On Mon, Feb 7, 2011 at 5:29 PM, Steven Watanabe <watanabesj@gmail.com> wrote:
On 2/7/2011 1:52 PM, Lorenzo Caminiti wrote:
void *result_ptr; // Plus append __LINE__... typedef typeof(*result_ptr) result_type;
That's almost what we did.
void deduce_result(); typedef boost::function_traits< typeof(&deduce_result)>::result_type result_type;
is a little better. (It can handle references.)
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... $ g++ --version g++ (GCC) 4.3.4 20090804 (release) 1 The same error raises even also if TYPEOF is used instead of TYPEOF_TPL. However, no error if g() is not a template so GCC compiles this just fine: #include <boost/typeof/typeof.hpp> int f() { return -1; } void g(int x) { // no template, no error -- why? int (deduce_func)(); typedef BOOST_TYPEOF(deduce_func) func_type; struct s { void call(func_type func) { func(); } } ss; ss.call(f); } int main() { g(1); return 0; } Both examples (template and not) compile instead just fine on MSVC. Do these examples compile on compilers other than GCC and MVSC? Thank you very much! -- Lorenzo