std :: set推导指南不符合我的期望

Question

我希望在下面的示例中演绎指南能够正确推断出类型，但它们don't：

#include <set>

struct Foo { };

bool cmp(const Foo&, const Foo& );

std::set my_set({Foo{}, Foo{}}, cmp);

编译器错误（gcc / clang都显示类似的诊断信息：

In file included from /opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/x86_64-linux-gnu/bits/c++allocator.h:33,

                 from /opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/bits/allocator.h:46,

                 from /opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/bits/stl_tree.h:64,

                 from /opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/set:60,

                 from <source>:1:

/opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/ext/new_allocator.h: In instantiation of 'class __gnu_cxx::new_allocator<bool(const Foo&, const Foo&)>':

/opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/bits/alloc_traits.h:634:11:   recursively required by substitution of 'template<class _Alloc> struct std::__is_allocator<_Alloc, std::__void_t<typename _Alloc::value_type, decltype (declval<_Alloc&>().allocate(long unsigned int{}))> > [with _Alloc = std::allocator<bool(const Foo&, const Foo&)>]'

/opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/bits/alloc_traits.h:634:11:   required by substitution of 'template<class _Alloc> using _RequireAllocator = typename std::enable_if<std::__is_allocator<_Alloc>::value, _Alloc>::type [with _Alloc = std::allocator<bool(const Foo&, const Foo&)>]'

/opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/bits/stl_set.h:938:5:   required by substitution of 'template<class _InputIterator, class _Compare, class _Allocator, class, class, class> std::set(_InputIterator, _InputIterator, _Compare, _Allocator)-> std::set<typename std::iterator_traits<_Iter>::value_type, _Compare, _Allocator> [with _InputIterator = bool (*)(const Foo&, const Foo&); _Compare = std::less<bool(const Foo&, const Foo&)>; _Allocator = std::allocator<bool(const Foo&, const Foo&)>; <template-parameter-1-4> = void; <template-parameter-1-5> = std::less<bool(const Foo&, const Foo&)>; <template-parameter-1-6> = <missing>]'

<source>:7:36:   required from here

/opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/ext/new_allocator.h:96:7: error: 'const _Tp* __gnu_cxx::new_allocator<_Tp>::address(__gnu_cxx::new_allocator<_Tp>::const_reference) const [with _Tp = bool(const Foo&, const Foo&); __gnu_cxx::new_allocator<_Tp>::const_pointer = bool (*)(const Foo&, const Foo&); __gnu_cxx::new_allocator<_Tp>::const_reference = bool (&)(const Foo&, const Foo&)]' cannot be overloaded with '_Tp* __gnu_cxx::new_allocator<_Tp>::address(__gnu_cxx::new_allocator<_Tp>::reference) const [with _Tp = bool(const Foo&, const Foo&); __gnu_cxx::new_allocator<_Tp>::pointer = bool (*)(const Foo&, const Foo&); __gnu_cxx::new_allocator<_Tp>::reference = bool (&)(const Foo&, const Foo&)]'

   96 |       address(const_reference __x) const _GLIBCXX_NOEXCEPT

      |       ^~~~~~~

/opt/compiler-explorer/gcc-9.2.0/include/c++/9.2.0/ext/new_allocator.h:92:7: note: previous declaration '_Tp* __gnu_cxx::new_allocator<_Tp>::address(__gnu_cxx::new_allocator<_Tp>::reference) const [with _Tp = bool(const Foo&, const Foo&); __gnu_cxx::new_allocator<_Tp>::pointer = bool (*)(const Foo&, const Foo&); __gnu_cxx::new_allocator<_Tp>::reference = bool (&)(const Foo&, const Foo&)]'

   92 |       address(reference __x) const _GLIBCXX_NOEXCEPT

      |       ^~~~~~~

Compiler returned: 1

对我来说，这似乎暗示编译器未使用构造函数获取初始化列表，而是尝试将其视为2迭代器版本（当然，也许我没有正确阅读它）。

这可能是什么原因？我是在误读扣除指南还是其他？

Answer 1

您必须提供分配器作为第三个参数，否则将应用带有两个参数的std::set的构造函数，并且将cmp视为分配器。

std::set my_set({Foo{}, Foo{}}, cmp, std::allocator<Foo>{});