Re: reference to a vector

From:
benben <benhongh@yahoo.com.au>
Newsgroups:
comp.lang.c++
Date:
Sat, 29 Apr 2006 12:58:43 +1000
Message-ID:
<4452d667$0$20112$afc38c87@news.optusnet.com.au>

struct bar {
  double variable;
  void init() { memset ( this, 0, sizeof ( bar ) ); }
  bar() { init(); }
  bool operator <( bar & f ) const
  { return ( variable > f.variable ) ; }
};

std::vector< bar >& run_it ()
{
  std::vector< bar > *ptr_f = new std::vector<bar >;
  bar f;

  f.variable = 99.;
  ptr_f->push_back ( f );
  return ( *ptr_f );
}

// later
int main()
{
  std::vector<bar > f = run_it();
  std::cout << f.size() << std::endl;
  std::cout << f[ 0 ].variable << std::endl;

}
For starters, the memset seems to make an assumption that bar is a pod
type and it's not,


The memory layout is most likely the same as a double so memsetting it
has the same effect of memsetting a double.

Agreed. It's horrible, and quite possibly UB.


I don't see it that way. I think it is just a simple matter made
complicated but the behavior is still well defined (but platform dependent.)

The definition could have been trivially the following:

     struct bar {
         double variable;
         bar():variable(0.0){}
         bool operator <( bar & f) const{
             return variable > f.variable;
         }
     };

No, ptr_f is a pointer to memory allocated from the free store. The
memory doesn't go away when run_it() exits, and run_it returns a
reference to the vector allocated on the free store. The problem is
that the memory leaks -- there's no way to delete the new'ed vector.


I agree with you that the memory in OP's code is leaked. But it is
possible, though, to delete that memory. Simply

     delete &f;

Regards,
Ben

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