Brian Dessent wrote:
Angelo Graziosi wrote:
... and in Fortran?
As long as you're using a recent gcc you can just use -mpc64.
How recent?
With GFortran 4.3.1 and
$ cat test_case.0.f90
program test_case
implicit none
integer :: k
integer, parameter :: DP = kind(1.D0),&
N = 29
real(DP), parameter :: A(0:N) = &
(/1.0000000000000D0,3.8860009363293D0,7.4167881843083D0,&
9.8599837415463D0,10.431383465276D0,9.4077161304727D0,&
7.5332956276775D0,5.4995844630326D0,3.7275104474917D0,&
2.3766149078263D0,1.4394444941337D0,0.8344437543616D0,&
0.4657058294147D0,0.2514185849207D0,0.1317920740387D0,&
0.0672995854816D0,0.0335554996094D0,0.0163785206816D0,&
0.0078368645385D0,0.0036846299276D0,0.0017020281304D0,&
0.0007756352209D0,0.0003469750783D0,0.0001544224626D0,&
0.0000666379875D0,0.0000295655909D0,0.0000118821415D0,&
0.0000057747681D0,0.0000017502652D0,0.0000014682034D0/)
real(DP), parameter :: BB(0:N-1) = &
(/(((k+2)*A(k+2)-A(k+1)*(A(1)+(k+1))),k = 0,N-2),&
(-A(N)*(A(1)+N))/)
real(DP) :: b(0:N-1),bbb(0:N-1)
do k = 0,N-2
b(k) = (k+2)*A(k+2)-A(k+1)*(A(1)+(k+1))
enddo
b(N-1) = -A(N)*(A(1)+N)
bbb(0:N-1) = (/(((k+2)*A(k+2)-A(k+1)*(A(1)+(k+1))),k = 0,N-2),&
(-A(N)*(A(1)+N))/)
do k = 0,N-1
print *, k,b(k)-BB(k),b(k)-bbb(k),BB(k)-bbb(k)
enddo
end program test_case
$ gfortran -mpc64 test_case.0.f90 -o test_case.0
$ ./test_case.0
0 -1.77635683940025046E-015 -1.77635683940025046E-015
0.0000000000000000
1 -3.55271367880050093E-015 -3.55271367880050093E-015
0.0000000000000000
2 -3.55271367880050093E-015 -3.55271367880050093E-015
0.0000000000000000
3 -7.10542735760100186E-015 -7.10542735760100186E-015
0.0000000000000000
4 0.0000000000000000 0.0000000000000000
0.0000000000000000
5 0.0000000000000000 0.0000000000000000
0.0000000000000000
...
Only 13 (over 28) 0. 0. 0.!
Angelo
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