问题
I would like to write Fortran code that works like the R function seq(). E.g.:
x <- seq(0,1,0.1)
will give the vector
x <- c(0, 0.1, 0.2, ..., 1)
I will run several simulations over which the length of the sequence will change, in R this is easily done, by just varying the second argument in seq(). I have tried to do something like this in Fortran with dynamical arrays and the function ALLOCATE to dynamically change the size of the array. This has not worked so far and lead to the error
Program received signal SIGSEGV: Segmentation fault - invalid memory reference.
Backtrace for this error:
#0 0x2B371ED7C7D7
#1 0x2B371ED7CDDE
#2 0x2B371F3B8FEF
#3 0x401BE9 in MAIN__ at test3D.f90:?
Segmentation fault (core dumped)
so I was wondering whether there is an easy way to mimic the behaviour of the R function seq() in Fortran.
For further reference see the program below
program ffl
implicit none
integer, parameter :: n = 2**12
integer :: m,j,l,o,num,r,posi
real(kind=8), dimension(n) :: results
real(kind=8) :: dt,dk,dp, dtt, laenge, basal, periode,c
real(kind=8), dimension(n,n) :: fitness, k_opt
real(kind=8) :: t0,t1,t2,t3
real(kind=8), dimension(:),allocatable :: t
real(kind=8), dimension(n) :: k,p, tt1
real(kind=8), dimension(6) :: x_new, res, q0
real(kind=8), dimension(6) :: k1,k2,k3,k4
real(kind=8) :: ts = 0.0
real(kind=8) :: ks = 0.0, ke = 1.0
real(kind=8) :: ps = 0.1, pe = 40.0
real(kind=8) :: tts = 0.0, tte = 1.0
real(kind=8), dimension(6) :: u0,f1,f2,f3,u1
external :: derivate
! computing the vectors
dk=(ke-ks)/real(n) ! calculating resolution
dp=(pe-ps)/real(n) ! calculating resolution
dtt=(tte-tts)/real(n) ! calculating resolution
k(1) = ks ! first value for k = 0.0
p(1) = ps ! first value for p = 0.001
tt1(1) = tts ! first value for tts = 0.0
num = 10
do m = 1,n
k(m) = k(m-1)+dk ! setting the basal expression vector with resolution dt
tt1(m) = tt1(m-1)+dtt
end do
do m = 1,n
p(m) = ps + 0.1
end do
do m = 1,n
periode = p(m)
do j = 1,n
laenge = tt1(j)
do l = 1,n
basal = k(l)
c = num * periode ! calculating the length of the simulation
dt=(c-ts)/real(n) ! calculating time resolution
r = 1
t(1) = ts ! setting first time value to t1 = 0
allocate(t(1)) ! Initialize array dimension
do while (ts + dt < c)
t(r) = ts
ts = ts + dt
r = r + 1
call resize_array
end do
! initial conditions
q0(1) = 0 ! x
q0(2) = basal ! y
q0(3) = 0 ! z
q0(4) = 0 ! a
q0(5) = 1 ! b
q0(6) = 0 ! w
x_new = q0 ! set initial conditions
! Solving the model using a 4th order Runge-Kutta method
do o = 1,n
call derivate(basal,periode,laenge,t(l),x_new,k1)
t1 = t(o) + dt/2
f1 = x_new + (dt*k1)/2
call derivate(basal,periode,laenge,t1,f1,k2)
t2 = t(o) + dt/2
f2 = x_new + (dt*k2)/2
call derivate(basal,periode,laenge,t2,f2,k3)
t3 = t(o) + dt
f3 = x_new + (dt*k3)/2
call derivate(basal,periode,laenge,t3,f3,k4)
res = x_new + (dt*(k1+2*k2+2*k3+k4))/6
if (res(2) < basal) then
res(2) = basal
endif
results(n) = res(6)
end do
fitness(j,l) = maxval(results)/c
end do
write(*,*) fitness
!posi = maxloc(fitness(:,j))
!k_opt(m,j) = k(posi) ! inputting that value into the optimal k matrix
end do
end do
!write(*,*) k_opt
!return k_opt
contains
! The subroutine increases the size of the array by 1
subroutine resize_array
real,dimension(:),allocatable :: tmp_arr
integer :: new
new = size(t) + 1
allocate(tmp_arr(new))
tmp_arr(1:new)=t
deallocate(t)
allocate(t(size(tmp_arr)))
t=tmp_arr
end subroutine resize_array
end program ffl
回答1:
Fortran 2003 has (re-)allocation upon assignment for allocatable arrays, and the program
program xgrid
implicit none
real, allocatable :: x(:)
integer :: i,n
do n=5,10,5
x = 0.1*[(i,i=0,n)]
write (*,"('x =',100(1x,f0.1))") x
end do
end program xgrid
compiled with gfortran 4.8.0, shows a Fortran one-liner equivalent to seq(), giving output
x = .0 .1 .2 .3 .4 .5
x = .0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0
回答2:
A simple implementation, if you really wanted the function, not wanting to always compute the n by hand. May need some clarification of the upper bound.
print *,seq(1.,10.,0.1)
contains
function seq(from, to, step)
real, allocatable :: seq(:)
real, intent(in) :: from, to, step
allocate(seq(0:int((to - from)/step)))
do i = 0, size(seq)
seq(i) = from + i * step
end do
end function
end
Regarding your program, when you use the fretures the compiler has, the backtrace would be much more helpful. Your resize_array should probably have tmp_arr(1:new-1)=t. The move_alloc() subroutine could make it little bit shorter.
来源：https://stackoverflow.com/questions/23245371/fortran-array-with-dynamic-size-as-easy-the-r-function-seq