Reputation: 413
When I try to compile my code using -fcheck=all
I get a runtime error since it seems I step out of bounds of my array dimension size. It comes from the part of my code shown below. I think it is because my loops over i,j only run from -ny
to ny
, -nx
to nx
but I try to use points at i+1,j+1,i-1,j-1
which takes me out of bounds in my arrays. When the loop over j
starts at -ny
, it needs j-1
, so it immediately takes me out of bounds since I'm trying to access -ny-1
. Similarly when j=ny, i=-nx,nx
.
My question is, how can I fix this problem efficiently using minimal code?
I need the array grad(1,i,j)
correctly defined on the boundary, and it needs to be defined exactly as on the right hand side of the equality below, I just don't know an efficient way of doing this. I can explicitly define grad(1,nx,j), grad(1,-nx,j), etc,
separately and only loop over i=-nx+1,nx-1,j=-ny+1,ny-1
but this causes lots of duplicated code and I have many of these arrays so I don't think this is the logical/efficient approach. If I do this, I just end up with hundreds of lines of duplicated code that makes it very hard to debug. Thanks.
integer :: i,j
integer, parameter :: nx = 50, ny = 50
complex, dimension (3,-nx:nx,-ny:ny) :: grad,psi
real, parameter :: h = 0.1
do j = -ny,ny
do i = -nx,nx
psi(1,i,j) = sin(i*h)+sin(j*h)
psi(2,i,j) = sin(i*h)+sin(j*h)
psi(3,i,j) = sin(i*h)+sin(j*h)
end do
end do
do j = -ny,ny
do i = -nx,nx
grad(1,i,j) = (psi(1,i+1,j)+psi(1,i-1,j)+psi(1,i,j+1)+psi(1,i,j-1)-4*psi(1,i,j))/h**2 &
- (psi(2,i+1,j)-psi(2,i,j))*psi(1,i,j)/h &
- (psi(3,i,j+1)-psi(3,i,j))*psi(1,i,j)/h &
- psi(2,i,j)*(psi(1,i+1,j)-psi(1,i,j))/h &
- psi(3,i,j)*(psi(1,i,j+1)-psi(1,i,j))/h
end do
end do
If I was to do this directly for grad(1,nx,j), grad(1,-nx,j)
, it would be given by
do j = -ny+1,ny-1
grad(1,nx,j) = (psi(1,nx,j)+psi(1,nx-2,j)+psi(1,nx,j+1)+psi(1,nx,j-1)-2*psi(1,nx-1,j)-2*psi(1,nx,j))/h**2 &
- (psi(2,nx,j)-psi(2,nx-1,j))*psi(1,nx,j)/h &
- (psi(3,nx,j+1)-psi(3,nx,j))*psi(1,nx,j)/h &
- psi(2,nx,j)*(psi(1,nx,j)-psi(1,nx-1,j))/h &
- psi(3,nx,j)*(psi(1,nx,j+1)-psi(1,nx,j))/h
grad(1,-nx,j) = (psi(1,-nx+2,j)+psi(1,-nx,j)+psi(1,-nx,j+1)+psi(1,-nx,j-1)-2*psi(1,-nx+1,j)-2*psi(1,-nx,j))/h**2 &
- (psi(2,-nx+1,j)-psi(2,-nx,j))*psi(1,-nx,j)/h &
- (psi(3,-nx,j+1)-psi(3,-nx,j))*psi(1,-nx,j)/h &
- psi(2,-nx,j)*(psi(1,-nx+1,j)-psi(1,-nx,j))/h &
- psi(3,-nx,j)*(psi(1,-nx,j+1)-psi(1,-nx,j))/h
end do
Upvotes: 2
Views: 122
Reputation: 8566
One possible way for you could be using an additional index variable for the boundaries, modified from the original index to avoid getting out-of-bounds. I mean something like this:
do j = -ny,ny
jj = max(min(j, ny-1), -ny+1)
do i = -nx,nx
ii = max(min(i, nx-1), -nx+1)
grad(1,i,j) = (psi(1,ii+1,j)+psi(1,ii-1,j)+psi(1,i,jj+1)+psi(1,i,jj-1)-4*psi(1,i,j))/h**2 &
- (psi(2,ii+1,j)-psi(2,ii,j))*psi(1,i,j)/h &
- (psi(3,i,jj+1)-psi(3,i,jj))*psi(1,i,j)/h &
- psi(2,i,j)*(psi(1,ii+1,j)-psi(1,ii,j))/h &
- psi(3,i,j)*(psi(1,i,jj+1)-psi(1,i,jj))/h
end do
end do
It's hard for me to write a proper code because it seems you trimmed part of the original expression in the code you presented in the question, but I hope you understand the idea and apply it correctly for your logic.
Opinions:
if
conditionals or function calls). Using "ghost cells", as suggested by @evets, could be even more performant. You should profile and compare.dimension(-nx:nx,-ny:ny,3)
instead. Fortran stores arrays in column-major order and, as you are accessing values on the neighborhood of the "x" and "y", they would be non-contiguous memory locations for a fixed "other" dimension is the leftest, and that could mean less cache-hits.Upvotes: 2
Reputation: 1026
In somewhat pseudo-code, you can do
do j = -ny, ny
if (j == -ny) then
p1jm1 = XXXXX ! Some boundary condition
else
p1jm1 = psi(1,i,j-1)
end if
if (j == ny) then
p1jp1 = YYYYY ! Some other boundary condition
else
p1jp1 = psi(1,i,j+1)
end if
do i = -nx, ny
grad(1,i,j) = ... term involving p1jm1 ... term involving p1jp1 ...
...
end do
end do
The j-loop isn't bad in that you are adding 2*2*ny conditionals. The inner i-loop is adding 2*2*nx conditionals for each j iteration (or 2*2*ny * 2*2*nx conditional). Note, you need a temporary for each psi with the triplet indices are unique, ie., psi(1,i,j+1)
, psi(1,i,j-1)
, and psi(3,i,j+1)
.
Upvotes: 1