark to try, notice that we are doing more 
work now but the point is that ICC is going to be the transformation 
and it not going to see that doit
is constant so it will not pull it out of the loop and it cannot unroll 
the loop into
just being a constant.


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

     double doit(double a)
     {
         double s = sin(a);
         double c = cos(a);

         return s * c;
     }

     int main(void)
     {
         double a = 1.0, r = 0.0;

         for (int i = 0; i < 100000000; ++i, a++)
             r += doit(a);

         printf("r = %f\n",r);
         return 0;
     }