🚀 go-pugleaf

In message <9512120442.AA00637@fafnir.fafnir.com> you write:
> gcc version 2.7.2
> NeXT 68040 Cube, NeXTSTEP 3.3 with patch 1
> 
> 1) gcc often fails to move function calls out of a loop, even when  
> they are declared "const"
> I hope I have done something stupid here; I was counting on GCC's  
> loop optimization.

Yep, you were 8-); your code declared the return value const, not the
function.  Please feast upon the following GNU extention (from the GNU
info pages):

`const'
     Many functions do not examine any values except their arguments,
     and have no effects except the return value.  Such a function can
     be subject to common subexpression elimination and loop
     optimization just as an arithmetic operator would be.  These
     functions should be declared with the attribute `const'.  For
     example,

          int square (int) __attribute__ ((const));

     says that the hypothetical function `square' is safe to call fewer
     times than the program says.

     The attribute `const' is not implemented in GNU C versions earlier
     than 2.5.  An alternative way to declare that a function has no
     side effects, which works in the current version and in some older
     versions, is as follows:

          typedef int intfn ();
          
          extern const intfn square;

     This approach does not work in GNU C++ from 2.6.0 on, since the
     language specifies that the `const' must be attached to the return
     value.

Hope this helps,
Rusty.
--
Rusty.Russell@adelaide.maptek.com.au           "Engineer? So you drive trains?"
Lies, damned lies, and out-of-date documentation.        #include <stddisclm.h>