[英]unordered_set<Foo> as data member of Foo?
如標題所述,我正在嘗試使用包含Foo類的對象的unordered_set作為Foo類的數據成員。 這在C ++中可能嗎?
我有以下代碼:
#include <unordered_set>
using namespace std;
struct FooHash;
class Foo {
public:
int id;
unordered_set<Foo, FooHash> foos; // error here
bool operator==(const Foo& foo) {
return id == foo.id;
}
};
struct FooHash {
size_t operator()(const Foo& foo) const {
return foo.id;
}
};
int main() {
Foo f;
unordered_set<Foo, FooHash> foos;
return 0;
}
但它引發以下錯誤:
In file included from /usr/include/c++/6.3.1/bits/hashtable.h:35:0,
from /usr/include/c++/6.3.1/unordered_set:47,
from main.cpp:1:
/usr/include/c++/6.3.1/bits/hashtable_policy.h: In instantiation of ‘struct std::__detail::__is_noexcept_hash<Foo, FooHash>’:
/usr/include/c++/6.3.1/type_traits:143:12: required from ‘struct std::__and_<std::__is_fast_hash<FooHash>, std::__detail::__is_noexcept_hash<Foo, FooHash> >’
/usr/include/c++/6.3.1/type_traits:154:38: required from ‘struct std::__not_<std::__and_<std::__is_fast_hash<FooHash>, std::__detail::__is_noexcept_hash<Foo, FooHash> > >’
/usr/include/c++/6.3.1/bits/unordered_set.h:95:63: required from ‘class std::unordered_set<Foo, FooHash>’
main.cpp:9:33: required from here
/usr/include/c++/6.3.1/bits/hashtable_policy.h:85:34: error: no match for call to ‘(const FooHash) (const Foo&)’
noexcept(declval<const _Hash&>()(declval<const _Key&>()))>
~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
In file included from /usr/include/c++/6.3.1/bits/move.h:57:0,
from /usr/include/c++/6.3.1/bits/stl_pair.h:59,
from /usr/include/c++/6.3.1/utility:70,
from /usr/include/c++/6.3.1/unordered_set:38,
from main.cpp:1:
/usr/include/c++/6.3.1/type_traits: In instantiation of ‘struct std::__not_<std::__and_<std::__is_fast_hash<FooHash>, std::__detail::__is_noexcept_hash<Foo, FooHash> > >’:
/usr/include/c++/6.3.1/bits/unordered_set.h:95:63: required from ‘class std::unordered_set<Foo, FooHash>’
main.cpp:9:33: required from here
/usr/include/c++/6.3.1/type_traits:154:38: error: ‘value’ is not a member of ‘std::__and_<std::__is_fast_hash<FooHash>, std::__detail::__is_noexcept_hash<Foo, FooHash> >’
: public integral_constant<bool, !_Pp::value>
向前聲明這兩個類並在之后聲明方法將產生以下兩個錯誤:
In file included from /usr/include/c++/6.3.1/unordered_set:44:0,
from main.cpp:1:
/usr/include/c++/6.3.1/ext/aligned_buffer.h: In instantiation of ‘struct __gnu_cxx::__aligned_buffer<Foo>’:
/usr/include/c++/6.3.1/bits/hashtable_policy.h:246:43: required from ‘struct std::__detail::_Hash_node_value_base<Foo>’
/usr/include/c++/6.3.1/bits/hashtable_policy.h:277:12: required from ‘struct std::__detail::_Hash_node<Foo, true>’
/usr/include/c++/6.3.1/bits/hashtable_policy.h:1894:60: required from ‘struct std::__detail::_Hashtable_alloc<std::allocator<std::__detail::_Hash_node<Foo, true> > >’
/usr/include/c++/6.3.1/bits/hashtable.h:170:11: required from ‘class std::_Hashtable<Foo, Foo, std::allocator<Foo>, std::__detail::_Identity, std::equal_to<Foo>, FooHash, std::__detail::_Mod_range_hashing, std::__detail::_Default_ranged_hash, std::__detail::_Prime_rehash_policy, std::__detail::_Hashtable_traits<true, true, true> >’
