auto_ptr question in c++ - c++

I am new here.
I am also new on C++
So here is the class and function i wrote.But i got the compiler error
My class:
class fooPlayer
{
public:
void fooPlayerfunc(){}//doing something here
char askYesNo(std::string question);
};
class fooPlayerFactory
{
public:
virtual std::auto_ptr<fooPlayer> MakePlayerX() const;
virtual std::auto_ptr<fooPlayer> MakePlayerO() const;
private:
std::auto_ptr<fooPlayer> MakePlayer(char letter) const;
std::auto_ptr<fooPlayer> my_player;
};
Implement my class:
auto_ptr<fooPlayer> fooPlayerFactory:: MakePlayer(char letter) const
{
my_player->fooPlayerfunc();
return my_player;
}
auto_ptr<fooPlayer> fooPlayerFactory::MakePlayerX() const
{
char go_first = my_player->askYesNo("Do you require the first move?");
MakePlayer(go_first);
return my_player;
}
auto_ptr<fooPlayer> fooPlayerFactory::MakePlayerO() const
{
return my_player;
}
My main() function here:
int main()
{
fooPlayerFactory factory;
factory.MakePlayerX();
factory.MakePlayerO();
}
I got the error:
error C2558: class 'std::auto_ptr<_Ty>' : no copy constructor available or copy constructor is declared 'explicit'
I do not know how to change it even after reading the document on this link:

The reason for the error is that you are calling the copy constructor of auto_ptr my_player in fooPlayerFactory::MakePlayerO() which is a const method. That means that is cannot modify its members.
However the copy constructor of auto_ptr DOES modify the right hand side so returning my_player trys to change its pointer to 0 (NULL), while assigning the original pointer to the auto_ptr in the return value.
The signature of the copy constuctor is
auto_ptr<T>::auto_ptr<T>(auto_ptr<T> & rhs)
not
auto_ptr<T>::auto_ptr<T>(const auto_ptr<T> & rhs)
The copy constructor of auto_ptr assigns ownership of the pointer to the left hand side, the right hand side then holds nothing.
I don't think you want to use auto_ptr here, you probably want boost::smart_ptr
It looks like you have mixed up two uses for auto_ptr
The first is as poor man's boost::scoped_ptr. This is to manage a single instance of a pointer in a class, the class manages the life time of the pointer. In this case you don't normally return this pointer outside your class (you can it is legal, but boost::smart_ptr / boost::weak_ptr would be better so clients can participate the life time of the pointer)
The second is its main purpose which is to return a newly created pointer to the caller of a function in an exception safe way.
eg
auto_ptr<T> foo() {
return new T;
}
void bar() {
auto_ptr<T> t = foo();
}
As I said I think you have mixed these two uses auto_ptr is a subtle beast you should read the auto_ptr docs carefully. It is also covered very well in Effective STL by Scott Meyers.

In your code:
auto_ptr<fooPlayer> fooPlayerFactory:: MakePlayer(char letter) const
{
my_player->fooPlayerfunc();
return my_player;
}
This is a const function, but fooPlayerfunc is not const - my compiler reports this error rather than the one you say you are getting. Are you posting the real code?

I don't think you actually want to constructing dynamic objects here.
A factory object creates and returns an object it normally does not keep a reference to it after creation (unless you are sharing it), and I don't actually see anywhere that you are creating the player.
If you only ever create one player internally in your (fooPlayerFactory). Then create an object and return references to it.

Edit: in response to the comment (which is correct, my bad), I left only the advice part.
Best practice is to have the factory methods just return a plain old pointer to the underlying object, and let the caller decide how to manage ownership (auto_ptr, scoped_ptr, or whatever).

Also your code is buggy, any class that implements virtual methods should have a virtual destructor.

I'm not seeing anywhere you construct my_player, so I have a feeling that some of the code is missing. Specifically, I think your constructor has this line:
my_player = new fooPlayer()
A fooPlayer object is not quite the same thing as an auto_ptr<fooPlayer> object, and auto_ptr is intentionally designed to prevent assigning from one to the other because, frankly, the alternative is worse. For the details, look up (1) conversion constructors, (2) the explicit keyword, and (3) copy constructors and destructive copy semantics.
You should change the constructor to either:
class fooPlayerFactory {
public:
fooPlayerFactory()
{
my_player = std::auto_ptr<fooPlayer>(new fooPlayer());
}
Or (using a member initializer list):
class fooPlayerFactory {
public:
fooPlayerFactory() : my_player(std::auto_ptr<fooPlayer>(new fooPlayer()) { }
The solution isn't pretty but, like I said, the alternative is worse due to some really arcane details.
As a bit of advice, though, you're making life harder than it needs to be; and may in fact be causing strange bugs. auto_ptr exists to manage the lifetime of an object, but the only reason you need to worry about the lifetime of my_player is that you've allocated it with new. But there's no need to call new, and in fact there's no need to keep my_player. And unless fooPlayerFactory is meant to be the base class for some other factory, there's no need to mark functions virtual.
Originally I thought you could get away with simply returning copies of the my_player object, but there's a problem: before returning my_player from MakePlayer() you call a method on it, and I assume that method changes the internal state of my_player. Further calls to MakePlayer() will change the state again, and I think you're going to eventually have my_player in the wrong state. Instead, return a different fooPlayer object with each request. Don't do memory management, just promise to construct the object. That way the user can decide on memory allocation:
fooPlayerFaclotry factory;
fooPlayer on_stack = factory.MakePlayerX();
fooPlayer* on_heap_raw_pointer = new fooPlayer(factory.MakePlayerO());
std::auto_ptr<fooPlayer> on_heap_managed_scope
= std::auto_ptr<fooPlayer>(factory.MakePlayerX());
I would change fooPlayerFactory to look like this:
class fooPlayerFactory
{
private:
fooPlayer MakePlayer(const char letter) const
{
fooPlayer result;
result.fooPlayerfunc();
return result;
}
public:
fooPlayer* MakePlayerX() const
{
char go_first = askYesNo("Do you require the first move?");
return MakePlayer(go_first);
}
fooPlayer MakePlayerO() const
{
return fooPlayer();
}
};

