Copy-assignment for a class with a reference member variable is a no-no because you can't reassign the reference. But what about move-assignment? I tried simply moveing it but, of course, that destroys the source object when I just want to move the reference itself:
class C
{
public:
C(X& x) : x_(x) {}
C(C&& other) : x_(std::move(other.x_)) {}
C& operator=(C&& other)
{
x_ = std::move(other.x_);
}
private:
X& x_;
};
X y;
C c1(y);
X z;
C c2(z);
c2 = c1; // destroys y as well as z
Should I just not be implementing move-assignment and sticking with move-construction only? That makes swap(C&, C&) hard to implement.
(Posted as an answer from comment as suggested by the OP)
In general if one wants to do non-trivial stuff with references in C++, one would be using reference_wrapper<T>, which is essentially a fancy value-semantic stand-in for a T&, then one would be done with it - it already provides (re)assignment and other operations. I'm sure that would make move constructor and assignment near-trivial, if not trivial (note not trivial as in per the is_trivially_* semantics).
"Reference wrapper" is added to C++03 as part of TR1, and is part of C++11.
Documentation: http://en.cppreference.com/w/cpp/utility/functional/reference_wrapper
A reference is, in some sense, a T *const with syntactic sugar on top to auto-derefence, auto-capture, and auto-lifetime extend temporaries. (note that this is not quite true, but often is in practice and practicality)
If you want a reseatable reference, C++ has those: they are called pointers. You can use an accessor to replace dereferencing with a function call if you like. The remaining feature (temporary lifetime extension) that is hard to emulate does not apply to struct members.
Related
I came across this answer to how to write C++ getters/setters and the author implies that when it comes to value-oriented properties, the setters in the standard library use the std::move() like this...
class Foo
{
X x_;
public:
X x() const { return x_; }
void x(X x) { x_ = std::move(x); }
}
(code taken directly from the aforementioned answer)
... to leverage the move operator, if it is specified, potentially improving performance.
That itself makes sense to me - the values are copied once when passed by value to the method, so there's no need to copy them a second time to the property if they can be moved instead. However, my knowledge of C++ isn't good enough to be certain that doing this is safe in all cases.
Does passing an argument by value always make a deep copy? Or does it depend on the object? Considering the std::move() supposedly "signals the compiler I don't care what happens to the moved object", this could have unexpected side effects if I intended the original object to stay.
Apologies if this is a well-known problem, I'm in the process of learning C++ and am really trying to get to the bottom of the language.
Yes there is.
Receiving a parameter by value and move is okay if you always send an rvalue to that parameter. It is also okay to send an lvalue, but will be slower than receiving by const ref, especially in a loop.
Why? It seem that instead of making a copy you simply make a copy and then move, in which the move in insignificant in term of perfomance.
False.
You assume that a copy assignment is as slow as a copy constructor, which is false.
Consider this:
std::string str_target;
std::string long1 = "long_string_no_sso hello1234";
std::string long2 = "long_string_no_sso goobye123";
str_target = long1; // allocation to copy the buffer
str_target = long2; // no allocation, reuse space, simply copy bytes
This is why for setter function, by default, you should receive by const ref by default, and add a rvalue ref to optimize for rvalues:
class Foo
{
X x_;
public:
X x() const { return x_; }
// default setter, okay in most cases
void x(X const& x) { x_ = x; }
// optionally, define an overload that optimise for rvalues
void x(X&& x) noexcept { x_ = std::move(x); }
};
The only place where this does not apply is on constructor parameter and other sink functions, since they always construct and there is no buffer to reuse:
struct my_type {
// No buffer to reuse, this->_str is a totally new object
explicit my_type(std::string str) noexcept : _str{std::move(str)}
private:
std::string _str;
};
A coworker and I are debating whether const or reference members are ever the right thing to do. const and reference members make a class noncopyable and unmovable unless you write your own copy and move operators which ignore the reference or const members. I can't find a case where ignoring copying or moving certain members just because they are references or const makes sense.
