Take a look a the following code example which uses class uncopiable similar to boost::noncopyable:
#include <vector>
class uncopiable {
using self = uncopiable;
protected:
uncopiable() {}
~uncopiable() {}
uncopiable(const self&) = delete;
self& operator=(const self&) = delete;
};
struct A {
struct B : uncopiable {
using self = B;
B() {
}
B(B&&) = default;
self& operator=(B&&) = default;
~B() {
}
};
A() { v.emplace_back(); }
~A() {}
private:
std::vector<B> v;
};
int main () {}
Since I wanted to make inner class move only I explicitly specified its move constructor and assignment operator to be default ones but also since I've heard that it's a good practice to specify all of the "special member functions" in such case I inherited it from uncopiable. The problem is that compilation fails with every compiler and something similar to the following error message is displayed (this message is excerpt from the clang one):
/usr/include/c++/v1/memory:1645:31: error: call to implicitly-deleted copy constructor of 'A::B'
...
main.cpp:26:10: note: in instantiation of function template specialization 'std::__1::vector >::emplace_back<>' requested here
main.cpp:19:3: note: copy constructor is implicitly deleted because 'B' has a user-declared move constructor
It could be fixed by removing inheritance (copy operations would still not be created). But writing copy operations to be explicitly deleted inside class after that is also okay.
My questions are: why does it happen? Could it be considered a deficiency of disabling constructors/assignment operators through inheritance of helper classes?
The problem is that your uncopiable class is not moveable. Therefore the default move constructor / assignment operator of the derived class try to use the deleted copy versions.
static_assert(std::is_move_constructible<uncopiable>::value, ""); // fails
static_assert(std::is_move_assignable<uncopiable>::value, ""); // fails
The reason for this is § 12.8 ¶ 9:
If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if
X does not have a user-declared copy constructor,
X does not have a user-declared copy assignment operator,
X does not have a user-declared move assignment operator, and
X does not have a user-declared destructor.
Declaring a copy operator or assignment operator as deleted still counts as declaring it.
The solution is of course to declare the move operations for uncopiable.
uncopiable(uncopiable&&) noexcept = default;
uncopiable& operator=(uncopiable&&) noexcept = default;
Note that the move operations should usually be declared noexcept. Especially if you want to use the type in a std::vector like in your example.
This compiles ok on MinGw:
#include <vector>
class uncopiable {
using self = uncopiable;
protected:
uncopiable() {}
~uncopiable() {}
uncopiable(const self&) = delete;
self& operator=(const self&) = delete;
};
struct A {
struct B : uncopiable {
using self = B;
B() {
}
B(B&&) {};
self& operator=(B&&) = default;
~B() {
}
};
A() { v.emplace_back(); }
~A() {}
private:
std::vector<B> v;
};
int main () {
A* a = new A();
}
Related
I have an std::multimap of the following type:
typedef std::multimap<std::pair<bool,uint32_t>, FooObject>
The FooObject has default move copy and assign constructors declared:
FooObject(FooObject&& v) = default;
FooObject& operator=(FooObject&& other)=default;
The copy/assign constructors are private to disable implicit copy.
So I should be able to emplace a pair into the map like this:
mymap.emplace(std::make_pair(false,32),FooObject());
This throws a list of errors with the one at the end:
error C2660: 'std::pair::pair': function does not take
2 arguments
If I declare move copy assign constructors without "default"
then it compiles ok.
FooObject(FooObject&& v){}
FooObject& operator=(FooObject&& other){}
Why is that? Does the compiler optimize away these constructors when marked with "default" keyword? I am using MSVC140
UPDATE:
Based on the comments below I found the reason - FooObject has a non-copiable member instance.
Here is the FooObject:
#define NO_COPY_ASSIGN(TypeName) \
TypeName (const TypeName &); \
void operator= (const TypeName &);
class FooObject
{
private:
NO_COPY_ASSIGN(FooObject)
public:
struct FooStruct
{
FooBuffer frameBuffer; //<--Here it is
}fooStruct;
FooObject(){}
/** Move constructor to allow insert into vector without copy */
FooObject(FooObject&& v) = default;
FooObject& operator=(FooObject&& other) = default;
};
*FooBuffer has also its copy/assign private.But I still don't get why replacing 'default' with {} fixes that.Please explain.
Your problem is that one of the member of FooObject is not move-able, which prevents the compiler from generating default move operations.
The {} versions of the move operations you implement yourself do no work (specifically: they don't actually do a move operation) on the members of FooObject and are thus legal.
