[英]unique_ptr to a derived class as an argument to a function that takes a unique_ptr to a base class
I'm trying to use a unique_ptr
to derived class in a function that takes a unique_ptr
to a base class. 我试图在一个函数中使用
unique_ptr
派生类,该函数将unique_ptr
带到基类。 Something like: 就像是:
class Base {};
class Derived : public Base {};
void f(unique_ptr<Base> const &base) {}
…
unique_ptr<Derived> derived = unique_ptr<Derived>(new Derived);
f(derived);
If I understand this answer correctly, this code should work, but it causes the following compile errors: 如果我正确理解了这个答案 ,那么这段代码应该可行,但它会导致以下编译错误:
error C2664: 'f' : cannot convert parameter 1 from 'std::unique_ptr<_Ty>' to 'const std::unique_ptr<_Ty> &'
错误C2664:'f':无法将参数1从'std :: unique_ptr <_Ty>'转换为'const std :: unique_ptr <_Ty>&'
IntelliSense: no suitable user-defined conversion from "std::unique_ptr<Derived, std::default_delete<Derived>>" to "const std::unique_ptr<Base, std::default_delete<Base>>" exists
IntelliSense:没有合适的用户定义转换,从“std :: unique_ptr <Derived,std :: default_delete <Derived >>”到“const std :: unique_ptr <Base,std :: default_delete <Base >>”存在
If I change f
to take unique_ptr<Derived> const &derived
, it works fine, but that's not what I want. 如果我改变
f
以获取unique_ptr<Derived> const &derived
,它可以正常工作,但这不是我想要的。
Am I doing something wrong? 难道我做错了什么? What can I do to work around this?
我该怎么做才能解决这个问题?
I'm using Visual Studio 2012. 我正在使用Visual Studio 2012。
You have three options: 你有三个选择:
Give up ownership. 放弃所有权。 This will leave your local variable without access to the dynamic object after the function call;
在函数调用之后,这将使您的局部变量无法访问动态对象; the object has been transferred to the callee:
对象已被转移到被调用者:
f(std::move(derived));
Change the signature of f
: 更改
f
的签名:
void f(std::unique_ptr<Derived> const &);
Change the type of your variable: 更改变量的类型:
std::unique_ptr<base> derived = std::unique_ptr<Derived>(new Derived);
Or of course just: 或者当然只是:
std::unique_ptr<base> derived(new Derived);
Or even: 甚至:
std::unique_ptr<base> derived = std::make_unique<Derived>();
Update: Or, as recommended in the comments, don't transfer ownership at all: 更新:或者,根据评论中的建议,根本不转让所有权:
void f(Base & b); f(*derived);
A possibile solution is to change the type of the argument to be a Base const*
, and pass derived.get()
instead. 可能的解决方案是将参数的类型更改为
Base const*
,然后传递derived.get()
。 There is no transfer of ownership with unique_ptr const<Base>&
(and the unique_ptr
is not being modified), so changing to a Base const*
does not change the meaning. 使用
unique_ptr const<Base>&
(并且未修改unique_ptr
)没有所有权转移,因此更改为Base const*
不会改变含义。
Herb Sutter discusses passing smart pointer arguments at length in Smart Pointer Parameters . Herb Sutter讨论了在智能指针参数中传递智能指针参数的过程 。 An extract from the linked article refers to this exact situation:
链接文章的摘录指的是这种情况:
Passing a
const unique_ptr<widget>&
is strange because it can accept only eithernull
or awidget
whose lifetime happens to be managed in the calling code via aunique_ptr
, and the callee generally shouldn't care about the caller's lifetime management choice.传递
const unique_ptr<widget>&
并且很奇怪,因为它只能接受null
或者生命周期恰好通过unique_ptr
在调用代码中管理的widget
,并且被调用者通常不应该关心调用者的生命周期管理选择。 Passingwidget*
covers a strict superset of these cases and can accept “null
or awidget
” regardless of the lifetime policy the caller happens to be using.传递
widget*
涵盖了这些情况的严格超集,并且可以接受“null
或widget
”,而不管调用者恰好使用的生命周期策略。
I had option #1 of the accepted answer and I still had the same compile error. 我有接受答案的选项#1,我仍然有相同的编译错误。 I banged my head on the wall for over an hour and I finally realized that I had
我把头撞在墙上一个多小时,我终于意识到我有了
class Derived : Base {};
instead of 代替
class Derived : public Base {};
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