I think having an unmovable object is very rarely logically sound, and is a choice that only has something to do with whether the class is location invariant. A noncopyable object is much more common, but the choice to make a class noncopyable has to do with whether it holds a resource which is logically noncopyable, if it represents sole ownership, etc. I can't think of a reason the sole trait of having a reference or const member or not would imply that the class should have either of these traits (noncopyable or unmovable)
Should they ever be used? Should one but not the other? What are examples?
I can't think of a reason the sole trait of having a reference or const member or not would imply that the class should have either of these traits (noncopyable or unmovable)
I think the premise is wrong.
Having a const data member or a reference data member does not make the class non-copyable: it simply makes it impossible to assign it or move from it in a destructive way.
That is because a destructive move operation is not sound for const objects or references: it would be against their semantics, since the state of a const object shall never be altered, and references cannot be unbound.
However, the fact that X is not copy-assignable or move-assignable does not compromise the possibility of constructing copies of those objects, which is - as you point out - a sound operation. The following program, for instance, will compile fine:
struct X
{
X() : x(0), rx(x) { }
const int x;
const int& rx;
};
int main()
{
X x;
X x1 = x; // Copy-initialization (copy-constructs x1 from x)
X x2 = X(); // Copy-initialization (copy-constructs x2 from a temporary)
}
The compiler will generate a copy-constructor implicitly for you, and moves will degenerate to copies. If this degeneration of move into copy is inappropriate for the semantics of your class, then you may consider not having const or reference members.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Can someone please explain move semantics to me?
Could someone point me to a good source or explain it here what are the move semantics?
Forget about C++0x for the moment. Move semantics are something that is language independent -- C++0x merely provides a standard way to perform operations with move semantics.
Definition
Move semantics define the behaviour of certain operations. Most of the time they are contrasted with copy semantics, so it would be useful to define them first.
Assignment with copy semantics has the following behaviour:
// Copy semantics
assert(b == c);
a = b;
assert(a == b && b == c);
i.e. a ends up equal to b, and we leave b unchanged.
Assignment with move semantics has weaker post conditions:
// Move semantics
assert(b == c);
move(a, b); // not C++0x
assert(a == c);
Note that there is no longer any guarantee that b remains unchanged after the assignment with move semantics. This is the crucial difference.
Uses
One benefit of move semantics is that it allows optimisations in certain situations. Consider the following regular value type:
struct A { T* x; };
Assume also that we define two objects of type A to be equal iff their x member point to equal values.
bool operator==(const A& lhs, const A& rhs) { return *lhs.x == *rhs.x; }
Finally assume that we define an object A to have sole ownership over the pointee of their x member.
A::~A() { delete x; }
A::A(const A& rhs) : x(new T(rhs.x)) {}
A& A::operator=(const A& rhs) { if (this != &rhs) *x = *rhs.x; }
Now suppose we want to define a function to swap two A objects.
We could do it the normal way with copy semantics.
void swap(A& a, A& b)
{
A t = a;
a = b;
b = t;
}
However, this is unnecessarily inefficient. What are we doing?
We create a copy of a into t.
We then copy b into a.
Then copy t into b.
Finally, destroy t.
If T objects are expensive to copy then this is wasteful. If I asked you to swap two files on your computer, you wouldn't create a third file then copy and paste the file contents around before destroying your temporary file, would you? No, you'd move one file away, move the second into the first position, then finally move the first file back into the second. No need to copy data.
In our case, it's easy to move around objects of type A:
// Not C++0x
void move(A& lhs, A& rhs)
{
lhs.x = rhs.x;
rhs.x = nullptr;
}
We simply move rhs's pointer into lhs and then relinquish rhs ownership of that pointer (by setting it to null). This should illuminate why the weaker post condition of move semantics allows optimisations.
With this new move operation defined, we can define an optimised swap:
void swap(A& a, A& b)
{
A t;
move(t, a);
move(a, b);
move(b, t);
}
Another advantage of move semantics is that it allows you to move around objects that are unable to be copied. A prime example of this is std::auto_ptr.
