Re: [Boost-users] [fusion] iterating over a sequence with non-const references

Subject: Re: [Boost-users] [fusion] iterating over a sequence with non-const references
From: paul Fultz (pfultz2_at_[hidden])
Date: 2012-04-01 15:56:11

>>  Hi,
>>  I would like to iterate over a sequence and preserve the 
>> 
>>  reference/const/non-const of the elements. That is say I have
>>  a sequence like this:
>> 
>>      tuple<int&, int> seq;
>> 
>>  I would like to this to be true:
>> 
>>      typeof(deref(begin(seq))) == int&
>>      typeof(deref(next(begin(seq)))) == int&&
>> 
>>  It seems fusion sequences always return a const reference to
>>  the element. Is there a workaround to get it to work like this?
> 
> Trust me, you don't want that, for roughly the same reason you don't
> want regular sequence rvalues to have rvalue-returning iterators:
> https://groups.google.com/d/msg/comp.lang.c++.moderated/1BwNTHCFiOs/L1T3WC9T_kMJ

I can see how that would be bad. Im using it to invoke functions which
wouldn't be a problem for that, but other use cases would cause problems.
Right now, it won't return a non-const reference from the iterator for tuples.

> 
>>  Or is this just not possible with Boost.Fusion?
> 
> I think you'd need to create your own iterator fusion-style iterator for
> this purpose.
> 
>>  Furthermore, I would like to use the transformation algorithms
>>  (especially push_back, and push_front) and still preserve the
>>  reference type. 
> 
> Sorry, I don't know what that means.  How could using an algorithm
> change a sequence's (or iterator's?) reference type?

Well for example in push_back, it adds const to the sequence like this:

    result_of::push_back<Sequence const, T>

Which will cause all references in the sequence to become const, as
well.

Thanks, for the reply, but im thinking about just implementing an invoke
that works just for tuples, but multiple tuples can be given. That way I won't
need to use push_back and push_front. I will just need to find a way to 
bring together all the tuples before the invoke.