
On 05/03/10 22:20, Sami wrote:
On Sat, May 1, 2010 at 9:08 AM, Rhys Ulerich
mailto:rhys.ulerich@gmail.com> wrote: Not as far as I know. You could accomplish this accumulating indices and strides relative to the origin (similar to the pseudocode for a(index_list) in Table 3 of http://www.boost.org/doc/libs/1_42_0/libs/multi_array/doc/reference.html).
One other option might be &a(index_list) - a.origin().
... since origin() is arr [0][0]...[0], neither of these is the answer in general. But thanks, good pointers there! Sad there isn't a multi_array::offset( indices ) or something that:
assert ( &arr( index_list ) == arr.data() + arr.*offset*( index_list ) )
this kind of stuff is useful when interacting with legacy libs or other languages.
Hmm... I puzzled why you think Sami's method won't work. For example, given the strides, s[ndim] and indices, i[ndim], and a.origin(), then the distance between a.origin() and the element at indices, i[0...ndim-1] is simply the sum of the vector product of s prepended with 1 and i. e.g., For a[2][2][2], the strides would be: s={2,4,8}, and for i={1,1,1}, then offset (of a[1]][1][1]) = 1*i[0]+s[0]*i[1]+s[1]*i[2] Wouldn't that work?