C++0x
C++0x allows move semantics through its rvalue reference feature. Specifically, operations of the kind:
a = b;
Have move semantics when b is an rvalue reference (spelt T&&), otherwise they have copy semantics. You can force move semantics by using the std::move function (different from the move I defined earlier) when b is not an rvalue reference:
a = std::move(b);
std::move is a simple function that essentially casts its argument to an rvalue reference. Note that the results of expressions (such as a function call) are automatically rvalue references, so you can exploit move semantics in those cases without changing your code.
To define move optimisations, you need to define a move constructor and move assignment operator:
T::T(T&&);
T& operator=(T&&);
As these operations have move semantics, you are free to modify the arguments passed in (provided you leave the object in a destructible state).
Conclusion
That's essentially all there is to it. Note that rvalue references are also used to allow perfect forwarding in C++0x (due to the specifically crafted type system interactions between rvalue references and other types), but this isn't really related to move semantics, so I haven't discussed it here.
Basically, rvalue references allow you to detect when objects are temporaries and you don't have to preserve their internal state. This allows for much more efficient code where C++03 used to have to copy all the time, in C++0x you can keep re-using the same resources. In addition, rvalue references enable perfect forwarding.
Have a look at this answer.
I read a ton of text explanations for about a year and didn't grasp everything about r-value references until I watch this excellent presentation by Scott Meyer : http://skillsmatter.com/podcast/home/move-semanticsperfect-forwarding-and-rvalue-references
He explain in a way that is funny and slow enough to understand each thing that happens in the processes.
I know, it 1h30 but really, it's the best explanation I've had in the last year.
After having read the articles (like the other answers), watching this video did melt it together in my mind in a consistent way and few days after I was able to explain it to some colleagues and explain how to use std::unique_ptr (as it is related - it only allow move semantics, not copy) because it requires understanding of std::move(), that requires understanding move semantics.
glad to see such a question and I'm happy to share my point. I think you are asking about a bug-fix on the designation of the C++ language itself, not just another C++ language feature. The "bug" has been there for tens of year. That is, the copy constructor.
Copy constructors seems very strange if you know in physics there are lots of things that can not be copied like energy and mass. That's just a joke, but in fact in the world of programming too, objects like exclusive file descriptors are not copyable. So C++ programmers and designers invented some tricks to deal with that. There are 3 famous: NRVO, boost::noncopyable and std::auto_ptr.
NRVO (Named Return Value Optimization) is a technic that lets a function returns an object by value without calling the copy constructor. But the problem with NRVO is that though the copy constructor is not actually called, a public copy constructor declaration is still needed, which means, objects of boost::noncopyable is not compatible with NRVO.
std::auto_ptr is another trial to bypass the copy constructor. You might have seen its "copy constructor" implemented like
template <typename _T>
auto_ptr(auto_ptr<_T>& source)
{
_ptr = source._ptr; // where _ptr is the pointer to the contained object
source._ptr = NULL;
}
This is not a copy at all, but a "move". You could consider this kind of behavior as the prototype of a move semantic.
But std::auto_ptr also has its own problem: it is not compatible with STL containers. So, unfortunately, anything about noncopyable is painful.
That was painful until, the C++0x move semantic is finally published and implemented by the compiler makers.
In simple way, you could just think of move semantic as something same as the "copy" behavior of std::auto_ptr, but with a full support by the language features so it works fine with containers and algorithm.
By the way in C++0x the std::auto_ptr is deprecated and a new template type std::unique_ptr is recommended.
My story will end now. Please refer to other posts if you want to know more about it like strange syntax and rvalue system.
Consider a class of which copies need to be made. The vast majority of the data elements in the copy must strictly reflect the original, however there are select few elements whose state is not to be preserved and need to be reinitialized.
Is it bad form to call a default assignment operator from the copy constructor?
The default assignment operator will behave well with Plain Old Data( int,double,char,short) as well user defined classes per their assignment operators. Pointers would need to be treated separately.
