Re: [boost] Re: unions

From: E. Gladyshev (egladysh_at_[hidden])
Date: 2003-10-08 14:33:25

--- David Abrahams <dave_at_[hidden]> wrote:
[...]
> > typedef variant< int, my_type > v1;
> > typedef variant< my_type, int > v2;
> >
> > I think that we should just realize that 
> > in the current variant, v1 and v2 have 
> > a vastly different behaviour.
> 
> Really?  What are the differences?

my_type m;

for( i = 0; i < 10000; ++i )
   v1 = m;

for( i = 0; i < 10000; ++i )
   v2 = m;

The first loop will be executed much faster
than the second one.

--- Another example

struct my_type
{
  my_type( const m_type& l )
  {
     throw;
  }
};

v1 = 123;
v2 = 123;

v1 = m;  //exception -> v1 = garbabe
v2 = m;  //exception -> v2 = 123

> 
> > It just goes against any conventional 
> > wisdom and intuition. People, please...
> >
> > I don't think that the *weak* exception 
> > safety (the way it is implement now) 
> 
> What is "weak" exception safety?

According to Eric,
the content *value* of a variant after an exception is not guaranteed at 
all. Only the content *type* is guaranteed, and for that matter it's 
only guaranteed to be one of the bounded types, not a particular one.

> IIUC the exception guarantees are the same whether or not the
> optimization takes effect.

IMHO, these "weak" guarantees worth nothing.

Eugene

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