Re: MPI Manifold

From: N2mopars@aol.com
Date: Sun Oct 24 1999 - 22:44:18 EDT


         for $i (1..10) {
             @list = somefunc($i);
             $LoL[$i] = @list; # WRONG!
         }

     That's just the simple case of assigning a list to a scalar
     and getting its element count. If that's what you really
     and truly want, then you might do well to consider being a
     tad more explicit about it, like this:

         for $i (1..10) {
             @list = somefunc($i);
             $counts[$i] = scalar @list;
         }

     Here's the case of taking a reference to the same memory
     location again and again:

         for $i (1..10) {
             @list = somefunc($i);
             $LoL[$i] = \@list; # WRONG!
         }

30/Jan/96 Last change: perl 3

PERLDSC(1) Perl Programmers Reference Guide PERLDSC(1)

     So, just what's the big problem with that? It looks right,
     doesn't it? After all, I just told you that you need an
     array of references, so by golly, you've made me one!

     Unfortunately, while this is true, it's still broken. All
     the references in @LoL refer to the _v_e_r_y _s_a_m_e _p_l_a_c_e, and
     they will therefore all hold whatever was last in @list!
     It's similar to the problem demonstrated in the following C
     program:

         #include <pwd.h>
         main() {
             struct passwd *getpwnam(), *rp, *dp;
             rp = getpwnam("root");
             dp = getpwnam("daemon");

             printf("daemon name is %s\nroot name is %s\n",
                     dp->pw_name, rp->pw_name);
         }

     Which will print

         daemon name is daemon
         root name is daemon

     The problem is that both rp and dp are pointers to the same
     location in memory! In C, you'd have to remember to
     _m_a_l_l_o_c() yourself some new memory. In Perl, you'll want to
     use the array constructor [] or the hash constructor {}
     instead. Here's the right way to do the preceding broken
     code fragments

         for $i (1..10) {
             @list = somefunc($i);
             $LoL[$i] = [ @list ];
         }

     The square brackets make a reference to a new array with a
     _c_o_p_y of what's in @list at the time of the assignment. This
     is what you want.

     Note that this will produce something similar, but it's much
     harder to read:

         for $i (1..10) {
             @list = 0 .. $i;
             @{$LoL[$i]} = @list;
         }

     Is it the same? Well, maybe so--and maybe not. The subtle
     difference is that when you assign something in square
     brackets, you know for sure it's always a brand new

30/Jan/96 Last change: perl 4

PERLDSC(1) Perl Programmers Reference Guide PERLDSC(1)

     reference with a new _c_o_p_y of the data. Something else could
     be going on in this new case with the @{$LoL[$i]}}
     dereference on the left-hand-side of the assignment. It all
     depends on whether $LoL[$i] had been undefined to start
     with, or whether it already contained a reference. If you
     had already populated @LoL with references, as in

         $LoL[3] = \@another_list;

     Then the assignment with the indirection on the left-hand-
     side would use the existing reference that was already
     there:

         @{$LoL[3]} = @list;

     Of course, this _w_o_u_l_d have the "interesting" effect of
     clobbering @another_list. (Have you ever noticed how when a
     programmer says something is "interesting", that rather than
     meaning "intriguing", they're disturbingly more apt to mean
     that it's "annoying", "difficult", or both? :-)

     So just remember to always use the array or hash
     constructors with [] or {}, and you'll be fine, although
     it's not always optimally efficient.

     Surprisingly, the following dangerous-looking construct will
     actually work out fine:

         for $i (1..10) {
             my @list = somefunc($i);
             $LoL[$i] = \@list;
         }

     That's because _m_y() is more of a run-time statement than it
     is a compile-time declaration _p_e_r _s_e. This means that the
     _m_y() variable is remade afresh each time through the loop.
     So even though it _l_o_o_k_s as though you stored the same
     variable reference each time, you actually did not! This is
     a subtle distinction that can produce more efficient code at
     the risk of misleading all but the most experienced of
     programmers. So I usually advise against teaching it to
     beginners. In fact, except for passing arguments to
     functions, I seldom like to see the gimme-a-reference
     operator (backslash) used much at all in code. Instead, I
     advise beginners that they (and most of the rest of us)
     should try to use the much more easily understood
     constructors [] and {} instead of relying upon lexical (or
     dynamic) scoping and hidden reference-counting to do the
     right thing behind the scenes.

     In summary:

30/Jan/96 Last change: perl 5

PERLDSC(1) Perl Programmers Reference Guide PERLDSC(1)

         $LoL[$i] = [ @list ]; # usually best
         $LoL[$i] = \@list; # perilous; just how my() was that list?
         @{ $LoL[$i] } = @list; # way too tricky for most programmers

CCCCAAAAVVVVEEEEAAAATTTT OOOONNNN PPPPRRRREEEECCCCEEEEDDDDEEEENNNNCCCCEEEE
     Speaking of things like @{$LoL[$i]}, the following are
     actually the same thing:

         $listref->[2][2] # clear
         $$listref[2][2] # confusing

     That's because Perl's precedence rules on its five prefix
     dereferencers (which look like someone swearing: $ @ * % &)
     make them bind more tightly than the postfix subscripting
     brackets or braces! This will no doubt come as a great
     shock to the C or C++ programmer, who is quite accustomed to
     using *a[i] to mean what's pointed to by the _i'_t_h element of
     a. That is, they first take the subscript, and only then
     dereference the thing at that subscript. That's fine in C,
     but this isn't C.

     The seemingly equivalent construct in Perl, $$listref[$i]
     first does the deref of $listref, making it take $listref as
     a reference to an array, and then dereference that, and
     finally tell you the _i'_t_h value of the array pointed to by
     $LoL. If you wanted the C notion, you'd have to write
     ${$LoL[$i]} to force the $LoL[$i] to get evaluated first
     before the leading $ dereferencer.

WWWWHHHHYYYY YYYYOOOOUUUU SSSSHHHHOOOOUUUULLLLDDDD AAAALLLLWWWWAAAAYYYYSSSS uuuusssseeee ssssttttrrrriiiicccctttt
     If this is starting to sound scarier than it's worth, relax.
     Perl has some features to help you avoid its most common
     pitfalls. The best way to avoid getting confused is to
     start every program like this:

         #!/usr/bin/perl -w
         use strict;

     This way, you'll be forced to declare all your variables
     with _m_y() and also disallow accidental "symbolic
     dereferencing". Therefore if you'd done this:

         my $listref = [
             [ "fred", "barney", "pebbles", "bambam", "dino", ],
             [ "homer", "bart", "marge", "maggie", ],
             [ "george", "jane", "alroy", "judy", ],
         ];

         print $listref[2][2];

     The compiler would immediately flag that as an error _a_t

30/Jan/96 Last change: perl 6

PERLDSC(1) Perl Programmers Reference Guide PERLDSC(1)

     _c_o_m_p_i_l_e _t_i_m_e, because you were accidentally accessing
     @listref, an undeclared variable, and it would thereby
     remind you to instead write:

         print $listref->[2][2]

DDDDEEEEBBBBUUUUGGGGGGGGIIIINNNNGGGG
     The standard Perl debugger in 5.001 doesn't do a very nice
     jowing issue that came up last spring, but
couldn't find anything current. Was that resolved?

3) Finally, does anyone know of any 1999/ V8's on dealer lots. Will consider
2 or 4 wd/ regular or extra cab. Nevada/ No. Calif/ Idaho would be great.
Does anyone know how to check dealer inventory? My local dealer only wants
to order a 2000 for me.

Thanks for your help. I can't wait to order my new toy!

-Vic



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