The rules for auto generating special move functions (constructor and assignment operator) in C++11 specify that no destructor can be declared. The logic is presumably that, if you need to do something special in destruction, that a move may not be safe.
However, for proper destructor calls in polymorphism, it is necessary to declare a base classes' destructor as virtual (otherwise deleting an instance of a sub class through a pointer of its base class will not properly chain the destructor).
I'm assuming, then, that even an empty destructor would prevent the compiler from automatically generating a special move functions. As in:
class Base {
virtual ~Base() { }
};
You can, however, default the destructor, as in:
class Base {
virtual ~Base() = default;
}
So question 1: Will this allow the compiler to auto generate special move functions?
There is a problem with the explicit default destructor, however. In at least the case of GCC 4.8.2, the signature is implicitly changed to noexcept. As in:
class Base {
virtual ~Base() = default; // compiler changes to:
// virtual ~Base() noexcept;
}
While I have no problem with noexcept in a destructor, this would break the following "client" code:
class Sub : public Base {
virtual ~Sub(); // this declaration is now "looser" because of no noexcept
}
So question 2 is more to the point: is there a way to allow auto generation of special move functions in C++11 and allow proper destructor chaining to sub classes (as described above), all without breaking subclass ("client") code?
No, a defaulted destructor is still considered user defined, so it will prevent the generation of move operations. Also declare the move operations default-ed to make the compiler generate them.
You need to only declare the move operations as default-ed in the base class. In the derived class, the destructor won't be user defined anymore (unless you explicitly say so), so the move operations won't be deleted.
So what I'd do is the following:
class Base
{
virtual ~Base() = default;
Base(Base&&) = default;
Base& operator=(Base&&) = default;
// probably need to think about copy operations also, as the move disables them
Base(const Base&) = default;
Base& operator=(const Base&) = default;
};
I highly recommend this talk by the person who contributed probably the most to the move semantics: http://www.slideshare.net/ripplelabs/howard-hinnant-accu2014
Or, if you can get your hands on, you should read the Item 17: Understand special member function generation from Scott Meyers' excellent book Effective Modern C++. This issue is excellently explained.
PS: I think you should think a bit more about your base classes. Most of the time, you should use abstract classes, so there will be no need to copy/move instances of them.
PSS: I think by default destructors are marked noexcept in C++11/14, so not explicitly specifying it shouldn't cause any problems:
Inheriting constructors and the implicitly-declared default
constructors, copy constructors, move constructors, destructors,
copy-assignment operators, move-assignment operators are all
noexcept(true) by default, unless they are required to call a function
that is noexcept(false), in which case these functions are
noexcept(false).
Related
I often see the code that deletes 1)copy ctor and 2)assign ctor.
What is the advantage to do so?
I append sample code.
class A{
...
private:
A(const A& copy) = delete;
A &operator=(const A &assign) = delete;
};
It prevents the class from being copied or assigned.
For example it useful when you have a A class that has a m_p pointer as a member variable which that class owns (A class is responsible for deallocation). In this case you can't safely copy or assign the A class. Because compiler generated copy constructor and assign operator will share the ownership of that pointer and in destructor when you will want to delete pointer you can get
a double-delete or access violation from dereferencing a freed pointer.
If you have a case where you don't want to create copies of your class then you restrict that by making the copy constructor and assignment operator private or mark them as delete.
This is useful when you want to create a config class or a cache where having multiple copies of your class don't make sense.
In standard library, strems are not copy constructable and copy-assignable.
The thing is, 4th edition of C++ Programming language says:
In this particular case, if you forgot to delete a copy or move operation, no harm is done. A
move operation is not implicitly generated for a class where the user has explicitly declared a destructor.
Furthermore, the generation of copy operations is deprecated in this case (§44.2.3). This
can be a good reason to explicitly define a destructor even where the compiler would have implicitly
provided one (§17.2.3).
I've tried this code:
#include <iostream>
class Foo {
public:
~Foo() {}
}
int main() {
Foo x;
Foo y(x);
return 0;
}
and there is no errors and exeptions here. I know that copy constructor should be generated implicitly in c++98, but 4th says that copy is deprecated. What does it mean?
My understanding is that an implicitly-declared constructor is not necessary implicitly-defined.
From cppreference :
Implicitly-declared copy constructor
If no user-defined copy constructors are provided for a class type (struct, class, or union), the compiler will always declare a copy constructor as a non-explicit inline public member of its class.
Implicitly-defined copy constructor
If the implicitly-declared copy constructor is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used.
The generation of the implicitly-defined copy constructor is deprecated if T has a user-defined destructor or user-defined copy assignment operator.
