Re: Help with constness and output operator<<

From:
Bart van Ingen Schenau <bart@ingen.ddns.info>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 23 Sep 2009 18:25:08 CST
Message-ID:
<29161981.ArONeeanhV@ingen.ddns.info>
marco.guazzone@gmail.com wrote:

Dear all,

Suppose you have a class, say "foobar", which let you store elements
in an array of predefined size, such that:
* only elements at even position are assignable [i.e., foo(i) = value
if (i % 2 == 0)]
* elements at odd position are set automatically to zero [i.e., foo(i)
= 0 if (i % 2 == 0)]

To do this I have two methods of foobar:
--- [snip] ---
template <typename T>
class foobar
{
...
reference operator()(std::size_t i)
{
     if (!(i % 2))
     {
         return v_[i];
     }

     throw std::runtime_error("Not assignable");
}

const_reference operator()(std::size_t i) const
{
     if (!(i % 2))
     {
         return v_[i];
     }

     return zero_; // The constant 0
}
};
--- [/snip] ---

The problem is that the "const" method is not called in the output
operator <<.
In fact when I execute the line:
   std::cout << foo(1) << std::endl
Instead of getting zero as output (resulting from the call to the
const method) I get the exception "Not assignable".
To make it to work I have to explicitly cast to type "foobar const&".

Where I am wrong?


You are wrong in thinking that the wider context where the
operator/function is used affects which overload is chosen.
For choosing between the const and non-const version, the only
consideration is if the operator/function is invoked on a const or non-
const object.

As foo is non-const, the non-const operator() overload is chosen in the
output statement.

To achieve the results you want, you need to use a proxy-object that can
detect the actual assignment. See below how this can be done.

[note: just for info, I'm using the GCC compiler 4.4.1 with flags -
Wall -Wextra -pedantic -ansi]

Here below is the entire code:

--- [code] ---
#include <cstddef>
#include <iostream>
#include <stdexcept>

template <typename T>
struct foobar
{
     typedef T value_type;
     typedef T& reference;
     typedef T const& const_reference;

     foobar(std::size_t n)
         : v_(new value_type[n/2])
     {
     }

     ~foobar() { delete[] v_; }


   private:
        struct proxy
        {
            proxy(value_type*& v, std::size_t i, const T& zero):
              v_(v), index(i), zero_(zero) {}
            operator const T&()
            {
                if (!(index % 2))
                {
                    return v_[index];
                }

                return zero_;
            }
            T& operator=(const T& val)
            {
                if (!(index % 2))
                {
                    v_[index] = val;
                    return v_[index];
                }
                else
                {
                    throw std::runtime_error("Not assignable");
                }
            }
        private:
            T*& v_;
            const T& zero_;
            std::size_t index;
        };
   public:
        proxy operator()(std::size_t i)
        {
            return proxy(v_, i, this->zero_);
        }

     const_reference operator()(std::size_t i) const
     {
         if (!(i % 2))
         {
             return v_[i];
         }

         return this->zero_;
     }

     private: value_type* v_;
     private: static const value_type zero_;

};
template <typename T>
const T foobar<T>::zero_ = 0;

int main()
{
     const std::size_t N = 5;
     foobar<int> foo(N);

     std::cout << "Populating..." << std::endl;;
     for (std::size_t i=0; i < N; i += 2)
     {
         foo(i) = i+1;
     }
     std::cout << "Querying..." << std::endl;;
     for (std::size_t i=0; i < N; ++i)
     {
         std::cout << "foo(" << i << ") ==> " << std::flush;
         std::cout << foo(i) << std::endl; // DON'T WORK
         //std::cout << (static_cast<foobar<int> const&>(foo))(i) <<
std::endl; // WORK
     }
}
--- [/code] ---

Thank you very much for any advice

Best,

-- Marco


Bart v Ingen Schenau
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