One drawback is that this method renders the assignment operator crippled since the extra reinitialization is not performed. It is also not possible to disable the use of the assignment operator thus opening up the option of the user to create a broken class by using the incomplete default assignment operator A obj1,obj2; obj2=obj1; /* Could result is an incorrectly initialized obj2 */ .
It would be good to relax the requirement that to a(orig.a),b(orig.b)... in addition to a(0),b(0) ... must be written. Needing to write all of the initialization twice creates two places for errors and if new variables (say double x,y,z) were to be added to the class, initialization code would need to correctly added in at least 2 places instead of 1.
Is there a better way?
Is there be a better way in C++0x?
class A {
public:
A(): a(0),b(0),c(0),d(0)
A(const A & orig){
*this = orig; /* <----- is this "bad"? */
c = int();
}
public:
int a,b,c,d;
};
A X;
X.a = 123;
X.b = 456;
X.c = 789;
X.d = 987;
A Y(X);
printf("X: %d %d %d %d\n",X.a,X.b,X.c,X.d);
printf("Y: %d %d %d %d\n",Y.a,Y.b,Y.c,Y.d);
Output:
X: 123 456 789 987
Y: 123 456 0 987
Alternative Copy Constructor:
A(const A & orig):a(orig.a),b(orig.b),c(0),d(orig.d){} /* <-- is this "better"? */
As brone points out, you're better off implementing assignment in terms of copy construction. I prefer an alternative idiom to his:
T& T::operator=(T t) {
swap(*this, t);
return *this;
}
It's a bit shorter, and can take advantage of some esoteric language features to improve performance. Like any good piece of C++ code, it also has some subtleties to watch for.
First, the t parameter is intentionally passed by value, so that the copy constructor will be called (most of the time) and we can modify is to our heart's content without affecting the original value. Using const T& would fail to compile, and T& would trigger some surprising behaviour by modifying the assigned-from value.
This technique also requires swap to be specialized for the type in a way that doesn't use the type's assignment operator (as std::swap does), or it will cause an infinite recursion. Be careful of any stray using std::swap or using namespace std, as they will pull std::swap into scope and cause problems if you didn't specialize swap for T. Overload resolution and ADL will ensure the correct version of swap is used if you have defined it.
There are a couple of ways to define swap for a type. The first method uses a swap member function to do the actual work and has a swap specialization that delegates to it, like so:
class T {
public:
// ....
void swap(T&) { ... }
};
void swap(T& a, T& b) { a.swap(b); }
This is pretty common in the standard library; std::vector, for example, has swapping implemented this way. If you have a swap member function you can just call it directly from the assignment operator and avoid any issues with function lookup.
Another way is to declare swap as a friend function and have it do all of the work:
class T {
// ....
friend void swap(T& a, T& b);
};
void swap(T& a, T& b) { ... }
I prefer the second one, as swap() usually isn't an integral part of the class' interface; it seems more appropriate as a free function. It's a matter of taste, however.
Having an optimized swap for a type is a common method of achieving some of the benefits of rvalue references in C++0x, so it's a good idea in general if the class can take advantage of it and you really need the performance.
With your version of the copy constructor the members are first default-constructed and then assigned.
With integral types this doesn't matter, but if you had non-trivial members like std::strings this is unneccessary overhead.
Thus, yes, in general your alternative copy constructor is better, but if you only have integral types as members it doesn't really matter.
Essentially, what you are saying is that you have some members of your class which don't contribute to the identity of the class. As it currently stands you have this expressed by using the assignment operator to copy class members and then resetting those members which shouldn't be copied. This leaves you with an assignment operator that is inconsistent with the copy constructor.
Much better would be to use the copy and swap idiom, and express which members shouldn't be copied in the copy constructor. You still have one place where the "don't copy this member" behaviour is expressed, but now your assignment operator and copy constructor are consistent.
class A
{
public:
A() : a(), b(), c(), d() {}
A(const A& other)
: a(other.a)
, b(other.b)
, c() // c isn't copied!
