How to get the Fortran SUM command result to exceed 2^24 for single precision arrays

问题

To check memory allocations we populate single precision arrays with unit values and interrogate with the SUM and DOT_PRODUCT commands. These intrinsics stop counting after 16777216 (= 2^24). How can we get these commands to count billions of elements? We prefer to avoid DO loops. This is not a problem with higher precision arrays.

program allocator

  use iso_fortran_env
  implicit NONE

  integer, parameter :: sp    = selected_real_kind ( REAL32 )
  integer, parameter :: xlint = selected_int_kind  ( INT64 )

  integer ( xlint )                            :: n = 100000000
  real    ( sp ), allocatable, dimension ( : ) :: array

  integer   ( xlint )     :: alloc_status = 0
  character ( len = 255 ) :: alloc_msg = ""

!   ALLOCATE
    allocate ( array ( 1 : n ), stat = alloc_status, errmsg = alloc_msg )
    if ( alloc_status /= 0 ) print *, 'allocation error on ', n, ' elements: stat = ', alloc_status, ', errmsg = ', alloc_msg

!   POPULATE
    array = 1.0_sp
    write ( *, '( "number of elements allocated = ", g0 )' )    n
    write ( *, '( "sum of elements              = ", g0 )' )    sum ( array )
    write ( *, '( "dot product                  = ", g0, / )' ) dot_product ( array, array )

!   DEALLOCATE
    deallocate ( array, stat = alloc_status, errmsg = alloc_msg )
    if ( alloc_status /= 0 ) print *, 'deallocation error on ', n, ' elements: stat = ', alloc_status, ', errmsg = ', alloc_msg

    write ( *, '( "compiler version = ", A )' ) compiler_version()
    write ( *, '( "compiler options = ", A )' ) trim ( compiler_options() )

end program allocator

Output:

number of elements allocated = 100000000
sum of elements              = 16777216.
dot product                  = 16777216.

compiler version = GCC version 4.6.2 20111019 (prerelease)
compiler options = -fPIC -mmacosx-version-min=10.6.8 -mtune=core2

回答1:

That's due to the limited precision with single precision reals...

Since you only have 24 bits for your significant "digits", your resolution is 1/2**24 = 1/16777216. In other words, you cannot resolve an addition of 1/1677721 to 1, or in your case

16777216 + 1 = 16777216

To be able to resolve this operations which is required for both sum and dot_product (even if calculated using simple loops), you would need (at least) another bit of precision:

program allocator

  use iso_fortran_env
  implicit NONE

  integer, parameter :: sp    = REAL32
  integer, parameter :: xlint = INT64

  integer ( xlint )                            :: n = 100000000
  real    ( sp ), allocatable, dimension ( : ) :: array
  real    ( REAL64 )                           :: s
  integer ( xlint )                            :: i

  integer   ( xlint )     :: alloc_status = 0
  character ( len = 255 ) :: alloc_msg = ""

!   ALLOCATE
    allocate ( array ( 1 : n ), stat = alloc_status, errmsg = alloc_msg )
    if ( alloc_status /= 0 ) print *, 'allocation error on ', n, ' elements: stat = ', alloc_status, ', errmsg = ', alloc_msg

!   POPULATE
    array = 1.0_sp
    write ( *, '( "number of elements allocated = ", g0 )' )    n
    write ( *, '( "sum of elements              = ", g0 )' )    sum ( array )
    write ( *, '( "dot product                  = ", g0, / )' ) dot_product ( array, array )

    ! Calculate the sum using a double precision float
    s = real( array(1), REAL64 )
    do i=2,n
      s = s + real( array(i), REAL64 )
    enddo ! i
    write ( *, '( "sum of elements              = ", g0 )' )    s
    ! Calculate the dot product using a double precision float
    s = real( array(1), REAL64 )**2
    do i=2,n
      s = s + real( array(i), REAL64 )**2
    enddo ! i
    write ( *, '( "dot product                  = ", g0, / )' ) s

!   DEALLOCATE
    deallocate ( array, stat = alloc_status, errmsg = alloc_msg )
    if ( alloc_status /= 0 ) print *, 'deallocation error on ', n, ' elements: stat = ', alloc_status, ', errmsg = ', alloc_msg

    write ( *, '( "compiler version = ", A )' ) compiler_version()
    write ( *, '( "compiler options = ", A )' ) trim ( compiler_options() )

end program allocator

Output:

number of elements allocated = 100000000
sum of elements              = 16777216.0
dot product                  = 16777216.0

sum of elements              = 100000000.00000000
dot product                  = 100000000.00000000

compiler version = GCC version 4.8.4 20140605 (prerelease)
compiler options = -cpp -iprefix /home/elias/opt/sde/bin/../lib/gcc/x86_64-unknown-linux-gnu/4.8.4/ -mtune=generic -march=x86-64 -O0 -Wall -Wextra

来源：https://stackoverflow.com/questions/25316371/how-to-get-the-fortran-sum-command-result-to-exceed-224-for-single-precision-ar