Re: locating an array

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 17 Feb 2009 05:44:26 -0800 (PST)
Message-ID:
<d4b95000-3acc-4e25-a7e2-9d45ee97fb8d@v42g2000yqj.googlegroups.com>
On Feb 16, 10:37 pm, r...@zedat.fu-berlin.de (Stefan Ram) wrote:

"Charles Arnett" <carn...@instrulab.com> writes:

I have a structure composedx of floats and strings. I want a
dummy array that is the same length starting at the same
address. I do this so I cann access the individual bytes of
the structure.


  I hope that there are not too many locations
  with undefined behavior below, and that it is
  always possible to align a =BBt*=AB at the address
  of an =BBs*=AB:

#include <iostream>
#include <ostream>

struct s { float a, b, c; };

struct t { char a[ sizeof( s )]; };

int main()
{ s s0;
  t * t0( reinterpret_cast< t* >( &s0 ));
  ::std::cout << sizeof( t0->a );
}


This seems more complicated than necessary, as well as
unnecissarily introducing undefined behavior. If the goal is
just to look at the individual bytes in an object:

    template< typename T >
    void
    dump( std::ostream& dest, T const& obj )
    {
        unsigned char const*p
                = reinterpret_cast< unsigned char const* >( &obj ) ;
        for ( size_t i = 0 ; i != sizeof( obj ) ; ++ i ) {
            if ( i != 0 ) {
                dest << ' ' ;
            }
            dest << HexFmt( 2 ) << static_cast< int >( p[ i ] ) ;
        }
    }

will do the trick, with no undefined behavior. (This can easily
be dressed up in a class, with an operator<<, and a free
function for type deduction, to make it easier to use.)

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