This is a bit complicated; I'd welcome any comments on how to improve the clarity of the question.
Ok, say I have an array:
real, allocatable :: A(:,:,:)
and I want to allocate it before I use it. Is it possible for the size of third dimension to depend on the size of the second dimension?
E.g.
do i=1,n
allocate(A(3,i,i**2))
end do
Obviously the above doesn't work. I'd like to end up with an array (or a set of arrays) with the shape(s)
(3,1,1), (3,2,4), (3,3,9), ... (3, n, n^2)
where the size of the third dimension is the square of the size of the second dimension.
My rule for the size of the dependent dimension is a bit more complicated, but if squaring is possible I can do the rest.
Is this possible? If so, how might I implement it in Fortran?
What would shape(A)
return? That would be interesting.
My other alternative is to allocate to the maximum size required, and be careful to only use certain elements in calculations, i.e.
allocate(A(3,n,n**2))
Even though I'm not hard up on memory at the moment, I'd like to have good programming practices. This is an interesting problem anyway.
Thank you.
EDIT:
What about having the size of a dimension depend on the value of an element in another dimension?
In the answer below the size of array in both dimensions depended on the index of B. I'd like something along the lines of
type myarray
real :: coord(3)
integer,allocatable :: lev(:)
integer, allocatable :: cell(:)
endtype myarray
type(myarray), allocatable :: data
allocate(data(m))
allocate(data%lev(n))
forall (j=1:n) !simple now, for argument's sake
lev(j)=j
endforall
! I was thinking of using a FORALL loop here, but the errors returned
! suggested that the compiler (gfortran) didn't expect IF blocks and ALLOCATE
! statements in a FORALL block
do i=1,m
do j=1,n
allocate(data(i)%cell(lev(j)**2))
enddo
enddo
You get what I mean? But the program falls over as it tries to allocate already allocated variables, e.g. when i=1
it allocates data(1)%cell(1)
, and then tries to allocate data(1)%cell(2)
...uh oh. What I want is something like:
Each data(i)
has an array lev(j)
of values, with j
running from 1 to n, and for each lev(j)
value we have a cell
of size lev
^2. Note that these cell
's are unique for each data(i)
and each lev
, and that the size of that particular cell
depends on the corresponding lev
value, and possibly the corresponding data(i)
too.
Would I have to use a derived type within a derived type?
A 2D array can be dynamically allocated in C using a single pointer. This means that a memory block of size row*column*dataTypeSize is allocated using malloc and pointer arithmetic can be used to access the matrix elements.
dynamically allocated arrays To dynamically allocate space, use calls to malloc passing in the total number of bytes to allocate (always use the sizeof to get the size of a specific type). A single call to malloc allocates a contiguous chunk of heap space of the passed size.
Allocate a new[] array and store it in a temporary pointer. Copy over the previous values that you want to keep. Delete[] the old array. Change the member variables, ptr and size to point to the new array and hold the new size.
Dynamic arrays in C++ are declared using the new keyword. We use square brackets to specify the number of items to be stored in the dynamic array. Once done with the array, we can free up the memory using the delete operator. Use the delete operator with [] to free the memory of all array elements.
Yes, you can use a derived type to accomplish this:
TYPE array
REAL,DIMENSION(:,:,:),ALLOCATABLE :: A
ENDTYPE array
INTEGER :: i
INTEGER,PARAMETER :: n=10
TYPE(array),DIMENSION(:),ALLOCATABLE :: B
ALLOCATE(B(n))
DO i=1,n
ALLOCATE(B(i)%A(3,i,i*i))
WRITE(*,*)SHAPE(B(i)%A)
ENDDO
END
This approach allows each element of array B to be a multi-dimensional array of a different shape.
The output of the program is as expected:
3 1 1
3 2 4
3 3 9
3 4 16
3 5 25
3 6 36
3 7 49
3 8 64
3 9 81
3 10 100
EDIT: To further answer OP's edited question. Yes, it seems like you would need to do something like this, use nested derived type (compare to your code example to figure out what you did wrong):
integer,parameter :: m=3,n=5
type cellarray
integer,dimension(:),allocatable :: c
endtype cellarray
type myarray
integer,allocatable :: lev(:)
type(cellarray),dimension(:),allocatable :: cell
endtype myarray
type(myarray),dimension(:),allocatable :: B
allocate(B(m))
! Allocate and assign lev and cell:
do i=1,m
allocate(B(i)%lev(n))
allocate(B(i)%cell(n))
do j=1,n
B(i)%lev(j)=j
enddo
enddo
! Based on value of lev, allocate B%cell%c:
do i=1,m
do j=1,n
allocate(B(i)%cell(j)%c(B(i)%lev(j)**2))
enddo
enddo
! Print out to check that it works:
do j=1,n
write(*,*)j,B(1)%lev(j),SIZE(B(1)%cell(j)%c)
enddo
end
Tried this with gfortran 4.6.2. It produces expected output:
1 1 1
2 2 4
3 3 9
4 4 16
5 5 25
I think you can do this simply by allocating/deallocating the array
Program test
Implicit none
Real, dimension(:,:,:), allocatable :: A
Integer :: i,N
Write(*,*)"Enter N"; Read(*,*)N
Do i = 1, N
if(Allocated(A)) then
deallocate(A);Allocate(A(i,i,i*i))
else
allocate(A(i,i,i*i))
end if
Write(*,*)Shape(A)
End do
end program test
Compiling the program using gfortran gives:
Enter N
5
1 1 1
2 2 4
3 3 9
4 4 16
5 5 25
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