One way to determine if tail-call is happening is to see if you can force a stack overflow. The following program does not produce a stack overflow using VC++ 2005 Express Edition and, even though its results exceed the capacity of long double rather quickly, you can tell that all of the iterations are being processed when TCO is happening:
/* FibTail.c 0.00 UTF-8 dh:2008-11-23
* --|----1----|----2----|----3----|----4----|----5----|----6----|----*
*
* Demonstrate Fibonacci computation by tail call to see whether it is
* is eliminated through compiler optimization.
*/
#include <stdio.h>
long double fibcycle(long double f0, long double f1, unsigned i)
{ /* accumulate successive fib(n-i) values by tail calls */
if (i == 0) return f1;
return fibcycle(f1, f0+f1, --i);
}
long double fib(unsigned n)
{ /* the basic fib(n) setup and return. */
return fibcycle(1.0, 0.0, n);
}
int main(int argc, char* argv[ ])
{ /* compute some fibs until something breaks */
int i;
printf("\n i fib(i)\n\n");
for (i = 1; i > 0; i+=i)
{ /* Do for powers of 2 until i flips negative
or stack overflow, whichever comes first */
printf("%12d %30.20LG \n", i, fib((unsigned) i) );
}
printf("\n\n");
return 0;
}
Notice, however, that the simplifications to make a pure tail-call in fibcycle is tantamount to figuring out an interative version that doesn't do a tail-call at all (and will work with or without TCO in the compiler.
It might be interesting to experiment in order to see how well the TCO can find optimizations that are not already near-optimal and easily replaced by iterations.