Re: Type deduction in template members

From:

Yechezkel Mett <ymett.on.usenet@gmail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Wed, 5 May 2010 12:21:26 CST

Message-ID:

<873dd469-19ff-4383-9731-4cb7c25e8897@24g2000yqy.googlegroups.com>

On May 5, 4:35 am, Boris Bralo <boris.br...@gmail.com> wrote:

Hi all;

Here's the code that doesn't compile on latest Comeau (online) and GCC
4.4.0. It works with VC++ 2005 :

Simplified code demonstrating the issue:

struct AA{
    int aaa(){ return 1; };
};

struct BB: public AA {};

template <typename R>
void f( BB* o, R (BB::*memfn)() ){
     (o->*memfn)();
}

int main(int , char**){
     BB bb;
     f(&bb, &BB::aaa);
     return 0;
}

Both GCC and Comeau try to instantiate A::meth(&bb, &AA::aaa) and
report error. Are they right?

I believe so. &BB::aaa is &AA::aaa and has the type int(AA::*)().
There is an implicit conversion from int(AA::*)() to int(BB::*)(), but
the parameter memfn is deduced (because the R is a template parameter)
and therefore does not undergo conversions.

The simplest solution seems to be to use a template parameter:

template <typename R, class T>
void f( BB* o, R (T::*memfn)() ){
     (o->*memfn)();
}

Yechezkel Mett

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