Re: [boost] [typeof] GCC error: expression not defined in this scope

Subject: Re: [boost] [typeof] GCC error: expression not defined in this scope
From: Vicente Botet (vicente.botet_at_[hidden])
Date: 2011-02-25 15:06:47

Lorenzo Caminiti wrote:
> 
> On Mon, Feb 7, 2011 at 5:29 PM, Steven Watanabe <watanabesj_at_[hidden]>
> wrote:
> 
> 
> Hello all,
> 
> Why the following TYPEOF code does not compile on GCC but it does on
> MSVC? Is this a GCC bug? If so, is there a workaround for it?
> 
> #include <boost/typeof/typeof.hpp>
> 
> int f() { return -1; }
> 
> template<typename T>
> void g(T x) {
>     int (deduce_func)();
>     typedef BOOST_TYPEOF_TPL(deduce_func) func_type; // error also if
> TYPEOF is used instead of TYPEOF_TPL
> 
>     struct s {
>         void call(func_type func) { func(); } // line 12
>     } ss;
>     ss.call(f); // line 14
> }
> 
> int main() {
>     g(1); // line 18
>     return 0;
> }
> 
> $ g++ -Wall -Werror -I../../.. r03.cpp
> r03.cpp: In member function ‘void g(T)::s::call(__typeof__
> (boost::type_of::ensure_obj(deduce_func))) [with T = int]’:
> r03.cpp:14:   instantiated from ‘void g(T) [with T = int]’
> r03.cpp:18:   instantiated from here
> r03.cpp:12: error: ‘deduce_func’ was not declared in this scope
> 
> What is this error? I don't understand it because with the local
> struct at line 12 I am just accessing the type generated by the
> typedef and not the actual expression that was used to deduce such a
> type...
> 
> 

If I understand deduce_func should be a pointer to a function returning int.
The declaration
    int (deduce_func)();

seems suspect to me. Have you tried with 
    int (*deduce_func)();

HTH,
Vicente

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