, d(other.d)
A& operator=(const A& other)
{
A tmp(other); // doesn't copy other.c
swap(tmp);
return *this;
}
void Swap(A& other)
{
using std::swap;
swap(a, other.a);
swap(b, other.b);
swap(c, other.c); // see note
swap(d, other.d);
}
private:
// ...
};
Note: in the swap member function, I have swapped the c member. For the purposes of use in the assignment operator this preserves the behaviour to match that of the copy constructor: it re-initializes the c member. If you leave the swap function public, or provide access to it through a swap free function you should make sure that this behaviour is suitable for other uses of swap.
Personally I think the broken assignment operator is killer. I always say that people should read the documentation and not do anything it tells them not to, but even so it's just too easy to write an assignment without thinking about it, or use a template which requires the type to be assignable. There's a reason for the noncopyable idiom: if operator= isn't going to work, it's just too dangerous to leave it accessible.
If I remember rightly, C++0x will let you do this:
private:
A &operator=(const A &) = default;
Then at least it's only the class itself which can use the broken default assignment operator, and you'd hope that in this restricted context it's easier to be careful.
I would call it bad form, not because you double-assign all your objects, but because in my experience it's often bad form to rely on the default copy constructor / assignment operator for a specific set of functionality. Since these are not in the source anywhere, it's hard to tell that the behavior you want depends on their behavior. For instance, what if someone in a year wants to add a vector of strings to your class? You no longer have the plain old datatypes, and it would be very hard for a maintainer to know that they were breaking things.
I think that, nice as DRY is, creating subtle un-specified requirements is much worse from a maintenance point of view. Even repeating yourself, as bad as that is, is less evil.
I think the better way is not to implement a copy constructor if the behaviour is trivial (in your case it appears to be broken: at least assignment and copying should have similar semantics but your code suggests this won't be so - but then I suppose it is a contrived example). Code that is generated for you cannot be wrong.
If you need to implement those methods, most likely the class could do with a fast swap method and thus be able to reuse the copy constructor to implement the assignment operator.
If you for some reason need to provide a default shallow copy constructor, then C++0X has
X(const X&) = default;
But I don't think there is an idiom for weird semantics. In this case using assignment instead of initialization is cheap (since leaving ints uninitialized doesn't cost anything), so you might just as well do it like this.
I have class foo that contains a std::auto_ptr member that I would like to copy construct but this does not appear to be allowed. There's a similar thing for the assignment. See the following example:
struct foo
{
private:
int _a;
std::string _b;
std::auto_ptr< bar > _c;
public:
foo(const foo& rhs)
: _a(rhs._a)
, _b(rhs._b)
, _c(rhs._c)
// error: Cannot mutate rhs._c to give up ownership - D'Oh!
{
}
foo& operator=(const foo& rhs)
{
_a = rhs._a;
_b = rhs._b;
_c = rhs._c;
// error: Same problem again.
}
};
I could just declare _c as mutable but I'm not sure this is correct. Does anyone have a better solution?
EDIT
OK, I'm not getting the kind of answer that I was expecting so I'll be a little more specific about the problem.
An object of type foo is created on the stack and passed by value into a container class (not stl) and then goes out of scope. I don't have any control over the container code. (It's actually an active queue implementation, with bugs.)
The bar class is a fairly heavyweight parser. It has very poor performance on new and delete so even if it was copy constructable, it would be way too expensive.
We can guarantee that when a bar object is created, it will only ever need to be owned in 1 place at a time. In this case it is being passed between threads and deleted when the transaction is completed. This is why I was hoping to use a std::autp_ptr.
I am very willing to consider boost smart pointers but I was hoping to guarantee this uniqueness if there is an alternative.
You might want to try following code:
foo(const foo& rhs)
: _a(rhs._a)
, _b(rhs._b)
, _c(_rhs._c.get() ? new bar(*_rhs._c.get()) : 0)
{
}
(Assignment operator is similar.)
