我想编写与R函数seq（）类似的Fortran代码。 E.g：
x <- seq(0,1,0.1)
将给出向量
x <- c(0, 0.1, 0.2, ..., 1)
我将运行几个模拟，序列的长度将在其中发生变化，在R中这很容易完成，只需改变seq（）中的第二个参数即可。我试图在Fortran中使用动态数组和函数ALLOCATE来动态地改变数组的大小。到目前为止，这还没有奏效，导致错误
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)
所以我想知道是否有一种简单的方法来模仿Fortran中R函数seq（）的行为。
如需进一步参考，请参阅以下程序
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
答案 0 :(得分：3)
Fortran 2003在分配可分配数组和程序
时进行了（重新）分配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
显示了一个等于seq（）的Fortran单行代码，给出了输出
x = .0 .1 .2 .3 .4 .5
x = .0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0
答案 1 :(得分：2)
一个简单的实现，如果你真的想要这个函数，不想总是手动计算n。可能需要澄清上限。
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
关于你的程序，当你使用编译器拥有的fretures时，回溯会更有帮助。你的resize_array应该有tmp_arr(1:new-1)=t。 move_alloc()子程序可以使它更短一些。