The difference between
FooObject(FooObject&& v) = default;
and
FooObject(FooObject&& v){}
Is that the former will emit a constructor that moves each member from v while the latter default constructs each member and does nothing with v.
Since FooBuffer is not movable that means that the compiler will delete FooObject(FooObject&& v) = default; as it would be ill-formed.
With FooObject(FooObject&& v){} you do not have that problem as you never try to move the members of v. Since there is no member initialization list the compiler will add one for you that just default constructs the members.
You can see this behavior more explicitly with this:
struct Moveable
{
Moveable() = default;
Moveable(Moveable&&) { std::cout << "in Moveable(Moveable&&)\n"; }
};
struct Foo
{
Foo() = default;
Foo(Foo&&) = default;
Moveable m;
};
struct Bar
{
Bar() = default;
Bar(Bar&&){}
Moveable m;
};
int main()
{
Foo f;
Bar b;
std::cout << "test_f\n";
Foo test_f(std::move(f));
std::cout << "test_b\n";
Bar test_b(std::move(b));
}
which outputs
test_f
in Moveable(Moveable&&)
test_b
Live Example
Showing that nothing is actually moved in Bar's move constructor.
I'm experiencing behavior which I don't understand in a copy constructor of derived class.
class A {
A(const A&);
public:
A() = default;
};
class B : public A {
friend class Factory;
B(const int v) : A(), m_test_val(v) {}
public:
int m_test_val;
B(const B&); // no implementation, just declaration
};
class Factory {
public:
static B create(const int v) {
return B(v);
}
};
int main() {
B b = Factory::create(2);
std::cout << b.m_test_val << '\n';
return 0;
}
The behavior I don't understand is a matter of a working copy constructor B::B(const B&); which, however, does not have any implementation.
When I use B::B(const B&) = default; instead, I get an error saying I'm using deleted function (implicitly deleted because of ill-formation) in the return statement of the Factory::create() function (The A::A(const A&) is private and without implementation on purpose).
And of course, when I use B::B(const B&) = delete;, compiler tells me I use a deleted function.
How is it possible that the copy constructor works with no implementation just with declaration?
Note: The example code is based on a much larger code that behaves the same way, hopefully I didn't leave something out.
The actual copy is elided by the compiler, which is allowed since the copy constructor is accessible. The compiler is of course under no obligation to elide this copy and if it didn't I would expect a linker error not finding the implementation of the copy constructor.
Looking at this question it mentions C++11 and later only:
The move constructor is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move constructor is valid (e.g. if it wouldn't need to assign constant members) (§12.8/10).
So if I have the following code:
class Y
{
public:
Y(const Y&) {}
};
struct hasY {
hasY() = default;
hasY(hasY&&) = default;
Y mem;
};
hasY hy, hy2 = std::move(hy); //this line fails as expected as Y has a user-defined copy constructor.
Now if I add the default constructor to Y:
Y() {}
The error goes away.
Where does it say that the default constructor causes the creation of the move constructor?
(using VS 2015 update 2)
class Y
{
public:
Y(const Y&) {}
};
This class has no default constructor, so
struct hasY {
hasY() = default;
hasY(hasY&&) = default;
Y mem; // << requires default ctor
};
the error you were getting had nothing to do with move constructors:
prog.cpp: In function 'int main()':
prog.cpp:13:7: error: use of deleted function 'hasY::hasY()'
hasY hy;
^
prog.cpp:8:5: note: 'hasY::hasY()' is implicitly deleted because the default definition would be ill-formed:
hasY() = default;
http://ideone.com/u46GWS
It is nothing related to the move constructor It is about the default constructor. Try this:
class Y
{
public:
Y(const Y&) {}
};
struct hasY {
hasY() = default;
hasY(hasY&&) = default;
Y mem;
};
hasY hy; // This will cause an error because there is no default constructor
Now if you add the default constructor: Y(){}, the error will go away.
As #M.M commented, The copy constructor will be called in a such case.
You may try this code:
class Y{
public:
Y(){std::cout << "Default constructor\n";}
Y(const Y&) {std::cout << "Copy constructor\n";}
};
struct hasY {
hasY() = default;
hasY(hasY&&) = default;
Y mem;
};
int main(){
hasY hy;
hasY h=std::move(hy);
}
It will print:
Default constructor
Copy constructor
If you want to make the class not moveable you should delete the move constructor by yourself Y(Y&&)=delete;
Where does it say that the default constructor causes the creation of the move constructor?
There is no relationship between the creation of the move constructor and the existence, or lack thereof, of the default constructor.