/usr/include/c++/6.3.1/bits/unordered_set.h:96:18: required from ‘class std::unordered_set<Foo, FooHash>’
main.cpp:12:37: required from here
/usr/include/c++/6.3.1/ext/aligned_buffer.h:85:34: error: invalid application of ‘sizeof’ to incomplete type ‘Foo’
: std::aligned_storage<sizeof(_Tp), std::alignment_of<_Tp>::value>
^
/usr/include/c++/6.3.1/ext/aligned_buffer.h:85:34: error: invalid application of ‘sizeof’ to incomplete type ‘Foo’
/usr/include/c++/6.3.1/ext/aligned_buffer.h: In instantiation of ‘void* __gnu_cxx::__aligned_buffer<_Tp>::_M_addr() [with _Tp = Foo]’:
/usr/include/c++/6.3.1/ext/aligned_buffer.h:110:41: required from ‘_Tp* __gnu_cxx::__aligned_buffer<_Tp>::_M_ptr() [with _Tp = Foo]’
/usr/include/c++/6.3.1/bits/hashtable_policy.h:250:34: required from ‘_Value* std::__detail::_Hash_node_value_base<_Value>::_M_valptr() [with _Value = Foo]’
/usr/include/c++/6.3.1/bits/hashtable_policy.h:1971:36: required from ‘void std::__detail::_Hashtable_alloc<_NodeAlloc>::_M_deallocate_node(std::__detail::_Hashtable_alloc<_NodeAlloc>::__node_type*) [with _NodeAlloc = std::allocator<std::__detail::_Hash_node<Foo, true> >; std::__detail::_Hashtable_alloc<_NodeAlloc>::__node_type = std::__detail::_Hash_node<Foo, true>]’
/usr/include/c++/6.3.1/bits/hashtable_policy.h:1984:22: required from ‘void std::__detail::_Hashtable_alloc<_NodeAlloc>::_M_deallocate_nodes(std::__detail::_Hashtable_alloc<_NodeAlloc>::__node_type*) [with _NodeAlloc = std::allocator<std::__detail::_Hash_node<Foo, true> >; std::__detail::_Hashtable_alloc<_NodeAlloc>::__node_type = std::__detail::_Hash_node<Foo, true>]’
/usr/include/c++/6.3.1/bits/hashtable.h:1901:7: required from ‘void std::_Hashtable<_Key, _Value, _Alloc, _ExtractKey, _Equal, _H1, _H2, _Hash, _RehashPolicy, _Traits>::clear() [with _Key = Foo; _Value = Foo; _Alloc = std::allocator<Foo>; _ExtractKey = std::__detail::_Identity; _Equal = std::equal_to<Foo>; _H1 = FooHash; _H2 = std::__detail::_Mod_range_hashing; _Hash = std::__detail::_Default_ranged_hash; _RehashPolicy = std::__detail::_Prime_rehash_policy; _Traits = std::__detail::_Hashtable_traits<true, true, true>]’
/usr/include/c++/6.3.1/bits/hashtable.h:1227:12: required from ‘std::_Hashtable<_Key, _Value, _Alloc, _ExtractKey, _Equal, _H1, _H2, _Hash, _RehashPolicy, _Traits>::~_Hashtable() [with _Key = Foo; _Value = Foo; _Alloc = std::allocator<Foo>; _ExtractKey = std::__detail::_Identity; _Equal = std::equal_to<Foo>; _H1 = FooHash; _H2 = std::__detail::_Mod_range_hashing; _Hash = std::__detail::_Default_ranged_hash; _RehashPolicy = std::__detail::_Prime_rehash_policy; _Traits = std::__detail::_Hashtable_traits<true, true, true>]’
/usr/include/c++/6.3.1/bits/unordered_set.h:126:7: required from here
/usr/include/c++/6.3.1/ext/aligned_buffer.h:99:36: error: using invalid field ‘__gnu_cxx::__aligned_buffer<_Tp>::_M_storage’
return static_cast<void*>(&_M_storage);
似乎標准中不允許使用不完整類型的容器。 我想嘗試使用指針,但是不能為指針操作數重載operator==
。 任何解決方法?
