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: Joel de Guzman (joel_at_[hidden])
Date: 2012-04-02 23:26:28

On 4/3/2012 2:45 AM, paul Fultz wrote:
> 
> 
>> push_back is designed to be non-mutating. That is why. This is
>> true of all the views returned from 'algos', by design. What
>> you can do is to make a container from the view through, e.g.
>> as_vector, as_list. It is *not* true that all sequences return
>> const refs to its elements. Containers definitely return references,
>> otherwise get<N>(c)= expr will not be possible.
> 
> Also, front and back return non-const reference, but thats not the point.
> During iteration, they return const reference. Thats why this is not 
> possible:
> 
> struct mutable_func
> {
>     template<class F>
>     struct result;
> 
>     template<class F, class T, class U>
>     struct result<F(T, U)>
>     {
>         typedef T type;
>     };
> 
>     template<class T, class U>
>     T operator()(T && x, U y) const
>     {
>         return x+=y;
>     }
> };
> 
> int i = 0;
> invoke(foo, tuple<int&, int>(i, 1)); //doesn't compile
> 
> here is the output:

What is 'foo'? Anyway, this (below) compiles fine with me with
g++ 4.6.1 and MSVC 10:

#include <boost/fusion/functional/invocation/invoke.hpp>
#include <boost/fusion/tuple.hpp>
#include <boost/fusion/sequence/io.hpp>
#include <iostream>

struct mutable_func
{
    template<class F>
    struct result;

    template<class F, class T, class U>
    struct result<F(T, U)>
    {
        typedef T type;
    };

    template<class T, class U>
    T operator()(T && x, U y) const
    {
        return x+=y;
    }
};

int main()
{
    using namespace boost::fusion;

    int i = 999;
    std::cout << invoke(mutable_func(), tuple<int&, int>(i, 1)) << std::endl;
    return 0;
}

Prints 1000. What am I missing?

Regards,
-- 
Joel de Guzman
http://www.boostpro.com
http://boost-spirit.com