Re: [Boost-users] Serializing a char *

From: Daryle Walker (darylew_at_[hidden])
Date: 2008-09-01 15:53:36

On Aug 28, 2008, at 10:47 PM, Diane wrote:

> Based on tutorial and reference material, and some previous posts  
> here, it seems
> like I should be able to serialize a char* by using a wrapper like  
> tracked_char:
>
> BOOST_STRONG_TYPEDEF(char, tracked_char)
>
> template<class Archive>
> void serialize(Archive &ar, tracked_char & tc){
>     ar & tc;
> }
>
> But when I try to serialize my struct that contains char*:
>
> typedef struct MYSTRUCT
> {
>    char chtest;
>    char *chstar;
> } MYSTRUCT;
>
> template<class Archive>
> void serialize(Archive & ar, MYSTRUCT & it, const unsigned int  
> version)
> {
>    ar & it.chtest;
>    tracked_char *tcp = it.chstar;  //****error on this line****//
>    ar & *tcp;
> }
>
> I get this compile error: cannot convert `char*' to `tracked_char*' in
> initialization
>
> What is the compiler complaining about here? How do I fix it? I'm  
> not sure if I
> have the correct usage of BOOST_STRONG_TYPEDEF, so I also tried:

You (generally?) can't convert one data pointer to another, unless  
the receiving type is "void" or an unambiguous base class type, and  
has at least as many cv-qualifiers.  If a BOOST_STRONG_TYPEDEF is  
what it says, then the pointer types have to be incompatible!  That's  
why the conversion fails.  I'm not sure that generated strong- 
typedefs are supposed to be compatible beyond the direct level.

> struct tracked_char {
>     char m_char;     // the real data
>     template<class Archive>
>     void serialize(Archive &ar, unsigned int version){
>         ar & m_char;
>     }
>     // casting operators - not compiled
>     char operator char () const {   //****error here****//
>         return m_char;
>     }
>     char & operator char() {        //****error here****//
>         return m_char;
>     }
> };
>
> I need someone to point me in the right direction.

Have you written a conversion operator before?  You leave out the  
return type; the type given in the name already tells the compiler  
what you want.

-- 
Daryle Walker
Mac, Internet, and Video Game Junkie
darylew AT hotmail DOT com