Write and read access=stream files in fortran - fortran

Question

I have a shell script from which I pass a binary file to a fortran program such that
Mth=$1
loop=1
it=1
while test $it -le 12
do
Mth=`expr $Mth + $loop`
file="DataFile"$Mth".bin"
./fort_exe ${Yr} ${nt} ${it}
# Increment loop
it=`expr $it + 1`
done
This script is used to pass 12 files within a do loop to the fortran program. In the fortran program, I read the binary file passed from the shell script and I am trying to write a 2nd file which would compile in a single file all the data that was read from the consecutive files e.g.
!Open binary file passed from shell script
open(1,file='Datafile'//TRIM{Mth)//.bin',action='read',form='unformatted',access='direct', &
recl=4*x*y, status='old')
! Open write file for t 1. The status is different in t 1 and t > 1 so I open it twice: I guess there is a more elegant way to do this...
open(2,file='Newfile.bin',action='write',form='unformatted', &
access='stream', position='append', status='replace')
irec = 0
do t = 1, nt
! Read input file
irec = irec + 1
read(1,rec=irec) val(:,:)
! write output file
irecW= irec + (imonth-1)*nt
if ( t .eq. 1) write(2,pos=irecW) val(:,:)
! Close file after t = 1, update the status to old and reopen.
if ( t .eq. 2) then
close (2)
open(2,file='Newfile.bin',action='write',form='unformatted', &
access='stream', position='append',status='old')
endif
if ( t .ge. 2) write(2,pos=irecW) val(:,:)
enddo
I can read the binary data from the first file no problem but when I try and read from another program the binary data from the file that I wrote in the first program such that
open(1,file='Newfile.bin',action='read',form='unformatted', &
access='stream', status='old')
irec=0
do t = 1, nt
! Read input file
irec = irec + 1
read(1,pos=irec) val(:,:)
write(*,*) val(:,:)
enddo
val(:,:) is nothing but a list of zeros. This is the first time I use access=stream which I believe is the only way I can use position='append'. I have tried compiling with gfortran and ifort but I do not get any error messages.
Does anyone have any idea why this is happening?

Answer 1

Firstly, I do not think you need to close and reopen your output file as you are doing. The status specifier is only relevant to the open statement in which it appears: replace will delete Newfile.bin if it exists at that time, before opening a new file with the same name. The status is implicitly changed to old, but this does not affect any operations done to the file.
However, since your Fortran code does not know you run it 12 times, you should have a way of making sure the file is only replaced the first time and opened as old afterwards; otherwise, Newfile.bin will only contain the information from the last file processed.
As for reading in the wrong values, this most likely occurs because of the difference between direct access (where you can choose a record length) and stream access (where you cannot). With stream access, data is stored as a sequence of "file storage units". Their size is in general compiler-dependent, but is available through the module iso_fortran_env as file_storage_size; it is usually 8 bits. This means that each entry will usually occupy multiple storage units, so you have to take care that a read or write with the pos = specifier does not access the wrong storage units.
Edit:
Some example code writing and reading with stream access:
program stream
use, intrinsic :: iso_fortran_env
implicit none
integer :: i, offset
real(real32), dimension(4,6) :: val, nval
open(unit=2, file='Newfile.bin', action='readwrite', form='unformatted', &
access='stream', status='replace')
do i = 1,2
call random_number(val)
write(2) val
enddo
! The file now contains two sequences of 24 reals, each element of which
! occupies the following number of storage units:
offset = storage_size(val) / file_storage_size
! Retrieve the second sequence and compare:
read(2, pos = 1 + offset*size(val)) nval
print*, all(nval == val)
close(2)
end program
The value true should be printed to the screen.
Note also that it's not strictly necessary to specify a pos while writing your data to the file, because the file will automatically be positioned beyond the last record read or written.
That said, direct or stream access is most beneficial if you need to access the data in a non-sequential manner. If you only need to combine input files into one, it could be easier to write the output file with sequential access, for which you can also specify recl and position = 'append'.

Answer 2

You can check for the existence of a file in standard Fortran, by using the inquire statement:
logical :: exist
inquire(file="test.dat", exist=exist)
if (exist) then
print *, "File test.dat exists"
else
print *, "File test.dat does not exist"
end if
Alternatively you can have a look at the modFileSys library which provides libc like file manipulation routines.
As for appending and streams: Appending files is also possible when you use "classical" record based fortran files, you do not have to use streams for that.