Re: [Boost-users] Container/Algorithm question

18 Jul 2005

      Hi Joaquin,

You're right, the code is simple, what could possible wrong, it must be
a communication error on my part regarding what I need it to do!  So
I'll try to explain what happens and why I need something else to
happen!

The insertion errors show up when you run it in a simple nested loop
(see test.cpp).  After the first iteration of the outer loop it fails,
because the values for pair.second (created by the inner loop) already
exist in the indices.  But every pair created by that simple nested loop
needs to be valid, it should only fail when a duplicate of the same
exact pair is inserted.  The requirement is that every pair in the range
(0..N, 0..M) be valid.  So that means there are N*M valid pairs (0..N,
0) (0..N, 1) (0..N, 2) etc.  But the code you graciously supplied me
doesn't do that!

Yours,

Elisha
...
-----Original Message-----
From: boost-users-bounces@lists.boost.org [mailto:boost-users-
bounces@lists.boost.org] On Behalf Of JOAQUIN LOPEZ MU?Z
Sent: Friday, July 15, 2005 11:20 PM
To: boost-users@lists.boost.org
Subject: Re: [Boost-users] Container/Algorithm question
----- Mensaje original -----
De: Elisha Berns 
Fecha: Sábado, Julio 16, 2005 3:15 am
Asunto: Re: [Boost-users] Container/Algorithm question
...
Joaquin,
Sorry to be the bearer of bad news, but for some reason the multi-
indexcontainer starts to fail for some larger number of pairs (a few
hundred).  The container doesn't/won't insert them, and to be
honest, I
don't have a clue at this point why.
The code is so simple that I fail to see how it can be failing,
so let's insert some traces:
std::pair insert(const value_type& x)
  {
    iterator it0=impl.find(boost::make_tuple(x.first));
    if(it0!=impl.end()){
      typename impl_type::nth_index_iterator<1>::type it1=
        impl.template get<1>().find(x.second);
      if(it1!=impl.template get<1>().end()){
        std::cout<
Can you insert the std::cout output sentences above to your code
and report what it's showing? If everything is working OK,
when an insertion fails, x.first==it0->first and
x.second=it1->second. Isn't it so? Can you provide the data
the container is fed with when the first (erroneous) insertion
failure happens?
...
If you're up for it you
might want
to stress test the code you sent me with thousands of unique pairs
and
redundant pairs.
I'll examine this closer next Monday if we
haven't located the problem before.
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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