Question

I can construct data on Line 19 from vvp.at(0) if I know the size of vf.

#include <iostream>
#include <vector>

typedef void* VoidPointer;
typedef std::vector< float > VF;
typedef std::vector< VoidPointer > VVP;

int main(int argc, char** argv)
{
    VF vf;
    vf.push_back(13.0f);
    vf.push_back(14.0f);
    vf.push_back(15.0f);
    vf.push_back(16.0f);

    VVP vvp;
    vvp.push_back( (VoidPointer)const_cast<float *>( &(vf.front()) ) );

    VF data ( static_cast< float* >( vvp.at(0) ),
              static_cast< float* >( vvp.at(0) ) + vf.size() );

    std::cout << "data.size() is " << data.size() << std::endl;

    for(VF::const_iterator i = data.begin(); i != data.end(); ++i)
    {
        std::cout << "data contains " << *i << std::endl;
    }

    return 0;
}

Leaving aside whether this is sensible (the example is contrived) I’d like to know how to cast vvp.at(0) to a std::vector<float> if I didn’t know the size of vf. I’m thinking along the lines of something like:

std::vector<float> data( *static_cast< std::vector<float>* >( vvp.at(0) ) );

But that causes the program to termintate with std::bad_alloc, I don’t mind copying if need be.

Answer 1

That is not a cast from vvp.at(0) to a vector, it’s a copy of an array of floats into a new vector. And you can’t copy it without knowing the length. You only saved a pointer to the first element, so the information was lost.

You could make std::vector<std::pair<VoidPointer, size_t> > VVP and save both &vf.front() and vf.size() (or start and end pointers, if you prefer).

You could make std::vector<VF *> and store pointers to vectors (i.e. vvp.push_back(&vf)) and now there’s no casting at all.

Edit: In case you didn’t realize: The pointer &vf is unrelated to &vf.front(). vf is a structure which contains the pointer &vf.front() (or a way to get it). There’s no information in just the address &vf.front() to let you find &vf.