問題描述
我有這段代碼不起作用,但我認(rèn)為意圖很明確:
I have this code that doesn't work, but I think the intent is clear:
testmakeshared.cpp
#include <memory>
class A {
public:
static ::std::shared_ptr<A> create() {
return ::std::make_shared<A>();
}
protected:
A() {}
A(const A &) = delete;
const A &operator =(const A &) = delete;
};
::std::shared_ptr<A> foo()
{
return A::create();
}
但是我編譯的時(shí)候出現(xiàn)這個(gè)錯(cuò)誤:
But I get this error when I compile it:
g++ -std=c++0x -march=native -mtune=native -O3 -Wall testmakeshared.cpp
In file included from /usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:52:0,
from /usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/memory:86,
from testmakeshared.cpp:1:
testmakeshared.cpp: In constructor ‘std::_Sp_counted_ptr_inplace<_Tp, _Alloc, _Lp>::_Sp_counted_ptr_inplace(_Alloc) [with _Tp = A, _Alloc = std::allocator<A>, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]’:
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:518:8: instantiated from ‘std::__shared_count<_Lp>::__shared_count(std::_Sp_make_shared_tag, _Tp*, const _Alloc&, _Args&& ...) [with _Tp = A, _Alloc = std::allocator<A>, _Args = {}, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:986:35: instantiated from ‘std::__shared_ptr<_Tp, _Lp>::__shared_ptr(std::_Sp_make_shared_tag, const _Alloc&, _Args&& ...) [with _Alloc = std::allocator<A>, _Args = {}, _Tp = A, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:313:64: instantiated from ‘std::shared_ptr<_Tp>::shared_ptr(std::_Sp_make_shared_tag, const _Alloc&, _Args&& ...) [with _Alloc = std::allocator<A>, _Args = {}, _Tp = A]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:531:39: instantiated from ‘std::shared_ptr<_Tp> std::allocate_shared(const _Alloc&, _Args&& ...) [with _Tp = A, _Alloc = std::allocator<A>, _Args = {}]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:547:42: instantiated from ‘std::shared_ptr<_Tp1> std::make_shared(_Args&& ...) [with _Tp = A, _Args = {}]’
testmakeshared.cpp:6:40: instantiated from here
testmakeshared.cpp:10:8: error: ‘A::A()’ is protected
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:400:2: error: within this context
Compilation exited abnormally with code 1 at Tue Nov 15 07:32:58
這條消息基本上是說,::std::make_shared
模板實(shí)例化堆棧中的一些隨機(jī)方法無法訪問構(gòu)造函數(shù),因?yàn)樗鞘鼙Wo(hù)的.
This message is basically saying that some random method way down in the template instantiation stack from ::std::make_shared
can't access the constructor because it's protected.
但我真的想同時(shí)使用 ::std::make_shared
并防止任何人創(chuàng)建一個(gè) ::std::shared_ptr 沒有指向的這個(gè)類的對象代碼>.有什么辦法可以做到這一點(diǎn)嗎?
But I really want to use both ::std::make_shared
and prevent anybody from making an object of this class that isn't pointed at by a ::std::shared_ptr
. Is there any way to accomplish this?
推薦答案
這個(gè)答案 可能更好,我可能會接受.但我也想出了一個(gè)更丑陋的方法,但仍然讓一切仍然是內(nèi)聯(lián)的并且不需要派生類:
This answer is probably better, and the one I'll likely accept. But I also came up with a method that's uglier, but does still let everything still be inline and doesn't require a derived class:
#include <memory>
#include <string>
class A {
protected:
struct this_is_private;
public:
explicit A(const this_is_private &) {}
A(const this_is_private &, ::std::string, int) {}
template <typename... T>
static ::std::shared_ptr<A> create(T &&...args) {
return ::std::make_shared<A>(this_is_private{0},
::std::forward<T>(args)...);
}
protected:
struct this_is_private {
explicit this_is_private(int) {}
};
A(const A &) = delete;
const A &operator =(const A &) = delete;
};
::std::shared_ptr<A> foo()
{
return A::create();
}
::std::shared_ptr<A> bar()
{
return A::create("George", 5);
}
::std::shared_ptr<A> errors()
{
::std::shared_ptr<A> retval;
// Each of these assignments to retval properly generates errors.
retval = A::create("George");
retval = new A(A::this_is_private{0});
return ::std::move(retval);
}
編輯 2017-01-06: 我對此進(jìn)行了更改,以明確表示此想法可以清楚且簡單地?cái)U(kuò)展到接受參數(shù)的構(gòu)造函數(shù),因?yàn)槠渌苏谔峁┻@些方面的答案并且似乎對這個(gè).
Edit 2017-01-06: I changed this to make it clear that this idea is clearly and simply extensible to constructors that take arguments because other people were providing answers along those lines and seemed confused about this.
這篇關(guān)于如何在只有受保護(hù)或私有構(gòu)造函數(shù)的類上調(diào)用 ::std::make_shared?的文章就介紹到這了,希望我們推薦的答案對大家有所幫助,也希望大家多多支持html5模板網(wǎng)!