Re: [Boost-users] [fusion] memory layout of fusion::vector vs boost::tuple/boost::array

9 Oct 2010


      On Sat, Oct 9, 2010 at 7:53 AM, OvermindDL1  wrote:
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On Sat, Oct 9, 2010 at 4:11 AM, alfC  wrote:
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On Oct 9, 2:24 am, alfC  wrote:
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On Oct 8, 8:00 pm, OvermindDL1  wrote:
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On Fri, Oct 8, 2010 at 6:57 PM, Joel de Guzman
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 wrote:
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On 10/9/2010 2:44 AM, alfC wrote:
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On Oct 6, 5:33 pm, Joel de Guzman  wrote:
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> On 10/7/2010 3:53 AM, alfC wrote:
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>> Hi,
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>>   I am trying to convert a boost::array into a fusion::vector.
>> reinterpreting the memory from boost::array. It works for tuples but
>> not fusion::vector, why is that. Is there a way to make it work? ( I
>> was hoping that the memory layout is the same to make conversion from
>> one to the other very easy.)
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> Don't do that. It will *never* be guaranteed to work even if it works
> now. The memory layout of fusion::vector is an internal implementation
> detail and can change anytime.
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The question is: is it guaranteed for boost::tuples?
(it seems so, just by the PODness of tuples of PODs in its stated
design.) If so, then it is a *feature* of boost::tuple.
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No it is not. 1) It is not documented 2) It is not a POD
3) Anything with a reinterpret_cast is not guaranteed to work
(you can only rely on reinterpret_cast only when casting
back from a lost type e.g. through type erasure).
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Disregarding any of that is playing with fire.
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The previous posts a few posts ago seems overly complicated, I would
just do this (suitably wrapped into a function to handle it, maybe
named copy):
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// Proper includes here...
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struct set1stTo2nd
{
    template<typename T>
    void operator()(T& t) const
    {
        using namespace boost::fusion;
        deref(begin(t)) = deref(next(begin(t)));
    }};
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typedef boost::array arrType;
typedef boost::fusion::vector vecType;
typedef boost::fusion::vector zipSeqType;
...
arrType arr(1,1,1);
vecType vec(5,5,5);
...
for_each(zipSeqType(arr,vec), set1stTo2nd());
assert(arr == make_vector(5,5,5));
FINALLY! I got it. I feel better now.
It turns out that the correct line of code is
for_each(
       boost::fusion::zip_view<zipSeqType>(zipSeqType(arr2,vec1)),
       set1stTo2nd()
);
you wrote the correct typedef but then didn't apply the zip_view
function.
(Although, related to the original question, I am still a bit worried
about too much unnecessary copy which seems unavoidable since the
memory layout of array and fusion::vector is
different.)
below is the full program that *works*:
#include
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include<iostream>
using std::clog; using std::endl;
using namespace boost::fusion;
struct set1stTo2nd
{
   template<typename T>
   void operator()(T t) const
   {
       deref(boost::fusion::begin(t)) =
deref(boost::fusion::next(boost::fusion::begin(t)));
   }
};
typedef boost::array arrType;
typedef boost::fusion::vector vecType;
typedef boost::fusion::vector zipSeqType;
int main(){
       boost::array arr1={{1.0, 2.1, 3.2}};
       clog << "arr1: " << arr1[0] << ", " <<  arr1[1] << ", " << arr1[2] <<
endl;
 boost::fusion::vector vec1(arr1);
       clog << "vec1: " << at_c<0>(vec1) << ", "<< at_c<1>(vec1) << ", " <<
at_c<2>(vec1) << endl;
       boost::array arr2;
       for_each(
               boost::fusion::zip_view<zipSeqType>(zipSeqType(arr2,vec1)),
               set1stTo2nd()
       );
       clog << "arr2: " << at_c<0>(arr2) << ", "<< at_c<1>(arr2) << ", " <<
at_c<2>(arr2) << endl;
       return 0;
}
prints:
arr1: 1, 2.1, 3.2
vec1: 1, 2.1, 3.2
arr2: 1, 2.1, 3.2
Have a good weekend,
Alfredo
Yeah, my mistake, busy at work and not paying attention, I meant to
have the fusion::zip there.  :)
another subtlety is that the correct signature is

struct set1stTo2nd
{
   template<typename T>
   void operator()(T t) const   // not T& t) const
    ...
...
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Re: [Boost-users] [fusion] memory layout of fusion::vector vs boost::tuple/boost::array

Alfredo Correa