8/5     In ANSI C, assume a typedef of Foo for an UnsignedInt16 ,
        then assume that I declare a multiarray Foo foo[a][b] = {..}.
        How do I assign a pointer to foo? Do I do a Foo **bar?
        If Foo **bar then bar=foo doesn't seem to work.
        \_ References to an array name only decay to pointers for one
           level. So do bar = &foo[0].x
           \_ huh?  what the hell is x?
              \_ x is the first element of type Foo
              \_ x is someone slipping up with emacs.
        \_ This doesn't make any sense. A multidimensional array does
           not contain actual pointers to any of its elements, so there
           is nothing for bar to point at. If you do foo[0], this
           standard notation. Also, foo *bar can be assigned to bar = foo
           if foo is Foo[a]
           produced only a temporary pointer; getting the address of
           if foo is Foo foo[a].
           this is not a sensible thing to do.
           \_How so? Main has char **argv, and I can reference it using
           standard notation. Also, Foo *bar can be assigned to bar = foo
           if foo is Foo foo[a]. So why wouldn't there be an analogous
           expression for a pointer to a multidimensional array? If
           what you say is true, then is it impossible to reference a
           multiarray using a pointer? How would I go about passing
           multiarrays to functions then which would like to store a
           pointer to the multiarray?
           \_ your C-fu is lacking.  The declaration T foo[a][b] allocates a
              contiguous chunk of memory that's a*b*sizeof(T) bytes long.
              argv, on the other hand, is an array of pointers and is not
                                             \_ sorry, I misspoke.  I meant
                                                to say that it is a pointer
                                                to an array of pointers.
              a true multidimensional array.  The fact that you can
              dereference elements to both using [] operators is syntactic
              convenience.  Go read the comp.lang.C FAQ, section 6 in
              particular:
                http://www.eskimo.com/~scs/C-faq/top.html
                \_Yes, I know that foo[a][b] allocates a contiguous chunk of
                memory, so does foo[a]. What I have a problem with is the fact
                that foo[a] can decay into a pointer but not foo[a][b]. I
                read the FAQ, and I think essentially the problem is is that
                ANSI C provides no method of recursive pointer decay from
                a multiarray. I don't know why not, perhaps it has something
                to do with  compiler performance issues. So I suppose i
                the short answer is
                "no, it can't be done" and I solved the issue by dynamically
                allocating a multiarray using a pointer pointer.
                  \_ Why should foo[a][b] decay into a pointer?
              \_ Hmm, I learned C 12 years ago and I still find this section
                 of the FAQ very useful.
        \_ What you can do is Foo (*bar)[b]; bar = foo;
           \_ shouldn't that be: Foo (*bar)(b); bar = foo;
              \_ uh, he's not declaring a function pointer.
        \_ If the OP has their answer from the above would you please delete
           this thread.  I'm losing 1d4+1 sanity points everytime my eyes pass
           over this part of the motd.  thanks.
           \_ the first "1" in 1d4+1 is redundant.  unless you're using it as
              a placeholder, remove it, or you threaten a segfault in the rest
              of us.
              \_ No, that's how it's done.  When you publish your own game
                 system you can drop the "1" in front.  Until then you take
                 1d30+7 silliness points and are stunned 1d4+1 motd threads.