C++ conversion rules ???

[Charles Galambos]

Hi,

I've just tried the pre-release egcs 2.95 compiler on my 
C++ code, and get thousands of warnings that follow the exact
same pattern as below.  Can I ask what the correct solution
to the problem is ?  I prefer to write code that is warning free
with the -Wall switch. As far as I can see I need to include both
functions, one to access a constant object one for non-constant use,
so how do I avoid the warning ?

> > I have trouble understanding why the following fragment:
> 
> I'm not sure what your question is; I'll try to answer it
> anyway. However, comp.std.c++ is a better place for this kind of
> problem.
> 
> > a.C:9: warning: choosing `A::operator char *()' over `A::operator const char *() const'
> > a.C:9: warning:   for conversion from `A' to `const char *'
> > a.C:9: warning:   because conversion sequence for the argument is better
> 
> The compiler is right in calling A::operator char *(). 
> 
> This is a case for overload resolution, according to 13.3
> [over.match]/2 (invocation of a conversion function for initialization
> of an object of a nonclass type from an expression of class type
> (13.3.1.5)).
> 
> According to 13.3.1.5, [over.match.conv]/1, both conversion functions
> are candidate functions. According to 13.3.2, [over.match.viable],
> both are viable functions. In particular, there is an implicit
> conversion sequence for the first (and only) argument.
> 
> The implicit argument is of type A. For one conversion function, the
> implicit parameter is of type A&, for the other, it is of type const
> A&. So there are two standard conversions to compare
> A -> A& (lvalue transformation)
> A -> const A& (lvalue transformation, qualification adjustment)
> 
> The first standard conversion sequence is better, so the compiler
> selects the conversion to char*. The resulting char* then gets
> qualification-adjusted to initialize "const char* b".

[...] text removed

Thanks,
Charles Galambos.