http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57749
--- Comment #18 from Steve Kargl <sgk at troutmask dot apl.washington.edu> --- On Tue, Jul 02, 2013 at 07:21:54AM +0000, zeccav at gmail dot com wrote: > http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57749 > > --- Comment #16 from Vittorio Zecca <zeccav at gmail dot com> --- > and that there are only two ifs to put in cpow to avoid the floating exception > and give the expected result(I am simplifying here, also because I do > not use C): > > if(base==0) > { > if(exponent>0) return 0; else raise hell; > } > > The actual code where the original issue occurred had the exponentiation > in the deep of nested loops, it would have been rather time consuming > to test base==0 > at the Fortran level It will be time consuming wherever it is tested. That's my entire point and why the C11 standard permits cpow(z,c) to be implemented as cexp(c*clog(z)) without trying to deal with all of the special cases (or accuracy issues). > And I still do not understand why if the exponent is integer no > exception is raised and > the expected result zero is delivered. > As in the following fragment (with option -ffpe-trap=zero,invalid): > complex x > x=cmplx(0e0,0e0) > i=2 > r=2e0 > print *,x**i ! no exception raised delivers zero The compiler knows that i is an integer, and the above case it is 2. The compiler evaluates x**2 as x*x. > print *,x**r ! exception raised > end > The Intel ifort and NAG nagfor compilers raise no exceptions and > deliver the expected result. While it may be possible for a compiler to determine that r in r=2e0 is an integral value of 2 and replace x**r by x*x, I suspect that it will fail in the general case. What does r = 8.125 x = (0.,0.) print *, x**r ! x**8 * sqrt(sqrt(sqrt(x))) = 0. or r = 1. / 3 x = (0.,0,) print *, x**r ! cube root of x? produce?
