Re: [boost] Re: in/out parameters, codingstylesandmaintenance

From: John Torjo (john_at_[hidden])
Date: 2003-05-02 02:55:45

Hi all,

Sorry, there have been way too many messages on this thread and I could not
have read them all. However, if the code for out, in_out has not been posted
yet, I guess a simple solution would be the following:

----------------------------
// works for gcc 3.2
#include <string>
#include <iostream>

template< class type>
struct out_t : public type
{
    template< class other_type> type & operator=( other_type val)
    {
        type::operator=( val);
        return *this;
    }
};

template< class type>
struct in_out_t : public type
{
    template< class other_type> type & operator=( other_type val)
    {
        type::operator=( val);
        return *this;
    }
};

template< class type>
inline out_t< type> & out( type & val)
{ return (out_t< type> &)val; }

template< class type>
inline in_out_t< type> & in_out( type & val)
{ return (in_out_t< type> &)val; }

void f( out_t< std::string> & a, const std::string & b, in_out_t<
std::string> & c)
{
    a = "a";
    c = "c";
}

int main()
{
    std::string x, y, z;
    std::cout << "x=" << x << std::endl;
    std::cout << "z=" << z << std::endl;
    f( out( x), y, in_out( z));
    // f( x, y, z); // ERROR - could not convert from ... to ...

    std::cout << "x=" << x << std::endl;
    std::cout << "z=" << z << std::endl;
    std::cin.get();
}
-----------------------------

(this would not work for fundamental types, but we could provide a
specialization for them)

Of course, it's not a full implementation, but it's a good start. What do
you think?

Best,
John