Is transpose costless?

[Rik]

jwe,

I had a question about optimization in the parser.  If I write

x = x.';

is that effectively a costless operation that simply swaps the first two
values of the dimension vector?  Or does Octave first create a temporary
intermediate (which is costless because of COW), but then the transpose
operator forces a deep copy to the intermediate, and then the final
assignment takes place (also potentially another deep copy).

--Rik