Matlab sub2ind in Fortran

7

Here is rather odd-looking way to do this in Fortran 2018 using reduce (and probably not very efficient based on what I saw in Compiler Explorer). A newer version of the Intel Fortran compiler is needed.

module fold
    implicit none
    public
contains
    function flatidx(extent,index) result(idx)
        integer, intent(in) :: extent(:), index(:)
        integer :: idx
        type :: pair
            integer :: e, i
        end type
        type(pair) :: tmp(size(extent))
        type(pair) :: res
        tmp%e = extent
        tmp%i = index
        res = reduce(tmp,op)
        idx = res%i
    contains
       pure function op(a,b) result(c)
        type(pair), intent(in) :: a, b
        type(pair) :: c
        c = pair(a%e*b%e, a%i + a%e*(b%i - 1))
       end function
    end function
end module

program test_fold
use fold, only: flatidx
implicit none
integer :: i, j

! 1-d
do i = 1, 8
    print *, i, flatidx([8],[i])
end do

! 2-d
do j = 1, 8
    do i = 1, 4
        print *, i, j, flatidx([4,8],[i,j])
    end do
end do
end program

The solution is inspired by the blog posts: