Why direct access I/O works incorrectly with Intel Visual Fortran

问题

This program

  program test

  real a(10)

  open(1, file='f1',access='direct', recl=20)
  do i=1, 10
      a(i) = i-1
  end do
  write(1, rec=1)(a(i),i=1,5)
  write(1, rec=2)(a(i),i=6,10)
  close(1)
  open(1, file='f1',access='direct',recl=8)
  read(1, rec=4)(a(i),i =5,9,4)
  print*,a
  end

works incorrect in visual Fortran (incorrect):

  0.0000000E+00   1.000000       2.000000       3.000000       9.000000
   5.000000       6.000000       7.000000      0.0000000E+00   9.000000

Result in WATCOM (correct):

  0.0000000   1.0000000       2.0000000       3.0000000       6.0000000
  5.0000000   6.0000000       7.0000000       7.0000000       9.0000000

Why?

回答1:

The default unit length of RECL in Visual Fortran is a word (4bytes). If you compile with option 'Use bytes as RECL= unit for unformatted file' (/assume:byterecl), you will get what you expect.

回答2:

You should not need to know when writing the code how exactly are the record long in whatever units. Setting recl=20 or recl=5 and depending on exact behaviour of your compiler and exact real size is a recipe for future problems.

You should ask the compiler for the size of the record in whichever units it uses:

integer :: recl5, recl2

inquire(iolength=recl5) (a(i),i=1,5)

open(1, file='f1',access='direct', recl=recl5)


...


inquire(iolength=recl2) (a(i),i =5,9,4)
open(1, file='f1',access='direct',recl=recl2)

Note: In Fortran 90 it could be much shorter by using subarray notation.

Note2: For some compilers your program will never work as the compiler is allowed to use record end marks into the file and you have to always open it with the same recl. But that is not common.

来源：https://stackoverflow.com/questions/37770912/why-direct-access-i-o-works-incorrectly-with-intel-visual-fortran

标签

compilation

fortran

intel-fortran

watcom