Re: Using std container to hold boost::shared_ptr with template parameter

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Mon, 22 Mar 2010 17:22:31 CST
On 22 Mrz., 19:04, Jardel Weyrich <> wrote:

Compiling the program below, gives these 2 errors:

1. test.cpp:10: error: type std::set<boost::shared_ptr<X>,
std::less<boost::shared_ptr<X> >, std::allocator<boost::shared_ptr<X>> > is not derived from type A<T>

2. test.cpp:10: error: expected ; before iterator

However, if I "typedef int type", it compiles fine. The Boost
documentation mentions the following:

"Every shared_ptr meets the CopyConstructible and Assignable
requirements of the C++ Standard Library, and so can be used in
standard library containers. Comparison operators are supplied so that
shared_ptr works with the standard library's associative containers.

The class template is parameterized on T, the type of the object
pointed to. shared_ptr and most of its member functions place no
requirements on T; it is allowed to be an incomplete type, or void.
Member functions that do place additional requirements (constructors,
reset) are explicitly documented below."

I thought the std allocator or the comparison operator could be the
cause, but I don't see a reasonable explanation for this. Any clue?

#include <boost/shared_ptr.hpp>
#include <set>

template <class T>
class A {
           typedef T type;
           typedef boost::shared_ptr<type> shared_type;
           typedef std::set<shared_type> container;
           typedef container::iterator iterator;

This needs to be:

       typedef typename container::iterator iterator;


Without any further context information the above
missing typename should be the only reason of
error. C++ requires the typename prefix here,
because container is a dependent type. And the
(rather simple) rule in C++ is that every dependent
name is assumed not to be a type. Of-course
container::iterator *is* a type, so you have
to attach the typename to make the compiler

HTH & Greetings from Bremen,

Daniel Kr?gler

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