From the C++11 Standard:
12.8 Copying and moving class objects
...
9 If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared
as defaulted if and only if
— X does not have a user-declared copy constructor,
— X does not have a user-declared copy assignment operator,
— X does not have a user-declared move assignment operator,
— X does not have a user-declared destructor, and
— the move constructor would not be implicitly defined as deleted.
I want to create a base class both noncopyable and nonmovable.
class noncopyable {
protected:
noncopyable() = default;
~noncopyable() = default;
noncopyable(noncopyable const &) = delete;
noncopyable &operator=(noncopyable const &) = delete;
};
class nonmovable {
protected:
nonmovable() = default;
~nonmovable() = default;
nonmovable(nonmovable &&) = delete;
nonmovable &operator=(nonmovable &&) = delete;
};
Is there any scenario that a class should be both noncopyable and nonmovable?
class non : public noncopyable, public nonmovable {
};
class foo : public non {
};
If there, what should be the convenient name for the "non" class here?
It has been argued, for instance here, that nonmoveable as demonstrated is a bad idea in the first place.
There are four reasonable options:
Type is naturally copyable and moveable.
Type is non-copyable, but moveable.
This happens often if it is managing a resource, and copying the resource is impossible, or so expensive as to be undesirable.
Type has no special moveability -- move ctor simply calls copy ctor.
This mostly happens if for technical reasons, you don't want to create an "empty state" for objects of this type, or there is no cheap / natural way to leave the moved-from object in any acceptable state.
Type is neither copyable nor moveable.
This may happen with a static singleton or something like this, where moving or copying it is something that should never happen in your program and you want to prevent from happening with a compile-time error.
If the type is copyable, but moving it fails, this is strictly a bad arrangement, and instead move should be made to fallback to copy. There's no advantage to making a move fail when a copy would be okay, it only inhibits generic programming.
So maybe you should only have "noncopyable" and "non", but not "nonmoveable" ?
While a "noncopyable" will work, an "nonmovable" base class will not provide what you expect:
#include <utility>
#include <iostream>
struct nonmovable
{
nonmovable() = default;
nonmovable(const nonmovable&) { std::cout << "copy\n"; }
nonmovable& operator = (const nonmovable&) { std::cout << "asign\n"; return *this; }
nonmovable(nonmovable&&) = delete;
nonmovable& operator = (nonmovable&&) = delete;
};
struct X : nonmovable {};
int main()
{
nonmovable n0;
nonmovable n1(n0);
// error: use of deleted function ‘nonmovable::nonmovable(nonmovable&&)’:
//nonmovable n2(std::move(n0));
X x0;
X x1(x0);
// However, X has a copy constructor not applying a move.
X x2(std::move(x0));
}
In addition, move construction and move assignment must be enabled explicitly after deletion of the copy constructor, if desiered:
struct noncopyable
{
noncopyable() = default;
// Deletion of copy constructor and copy assignment makes the class
// non-movable, too.
noncopyable(const noncopyable&) = delete;
noncopyable& operator = (const noncopyable&) = delete;
// Move construction and move assignment must be enabled explicitly, if desiered.
noncopyable(noncopyable&&) = default;
noncopyable& operator = (noncopyable&&) = default;
};
The, names "noncopyable" and "nonmovable" itself are good descriptive names. However, "boost::noncopyable" is both (non copyable and non movable), which might be a better (historical) design decision.
as an example - signleton pattern. Also, if u defined copy-constructor/assignment operator/destructor, move-constructor/assignment wont be generated.
Is this
struct Example {
string a, b;
Example(Example&& mE) : a{move(mE.a)}, b{move(mE.b)} { }
Example& operator=(Example&& mE) { a = move(mE.a); b = move(mE.b); return *this; }
}
equivalent to this
struct Example {
string a, b;
Example(Example&& mE) = default;
Example& operator=(Example&& mE) = default;
}
?
Yes both are the same.
But
struct Example {
string a, b;
Example(Example&& mE) = default;
Example& operator=(Example&& mE) = default;
}
This version will permits you to skip the body definition.
However, you have to follow some rules when you declare explicitly-defaulted-functions :
8.4.2 Explicitly-defaulted functions [dcl.fct.def.default]
A function definition of the form:
attribute-specifier-seqopt decl-specifier-seqopt declarator virt-specifier-seqopt = default ;
is called an explicitly-defaulted definition. A function that is explicitly defaulted shall
be a special member function,
have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T”, where T is the name of the member function’s class) as if it had been implicitly declared,
not have default arguments.
