Does MINLOC work for arrays beginning at index 0? (Fortran 90/95)

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轻奢々 2020-12-11 08:48

After using C for a while, I went back to Fortran and allocated the arrays in my code from index 0 to N:

real(kind=dp), dimension(:), allocatable :: a 
alloc         


        
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  • 2020-12-11 09:15

    Your array a indeed starts at index 0, but you did not use that. You searched for a minimum of array abs(a(:)). This anonymous array expression starts at 1 as all arrays do by default.

    But even if you used a the result would be the same and is consistent with how array argument passing works in Fortran.

    The Fortran standard clearly states:

    The i subscript returned lies in the range 1 to ei , where ei is the extent of the idimension of ARRAY. If ARRAY has size zero, all elements of the result are zero.

    Lower bounds are not automatically passed with the array if you used assumed shape arguments. For example if you have your own function

      function f(arg)
        real :: arg(:)
    

    arg starts always at 1 no matter where the actual argument started in the calling code.

    You can change it to start at some other value

      function f(arg)
        real :: arg(-42:)
    

    and it would be indexed starting from that value.

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  • 2020-12-11 09:38

    There are two simple methods to handle complication of adjusting the indices obtained from minloc(): One is simply adding lbound() - 1 for all indices, and the other is using an array pointer with 1-based indices. An example code may look like this:

    program test
    implicit none
    integer, allocatable, target :: a(:,:)
    integer, pointer :: anew(:,:)
    integer :: loc(2)
    
    allocate( a( 0:4, 2:5 ), source= 10 )  !! make an array filled with 10
    
    a( 2, 3 ) = -700                       !! set the minimum value
    
    loc(:) = minloc( a )                   !! minloc() receives "a" with 1-based indices
    print *, loc(:)                        !! so we get [3,2]
    print *, a( loc(1), loc(2) )           !! 10 (wrong result...)
    
    !! Method (1) : adjust indices manually
    
    loc(:) = loc(:) + lbound( a ) - 1
    print *, a( loc(1), loc(2) )           !! -700 (now a correct result)
    
    !! Method (2) : use array pointer with 1-based indices
    
    anew( 1:, 1: ) => a
    
    loc(:) = minloc( anew )
    print *, loc(:)                        !! we get [3,2] again
    print *, anew( loc(1), loc(2) )        !! -700  (this time, no need to adjust indices)
    
    end program
    
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