如何使用CUDA Fortran在结构中分配数组的数组？

Question

With CUDA, I'm trying to allocate arrays in a structure, but I'm having an issue and I don't know why. So here is a short code (stored in a file called struct.cuf ) that describe my problem. I'm compiling with the PGI 16.10 version , and I'm using the following options : -O3 -Mcuda=cc60 -tp=x64 struct.cuf -o struct_out

module structure
contains

type mytype
 integer :: alpha,beta,gamma
 real,dimension(:),pointer :: a
end type mytype

type mytypeDevice
 integer :: alpha,beta,gamma
 real,dimension(:),pointer,device :: a
end type mytypeDevice

end module structure

program main
 use cudafor
 use structure

 type(mytype) :: T(3)
 type(mytypeDevice),device :: T_Device(3)

 ! For the host
 do i=1,3
  allocate(T(i)%a(10))
 end do
 T(1)%a=1; T(2)%a=2; T(3)%a=3

 ! For the device
 print *, 'Everything from now is ok'
 do i=1,3
  allocate(T_Device(i)%a(10))
 end do
 !do i=1,3
 ! T_Device(i)%a=T(i)%a
 !end do

end program main

The output error :

 Everything from now is ok
Segmentation fault     

What I am doing wrong here ?

The only solution I found (and working) is to stored the values in differents arrays and transfers them to the GPU, but it's very "Heavy". Mostly if I use a lot of structures like mytype.

EDIT : Code has been modified to use Vladimir F's solution. If I remove the device attribute from T_Device(3) declaration, then allocation seems ok and giving values too (commented lines below allocation). But I need that device attribute for T_Device(3) , because I'm gonna use it in kernels.

Thanks !

Answer 1

I think you need a device pointer

type mytype_device
 ...
 real,dimension(:),pointer, device :: a
end type

Never used CUDA Fortran in my life, but it seems obvious enough to wager.

Answer 2

The problem here is how you have declared T_Device . To use host side allocation you first populate a host memory copy of the device structure, and then copy it to device memory. This:

type(mytypeDevice) :: T_Device(3)

do i=1,3
  allocate(T_Device(i)%a(10))
 end do

will work correctly. This is a very standard design pattern in C++ based CUDA code, and the principle here is identical.