Is a =default move constructor equivalent to a member-wise move constructor?
Yes. Update: Well, not always. Look at this example:
#include <iostream>
struct nonmovable
{
nonmovable() = default;
nonmovable(const nonmovable &) = default;
nonmovable( nonmovable &&) = delete;
};
struct movable
{
movable() = default;
movable(const movable &) { std::cerr << "copy" << std::endl; }
movable( movable &&) { std::cerr << "move" << std::endl; }
};
struct has_nonmovable
{
movable a;
nonmovable b;
has_nonmovable() = default;
has_nonmovable(const has_nonmovable &) = default;
has_nonmovable( has_nonmovable &&) = default;
};
int main()
{
has_nonmovable c;
has_nonmovable d(std::move(c)); // prints copy
}
It prints:
copy
http://coliru.stacked-crooked.com/a/62c0a0aaec15b0eb
You declared defaulted move constructor, but copying happens instead of moving. Why? Because if a class has even a single non-movable member then the explicitly defaulted move constructor is implicitly deleted (such a pun). So when you run has_nonmovable d = std::move(c), the copy constructor is actually called, because the move constructor of has_nonmovable is deleted (implicitly), it just doesn't exists (even though you explicitly declared the move constructor by expression has_nonmovable(has_nonmovable &&) = default).
But if the move constructor of non_movable was not declared at all, the move constructor would be used for movable (and for every member that has the move constructor) and the copy constructor would be used for nonmovable (and for every member that does not define the move constructor). See the example:
#include <iostream>
struct nonmovable
{
nonmovable() = default;
nonmovable(const nonmovable &) { std::cerr << "nonmovable::copy" << std::endl; }
//nonmovable( nonmovable &&) = delete;
};
struct movable
{
movable() = default;
movable(const movable &) { std::cerr << "movable::copy" << std::endl; }
movable( movable &&) { std::cerr << "movable::move" << std::endl; }
};
struct has_nonmovable
{
movable a;
nonmovable b;
has_nonmovable() = default;
has_nonmovable(const has_nonmovable &) = default;
has_nonmovable( has_nonmovable &&) = default;
};
int main()
{
has_nonmovable c;
has_nonmovable d(std::move(c));
}
It prints:
movable::move
nonmovable::copy
http://coliru.stacked-crooked.com/a/420cc6c80ddac407
Update: But if you comment out the line has_nonmovable(has_nonmovable &&) = default;, then copy will be used for both members: http://coliru.stacked-crooked.com/a/171fd0ce335327cd - prints:
movable::copy
nonmovable::copy
So probably putting =default everywhere still makes sense. It doesn't mean that your move expressions will always move, but it makes chances of this higher.
One more update: But if comment out the line has_nonmovable(const has_nonmovable &) = default; either, then the result will be:
movable::move
nonmovable::copy
So if you want to know what happens in your program, just do everything by yourself :sigh:
Yes, a defaulted move constructor will perform a member-wise move of its base and members, so:
Example(Example&& mE) : a{move(mE.a)}, b{move(mE.b)} { }
is equivalent to:
Example(Example&& mE) = default;
we can see this by going to the draft C++11 standard section 12.8 Copying and moving class objects paragraph 13 which says (emphasis mine going forward):
A copy/move constructor that is defaulted and not defined as deleted
is implicitly defined if it is odrused (3.2) or when it is explicitly
defaulted after its first declaration. [ Note: The copy/move
constructor is implicitly defined even if the implementation elided
its odr-use (3.2, 12.2). —end note ][...]
and paragraph 15 which says:
The implicitly-defined copy/move constructor for a non-union class X
performs a memberwise copy/move of its bases and members. [ Note:
brace-or-equal-initializers of non-static data members are ignored.
See also the example in 12.6.2. —end note ] The order of
initialization is the same as the order of initialization of bases and
members in a user-defined constructor (see 12.6.2). Let x be either
the parameter of the constructor or, for the move constructor, an
xvalue referring to the parameter. Each base or non-static data member
is copied/moved in the manner appropriate to its type:
if the member is an array, each element is direct-initialized with the corresponding subobject of x;
if a member m has rvalue reference type T&&, it is direct-initialized with static_cast(x.m);
otherwise, the base or member is direct-initialized with the corresponding base or member of x.
Virtual base class subobjects shall be initialized only once by the
implicitly-defined copy/move constructor (see 12.6.2).
apart very pathological cases ... YES.
To be more precise, you have also to considered eventual bases Example may have, with exact same rules. First the bases -in declaration order- then the members, always in declaration order.