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When allocating zero-sized arrays in Fortran, I am getting counterintuitive behavior.

This code: 

program test_zerosized
  implicit none
  integer, allocatable :: a(:),b(:)
  allocate(a(0))
  print *, ' a lower bound = ',lbound(a,1)
  print *, ' a upper bound = ',ubound(a,1)

  allocate(b(0:0))
  print *, ' b lower bound = ',lbound(b,1)
  print *, ' b upper bound = ',ubound(b,1)
  return
end program test_zerosized

Produces the following output:

  a lower bound =            1
  a upper bound =            0
  b lower bound =            0
  b upper bound =            0

Is my compiler (gcc/gfortran 6.2.0) standard conforming? I don't get why lbound(a,1)==1 instead of lbound(a,1)==0, since the total total array size is of zero elements. Thanks!

Federico Perini
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    Well if Lbound( a ) = 0 and Ubound( a ) = 0 it's not zero sized, there's 1 element, namely a( 0 ) – Ian Bush Nov 20 '18 at 13:50

1 Answers1

4

The result you observe is the correct behaviour.

The array a is zero-sized, and lbound works on such arrays (F2008, 13.7.90) (my emphasis):

If ARRAY is a whole array and either ARRAY is an assumed-size array of rank DIM or dimension DIM of ARRAY has nonzero extent, LBOUND (ARRAY, DIM) has a value equal to the lower bound for subscript DIM of ARRAY. Otherwise the result value is 1.

ubound works in a complementary way.

Compare this with the size-1 array b with lower bound zero and upper bound zero.

The allocatable nature of a is not relevant, and you would see the same result with an explicit shape array of zero size.

francescalus
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    got it, thanks @francescalus, so in some sense the standard wants bound-based size checking to be consistent: for array a, `size(a) == ubound(a)-lbound(a)+1 == 0-1+1 == 0`, while for array b, `size(b) == ubound(b)-lbound(b)+1 == 0-0+1 == 1`, which contains 1 element as @Ian Bush pointed out – Federico Perini Nov 21 '18 at 14:05
  • Indeed (taking care to consider the extents/bounds/sizes in terms of individual ranks). – francescalus Nov 21 '18 at 14:10