Re: [boost] [xpressive] semantic actions: is ref() necessary

From: Eric Niebler (eric_at_[hidden])
Date: 2008-02-06 10:58:18

Thorsten Ottosen wrote:
> Hi Eric,
> 
> My semantic actions might look like
> 
>   +(name[push_back(ref(causeNames),as<std::string>(_) )]);
> 
> My first question here is whether ref() is really necessary?
> 
> In your example from the documentation, I can better understand it:
> 
>    [ ref(result)[s1] = as<int>(s2) ];
> 
> because you apply the [] operator to the result.
> 
> But otherwise, is there any time you not want a reference stored?
> In my case above, couldn't the destructor of push_back() take a T& 
> argumentment and construct a lazy reference from that?

The issue is one of lifetime management. Consider:

sregex make_regex( int &i, int j )
{
     sregex rx = (some >> regex)[ ref(i) += j ]; // TODO
     return rx;
}

What happens to 'j'? If we follow your suggestion, 'j' should be held by 
reference. Then when make_regex returns, the resulting regex object is 
left holding a dangling reference to an int. That's why anything not 
wrapped in ref() is held *by value* by the regex. So that's why ref() is 
needed in your case.

Have a look at xpressive::local<> and xpressive::reference<> as ways of 
decorating an object so that xpressive knows how to store it in an action.

> I also have another question: is it really necessary to write the
> as<std::string>(_) part of the semantic action?
> 
> Can this type not be deduced from the template argument to push_back()?
> In my case it would simply be Container::value_type.
> 
> If so, we would end up with the really slick looking
> 
>   +(name[push_back(causeNames)]);

No, push_back takes two arguments, and I don't want to get into the game 
of making xpressive semantic actions too clever by a half by trying to 
guess what you mean, or "know" the semantics of push_back, etc.. 
However, in a semantic action, _ stands in for a sub_match object, which 
happens to have an implicit conversion to std::string. So this should work:

// Untested.
local<std::vector<std::string> > causeNames;
sregex rx = +(name[ push_back(causeNames, _) ]);

HTH,

-- 
Eric Niebler
Boost Consulting
www.boost-consulting.com