On Wed, 8 Dec 2004 16:15:47 -0500, Arkadiy Vertleyb <vertleyb@hotmail.com> wrote:
"Peder Holt" <peder.holt@gmail.com> wrote
Then again, the following compiles with VC7.1, so maybe there is hope after all :)
Wow...
The reason it compiles is that you are using "class" where I used "typename". The following does not work:
Correct. I first tried with typename.
template<typename A0> void deduce_container(*typename* std::_Tree<A0>::iterator) { std::cout << "map" << "\n"; }
I don't think this is a legal standard usage, though... Probably a Microsoft - specific "feature"... Any comments from language experts? Should the following work according to the Standard?
I tried a modified version with comeau online compiler, and it did not compile. I like the syntax, though. There is a difference between nested classes and typedefs, and this is a nice way to show the difference explicitly in the source :) I have written a simple test in order to discover if other compilers support it: template<typename T> struct outer { class nested{}; }; template<typename T> void deduce(class outer<T>::nested) {} int main() { outer<int>::nested a; deduce(a); return 0; } Successfully compiles on VC6.5 and VC7.1 Fails to compile on Comeau C/C++ 4.3.3 I propose to define the macro BOOST_HAS_TEMPLATE_NESTED_CLASS_DEDUCTION for the VC compilers (and possibly other compilers supporting this) and supplying BOOST_TYPEOF_REGISTER_TEMPLATE_NESTED if the above is set. Peder
#pragma warning (disable:4786) #include <map> #include <list> #include <iostream>
template<typename A0,typename A1> void deduce_container(class std::list<A0,A1>::iterator) { std::cout << "list" << "\n"; }
template<typename A0> void deduce_container(class std::_Tree<A0>::iterator) { std::cout << "map" << "\n"; }
int main() { std::map<int,double>::iterator a; std::list<int*>::iterator b; deduce_container(a); deduce_container(b); return 0; }
Arkadiy
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