Re: Compile time executed functions

[Jean-Marc Bourguet <jbourguet@free.fr>]

Ken Garlington wrote:
> 
> "Jean-Marc Bourguet" <jbourguet@free.fr> wrote in message
> news:3AC84628.D8B70C7C@free.fr...
[...]
> : I think the success of "template meta programming" (as this abuse of the
> : template system is called by some) is the fact that there is no need to
> : count on compiler optimization to get the effect,
> 
> This is a particularly puzzling statement. Why do you believe you are not
> counting on a compiler optimization for the case you presented?

You keep calling evaluating these expressions an optimization.  I think
nobody familiar with C++ will.  There is no formal guarantee but the
trouble would be to make a compiler not compute these at compile time
and still respect the other aspect of the standard.
 
> : and you may also do
> : some other things like computing type.  So running this
> :
> : #include <iostream>
> :
> : template <int F>
> : struct factorial {
> :    enum { RET = factorial<F-1>::RET*F };
> : };
> :
> : template <>
> : struct factorial<1> {
> :    enum { RET = 1 };
> : };
> :
> : template <int F>
> : struct ken {
> :   typedef int T;
> : };
> :
> : template <>
> : struct ken<3628800> {
> :   typedef double T;
> : };
> :
> : int main() {
> :   ken<factorial<10>::RET >::T x10 = 0.5;
> :   ken<factorial<11>::RET >::T x11 = 0.5;
> :   std::cout << "x10=" << x10
> :     << "\nx11=" << x11 << std::endl;
> : }
> :
> : produces
> :
> : x10=0.5
> : x11=0
> :
> : because x10 is a double while x11 is an int.  But I agree, that is still
> : not a formal guarantee that the compiler does not compute the type of
> : x10 and x11 at run time :-)
> 
> I'm not sure what it means to "compute a type." 

The type of x10 is double because factorial<10>::RET evaluates to
3628800 and the type of x11 is int because factorial<11>::RET evaluates
to something different than 3628800.  The above program is exactly the
same as

#include <iostream>

int main() {
  double x10 = 0.5;
  int    x11 = 0.5;
  std::cout << "x10=" << x10
            << "\nx11=" << x11 << std::endl; 
}

-- Jean-Marc