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为什么不编译this code？自定义分配器＆默认成员
#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;
}
这似乎需要有一个指针，参考，pointer_const & reference_const类型。但是，根据cppreference这些成员都是可选项。这似乎是如果STL没有使用allocator_trait（我正在编译-std = C++ 11，所以它应该是好的）。
有什么想法？
[编辑]在铛，错误是：
[email protected]/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.
什么是错误？ – Nick
@Nick我用铿锵输出（gcc失败了）更新了我的问题 –
请注意，基本上所有编译器的C++ 11支持仍然是实验/不完整的。因此，它可能只是一个实现错误（现在无法查看标准的相关部分，所以这只是一个猜测，如果这些成员确实是可选的） – Grizzly