I've written a rudimentary algorithm in Fortran 95 to calculate the gradient of a function (an example of which is prescribed in the code) using central differences augmented with a procedure known as Richardson extrapolation.
function f(n,x)
! The scalar multivariable function to be differentiated
integer :: n
real(kind = kind(1d0)) :: x(n), f
f = x(1)**5.d0 + cos(x(2)) + log(x(3)) - sqrt(x(4))
end function f
!=====!
!=====!
!=====!
program gradient
!==============================================================================!
! Calculates the gradient of the scalar function f at x=0using a finite !
! difference approximation, with a low order Richardson extrapolation. !
!==============================================================================!
parameter (n = 4, M = 25)
real(kind = kind(1d0)) :: x(n), xhup(n), xhdown(n), d(M), r(M), dfdxi, h0, h, gradf(n)
h0 = 1.d0
x = 3.d0
! Loop through each component of the vector x and calculate the appropriate
! derivative
do i = 1,n
! Reset step size
h = h0
! Carry out M successive central difference approximations of the derivative
do j = 1,M
xhup = x
xhdown = x
xhup(i) = xhup(i) + h
xhdown(i) = xhdown(i) - h
d(j) = ( f(n,xhup) - f(n,xhdown) ) / (2.d0*h)
h = h / 2.d0
end do
r = 0.d0
do k = 3,M r(k) = ( 64.d0*d(k) - 20.d0*d(k-1) + d(k-2) ) / 45.d0
if ( abs(r(k) - r(k-1)) < 0.0001d0 ) then
dfdxi = r(k)
exit
end if
end do
gradf(i) = dfdxi
end do
! Print out the gradient
write(*,*) " "
write(*,*) " Grad(f(x)) = "
write(*,*) " "
do i = 1,n
write(*,*) gradf(i)
end do
end program gradient
In single precision it runs fine and gives me decent results. But when I try to change to double precision as shown in the code, I get an error when trying to compile claiming that the assignment statement
d(j) = ( f(n,xhup) - f(n,xhdown) ) / (2.d0*h)
is producing a type mismatch real(4)/real(8)
. I have tried several different declarations of double precision, appended every appropriate double precision constant in the code with d0
, and I get the same error every time. I'm a little stumped as to how the function f
is possibly producing a single precision number.
The problem is that f is not explicitely defined in your main program, therefore it is implicitly assumed to be of single precision, which is the type real(4) for gfortran.
I completely agree to the comment of High Performance Mark, that you really should use implicit none
in all your fortran code, to make sure all object are explicitely declared. This way, you would have obtained a more appropriate error message about f not being explicitely defined.
Also, you could consider two more things:
Define your function within a module and import that module in the main program. It is a good practice to define all subroutines/functions within modules only, so that the compiler can make extra checks on number and type of the arguments, when you invoke the function.
You could (again in module) introduce a constant for the precicision and use it everywhere, where the kind of a real must be specified. Taking the example below, by changing only the line
integer, parameter :: dp = kind(1.0d0)
into
integer, parameter :: dp = kind(1.0)
you would change all your real variables from double to single precision. Also note the
_dp
suffix for the literal constants instead of thed0
suffix, which would automatically adjust their precision as well.module accuracy implicit none integer, parameter :: dp = kind(1.0d0) end module accuracy module myfunc use accuracy implicit none contains function f(n,x) integer :: n real(dp) :: x(n), f f = 0.5_dp * x(1)**5 + cos(x(2)) + log(x(3)) - sqrt(x(4)) end function f end module myfunc program gradient use myfunc implicit none real(dp) :: x(n), xhup(n), xhdown(n), d(M), r(M), dfdxi, h0, h, gradf(n) : end program gradient
来源:https://stackoverflow.com/questions/16446397/data-type-mismatch-in-fortran