Re: [boost] [Atomic] Rationale for preventing copy construction/assignment?

Subject: Re: [boost] [Atomic] Rationale for preventing copy construction/assignment?
From: Collin Dauphinee (collin_at_[hidden])
Date: 2013-05-30 00:59:41

On 5/29/2013 8:49 PM, Steven Watanabe wrote:
> AMDG
>
> On 05/29/2013 01:22 PM, Collin Dauphinee wrote:
>> I've recently noticed that boost::atomic<T> prevents assignment from
>> boost::atomic<T>, as well as copy construction. I'm having trouble
>> understanding why this decision was made, as it makes boost::atomic<T> (and
>> classes with a boost::atomic<T> member) very hard to store in containers,
>> and I can't immediately see any safety issues that would be caused by
>> allowing copy operations.
>>
>> What is the rationale for not allowing assignment or copy construction from
>> another boost::atomic<T>?
>>
> How would you make it sequentially consistent
> without using a mutex?
>
> In Christ,
> Steven Watanabe
>

I don't view an assignment as a single operation, so I wouldn't expect 
both reading the source and storing it's value in the destination to be 
sequentially consistent. I would expect an assignment to be a 
'convenience' function wrapping store and load, because as far as I'm 
aware, there is no way to do this sequentially without a mutex.

My issue is stemming from wanting to store an object containing a few 
member variables, one being an atomic counter, in an std::map. The work 
around is to either implement the assignment operator just to accomodate 
the counter, or wrapping the counter in a class providing the same 
interface. Both result in bad code, and I think this is something an 
average user would run into fairly often.

I imagine that the situation where a user actually cares about the 
sequential consistency of assignment would be very rare, and in the 
cases where the user does care, they would realize the cost of 
guaranteeing sequential consistency for an assignment and check the 
documentation. Again, this would be completely for convenience.

I could definitely be wrong on all of this, though.