自定義分配器和默認成員

Question

為什么這段代碼沒有編譯？

#include <cstdlib>
#include <list>

template < typename Type >
class Allocator {
public:
    using value_type = Type;
public:
    template < typename Other >
    struct rebind { using other = Allocator< Other >; };
public:
    Type * allocate( std::size_t n ) { return std::malloc( n ); }
    void deallocate( Type * p, std::size_t ) throw ( ) { std::free( p ); }
};

int main( void ) {
    std::list< void *, Allocator< void * > > list;
    return 0;
}

它似乎需要指針，引用，pointer_const和reference_const類型。 但是，根據cppreference，這些成員都是可選項。 似乎STL沒有使用allocator_trait（我正在使用-std = c ++ 11進行編譯，所以它應該是好的）。

任何的想法 ？

[編輯]在clang上，錯誤是：

user@/tmp > clang++ -std=c++11 test.cc
In file included from test.cc:2:
In file included from /usr/lib/gcc/i686-pc-linux-gnu/4.7.1/../../../../include/c++/4.7.1/list:63:
/usr/lib/gcc/i686-pc-linux-gnu/4.7.1/../../../../include/c++/4.7.1/bits/stl_list.h:449:40: error: no type named 'pointer' in 'Allocator<void *>'
      typedef typename _Tp_alloc_type::pointer           pointer;
              ~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~
test.cc:17:46: note: in instantiation of template class 'std::list<void *, Allocator<void *> >' requested here
    std::list< void *, Allocator< void * > > list;
                                             ^
In file included from test.cc:2:
In file included from /usr/lib/gcc/i686-pc-linux-gnu/4.7.1/../../../../include/c++/4.7.1/list:63:
/usr/lib/gcc/i686-pc-linux-gnu/4.7.1/../../../../include/c++/4.7.1/bits/stl_list.h:450:40: error: no type named 'const_pointer' in 'Allocator<void *>'
      typedef typename _Tp_alloc_type::const_pointer     const_pointer;
              ~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~
/usr/lib/gcc/i686-pc-linux-gnu/4.7.1/../../../../include/c++/4.7.1/bits/stl_list.h:451:40: error: no type named 'reference' in 'Allocator<void *>'
      typedef typename _Tp_alloc_type::reference         reference;
              ~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~
/usr/lib/gcc/i686-pc-linux-gnu/4.7.1/../../../../include/c++/4.7.1/bits/stl_list.h:452:40: error: no type named 'const_reference' in 'Allocator<void *>'
      typedef typename _Tp_alloc_type::const_reference   const_reference;
              ~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~
4 errors generated.

Answer 1

這是GCC的C ++標准庫中的一個錯誤。

使用列表時，它們沒有通過allocator_traits正確地包裝對分配器的訪問。

但是，它們確實正確地實現了向量。 如果您使用std::vector而不是std::list則會編譯此代碼。