在將它們用作unordered_set
的模板參數之前,需要完全定義您的類。 雖然Foo
在其自身的定義期間無法完全定義,但其指針類型卻是。 您可以很容易地用unique_ptr<Foo>
替換Foo
,以解決此問題。
#include <memory>
#include <unordered_set>
// Forward declaration
class Foo;
// Declare classes
struct FooHash {
size_t operator()(const std::unique_ptr<Foo>& foo) const;
};
class Foo {
public:
int id;
std::unordered_set<std::unique_ptr<Foo>, FooHash> foos;
};
// Method implementations
size_t FooHash::operator()(const std::unique_ptr<Foo>& foo) const {
return foo->id;
}
正如FrançoisAndrieux在其答案中指出的那樣,std :: unordered_set不能使用不完整的類型。 他使用std::unique_ptr
給出了可能的解決方案。
使用std::unique_ptr
一個缺點可能是您的類現在需要顯式的動態內存分配才能插入Foo::foos
。 我認為這是應該隱藏的實現細節。
另一個(可能更大)的問題可能是您松散了值語義 ,因為std :: unique_ptr只能移動,不能復制。 您將必須實現自定義副本構造函數和賦值運算符,以獲取值語義。
使用boost :: recursive_wrapper ,還有另一種可能隱藏指針的解決方案。 盡管boost::recursive_wrapper
是為boost::variant
發明的,但它可以用於不允許使用不完整類型的其他情況。
使用recursive_wrapper,Foo類變為:
class Foo {
public:
using Wrapper = boost::recursive_wrapper<Foo>;
Foo() = default;
explicit Foo( int id ) : id( id ) {}
struct Hash {
size_t operator()(const Wrapper& foo) const { return foo.get().id; }
};
struct KeyEqual {
bool operator()(const Wrapper& foo1, const Wrapper& foo2) const {
return foo1.get().id == foo2.get().id;
}
};
using Set = std::unordered_set< Wrapper, Hash, KeyEqual >;
Set foos;
int id = 0;
};
我通過使用std::unordered_set
的第三個模板參數KeyEqual進行了另一個更改,以獨立於Foo::operator==
定義鍵的相等性。 現在可以通過比較Foo的所有成員而不只是id來按原樣實現Foo::operator==
。 這留給讀者練習。
要在Foo::foos
插入和查找項目,您可以定期創建Foo實例,而不必使用動態內存分配。 在幕后, boost::recursive_wrapper
會動態分配內存。
您唯一會注意到使用包裝器的地方是當您通過Foo::Set
的迭代器訪問Foo時,因為這時您必須調用boost::reference_wrapper::get()
。 這並不比使用std::unique_ptr
差,后者需要使用雙重間接std::unique_ptr
(即(*it)->id
)。
用法示例:
int main() {
Foo f1;
// No explicit dynamic memory allocation here!
f1.foos.insert( Foo( 1 ) );
f1.foos.insert( Foo( 1 ) );
f1.foos.insert( Foo( 2 ) );
// Value semantics are kept.
Foo f2 = f1;
f2.foos.erase( Foo( 1 ) );
std::cout << "f1.size: " << f1.foos.size() << std::endl;
std::cout << "f2.size: " << f2.foos.size() << std::endl;
// Find a Foo, still clean syntax.
auto it = f1.foos.find( Foo( 2 ) );
if( it != f1.foos.end() )
{
// Only when accessing Foo through the iterator we notice
// the recursive_wrapper because we must call its get() method.
std::cout << "found a Foo with id: " << it->get().id << std::endl;
}
return 0;
}
完整演示: http : //coliru.stacked-crooked.com/a/00d2a48106f2cc99
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