So in your case, the copy constructor is implicitly-declared but not implicitly-defined if not odr-used, which basically means it is not defined unless required somewhere.
See also : What is the distinction between implicitly-declared and implicitly-defined copy constructors?
Following code does not compile with clang-700.1.81 and it's standard library:
#include <memory>
class something;
std::unique_ptr<something> external_function();
std::unique_ptr<something> local_function()
{
auto thing = external_function();
return thing;
}
The diagnostics by clang:
......./include/c++/v1/memory:2626:46: note: in instantiation of member function 'std::__1::unique_ptr.....requested here
_LIBCPP_INLINE_VISIBILITY ~unique_ptr() {reset();}
^
test.cc:10:18: note: in instantiation of member function 'std::__1::unique_ptr<something, std::__1::default_delete<something> >::~unique_ptr' requested here
auto thing = external_function();
^
test.cc:4:7: note: forward declaration of 'something'
class something;
^
I guess it is trying to destroy the unique_ptr after copying it as return value, but is this really necessary? It is going be moved anyways, does it need to check if it can copy before realizing that it is easier to move it?
I could of course do this easily with a naked pointer.
Is there some other way to allow a uniqe_ptr to just "pass through" a translation unit, as shown in the example, without including extra header to get the definition of class something?
------EDIT--------
Also tried with GCC 5.3.0 and gnu libstdc++
Does not compile as well, with similar error messages.
------EDIT----
I think it is just trying to destroy the original thing object.
Thanks to Rudolf for the deleter idea ( a bit messy, but only option for this )
Looking at the library code I found this in unique_ptr's code:
if (__tmp)
__ptr_.second()(__tmp);
where second(_tmp) destructs the object pointed to. Even though it is never called, the compiler needs a definition to compile it. This is silly, but apparently gotta live with it.
From cppreference.com:
std::unique_ptr may be constructed for an incomplete type T, such as to facilitate the use as a handle in the Pimpl idiom. If the default deleter is used, T must be complete at the point in code where the deleter is invoked, which happens in the destructor, move assignment operator, and reset member function of std::unique_ptr. (Conversely, std::shared_ptr can't be constructed from a raw pointer to incomplete type, but can be destroyed where T is incomplete).
Thus, with a custom deleter you can use a forward declared class if the full declaration is available for the deleter:
#include <memory>
class Foo;
class FooDeleter
{
public:
void operator()(Foo* pInstance);
};
std::unique_ptr<Foo, FooDeleter> pFoo;
class Foo
{
};
void FooDeleter::operator()(Foo* pInstance)
{
delete pInstance;
}
The question titles says it all. I need to know if the default copy/move assignment/ctors implemented implicitly by the compiler are declared noexcept.
The standard says:
An inheriting constructor (12.9) and an implicitly declared special
member function (Clause 12) have an exception-specification. If f is
an inheriting constructor or an implicitly declared default
constructor, copy constructor, move constructor, destructor, copy
assignment operator, or move assignment operator, its implicit
exception-specification specifies the type-id T if and only if T is
allowed by the exception-specification of a function directly invoked
by f’s implicit definition; f allows all exceptions if any function it
directly invokes allows all exceptions, and f has the
exception-specification noexcept(true) if every function it directly
invokes allows no exceptions.
So if the implicitly declared copy/move assignment/ctors of your class do not need to call anything that is marked noexcept(false) then they will have the noexcept(true) specifier. The functions that need to be called will be the copy/move assignment/ctors of base class and non-static data members.
Obviously they can't just be unconditionally noexcept, that would be both silly and wrong (for example, the implicit copy constructor for a class containing a std::string member might need to allocate memory, so it can't sensibly be noexcept).
They're noexcept if they only call functions that are noexcept, and they're not noexcept if they call any functions that are not noexcept.
In C++0x, is it legal / expected that some classes can be copied but not moved? I'm looking at implementing a heterogenous class that resizes, and I'm not sure I could handle it if some classes needed copying and some needed moving.
Yes, it's legal for a class to be copyable but not movable:
class MyClass {
public:
/* Copyable... */
MyClass(const MyClass&);
MyClass& operator= (const MyClass&);
/* ... but not movable. */
MyClass(MyClass&&) = delete;
MyClass& operator= (MyClass&&) = delete;
};
However, I can't think of a good reason as to why anyone would want to do this. Knowing C++ coders (like me!) though, I think that you should anticipate that this might come up.
Out of curiosity, what code are you relying on that would break if a class was copyable but not movable?
I'm quite sure that making the move constructor protected/private is a way of limiting the ability to move objects of that class. If you can expect that in a template? I don't think so. You'd need std::enable_if for that kind of check.