However this will only work if bar is CopyConstructible and if this indeed does what you want. The thing is that both foo objects (_rhs and constructed one) will have different pointers in _c.
If you want them to share the pointer then you must not use auto_ptr as it does not support shared ownership. Consider in such case use of shared_ptr from Boost.SmartPtr for example (which will be included in new C++ standard). Or any other shared pointer implementation as this is such a common concept that lots of implementations are available.
As you have discovered you can't copy a std::auto_ptr like that. After the copy who owns the object pointed to? Instead you should use a reference counted smart pointer. The Boost library has a shared_ptr you could use.
First, I'd avoid auto_ptr
Transfer of ownership is good in some scenarios, but I find they are rare, and "full fledged" smart pointer libraries are now available easily. (IIRC auto_ptr was a compromise to include at least one example in the standard library, without the delays that a good implementation would have required).
See, for example here
or here
Decide on semantics
Should the copy of foo hold a reference to the same instance of bar? In that case, use boost::shared_ptr or (boost::intrusive_ptr), or a similar library.
Or should a deep copy be created?
(That may sometimes be required, e.g. when having delay-created state). I don't know any standard implementation of that concept, but it's not to complex to build that similar to existing smart pointers.
// roughly, incomplete, probably broken:
template <typename T>
class deep_copy_ptr
{
T * p;
public:
deep_copy_ptr() : p(0) {}
deep_copy_ptr(T * p_) : p(p_) {}
deep_copy_ptr(deep_copy_ptr<T> const & rhs)
{
p = rhs.p ? new T(*rhs.p) : 0;
}
deep_copy_ptr<T> & operator=(deep_copy_ptr<T> const & rhs)
{
if (p != rhs.p)
{
deep_copy_ptr<T> copy(rhs);
swap(copy);
}
}
// ...
}
The std::auto_ptr is a good tool for managing dynamic object in C++ but in order to use it effectivelly it's important to unserstand how auto_ptr works. This article explains why, when and where this smart pointer should be used.
In your case, first of all your should decide what you want to do with the object inside your auto_ptr. Should it be cloned or shared?
If it should be cloned, make sure it has a copy constructor and then your create a new auto_ptr which contains a copy of your the object see Adam Badura's answer.
If it should shared, you should use boost::shared_ptr as Martin Liversage suggested.
If I have class containing an auto_ptr, and want deep-copy semantics, I generatally only do this for classes that have a virtual copy operator, i.e. clone().
Then, within the copy constructor, I initialize the auto_ptr to a clone() of the other; e.g.
class Foo
{
public:
Foo(const Foo& rhs) : m_ptr(rhs.m_ptr->clone());
private:
std::auto_ptr<T> m_ptr;
};
clone() is typically implemented as follows:
class T
{
std::auto_ptr<T> clone() const
{
return std::auto_ptr<T>(new T(*this));
}
};
We are imposing the condition that T is clonable, but this condition is essentially imposed by having a copiable class with an auto_ptr member.
The whole idea of the auto_ptr is that there's only one owner of the referred to object. This implies you cannot copy the pointer without removing the original ownership.
Since you cannot copy it, you also can't copy an object containing an auto_ptr.
You might try to use move-semantics by e.g. using std::swap instead of copy.
My first choice would be to avoid auto_ptr in this situation altogether. But if I were backed against a wall, I might try to use the keyword mutable in the declaration of _c - this will allow it to be modified even from a const reference.
Given the edit, then it appears you want tranfer of ownership semantics.
In that case, then you'll want to have your copy constructor and assignment operator accept non-const references to their arguments, and perform the initialization/assignment there.
You can't use const references in a copy constructor or assignment operator that involves an auto_ptr<>. Remove the const. In other words, use declarations like
foo(foo & rhs);
foo & operator=(foo & rhs);
These forms are explicitly mentioned in the Standard, primarily in section 12.8. They should be usable in any standard-conforming implementation. In fact, paragraphs 5 and 10 of 12.8 says that the implicitly defined copy constructor and assignment operator (respectively) will take a non-const reference if any of the members require it.