Related

How to use getters and setters without generating a copy?

I would like to know how to use getters and setters for a member variable which takes a lot of memory. Usualy i would do as bellow:
class A
{
private:
BigObject object;
public:
BigObject getObject() const
{
return object;
}
void setObject(const BigObject& object)
{
this->object = object;
}
};
However this getter and setter i believe will copy the BigObject which i do not want. Is there a better way to do this?
I thought of doing it this way but i read on the internet that it's not a good idea because it can lead to a segmentation fault if used badly:
BigObject& getObject()
{
return object
}
(If you do not care about encapsulation in this case, meaning the A::object member should be modifiable by anyone without restriction, then look at SergeyA's answer).
Return by const reference in order to avoid copying and still maintain encapsulation (meaning the caller can't modify the member by mistake):
const BigObject& getObject() const
{
return object;
}
If the caller actually wants a copy, they can do so easily themselves.
If you want to prevent dangling references (the segfault you mentioned) when the getter is used on a temporary, you can only return a copy when the getter is actually called on a temporary:
BigObject getObject() const &&
{
return object;
}
const BigObject& getObject() const &
{
return object;
}
This will return a copy when calling getObject() on a temporary. Alternatively, you can completely prevent calling the getter on a temporary by deleting that particular overload:
BigObject getObject() const && = delete;
const BigObject& getObject() const &
{
return object;
}
Keep in mind that this is not a guaranteed safety net. It prevents some mistakes, but not all of them. The caller of the function should still be aware about object lifetimes.
You can also improve your setter, btw. Right now, it will always copy the object regardless how the caller passes the argument. You should take it by value instead and move it into the member:
void setObject(BigObject object)
{
this->object = std::move(object);
}
This requires that BigObject is movable though. If it's not, then this will be even worse than before.
Best solution: make code of your class exactly that:
struct A
{
BigObject object;
};
Explanation - avoid trivial setters and getters. If you find yourself putting those into your classes, expose the member directly and be done with it.
Do not ever listen to people who'd say "But what if in the future we add non-trivial logic"? I have seen more than a healthy dose of trivial setters and getters, been around for decades, and never replaced with something non-trivial.
The common practice is to:
class A
{
public:
// this method is const
const& BigObject getObject() const
{
return object;
}
// this method is not const
void setObject(const BigObject& object)
{
object = object;
}
private:
BigObject object;
};
If you need to get a read-only object - it's perfectly fine. Otherwise, consider changes in architecture.
An alternative would be to store a std::shared_ptr and return a std::shared_ptr or std::weak_ptr.
Instead of returning a copy of the member, you can return a reference to it. This way there is no need to copy the member.
I thought of doing it this way but i read on the internet that it's not a good idea because it can lead to a segmentation fault if used badly
The solution is to not use it badly.
Returning a reference to a member is fairly common pattern and not generally discouraged. Although, for types that are fast to copy, returning a copy is generally superior when there is no need to refer to the member itself.
There is a solution that avoids both copying and breakage of encapsulation: Use a shared pointer, and return a copy of that shared pointer from the getter. However, this approach has a runtime cost and requires dynamic allocation, so it is not ideal for all use cases.
In case of setter, you may use move assignment instead, which is more efficient than copy assignment for some types.

C++ force dynamic allocation with unique_ptr?

I've found out that unique_ptr can point to an already existing object.
For example, I can do this :
class Foo {
public:
Foo(int nb) : nb_(nb) {}
private:
int nb_;
};
int main() {
Foo f1(2);
Foo* ptr1(&f1);
unique_ptr<Foo> s_ptr1(&f1);
return 0;
}
My question is :
If I create a class with unique_ptr< Bar > as data members (where Bar is a class where the copy constructor was deleted) and a constructor that takes pointers as argument, can I prevent the user from passing an already existing object/variable as an argument (in that constructor) (i.e. force him to use the new keyword) ?
Because if he does, I won't be able to guarantee a valide state of my class objects (the user could still modify data members with their address from outside of the class) .. and I can't copy the content of Bar to another memory area.
Example :
class Bar {
public:
Bar(/* arguments */) { /* data members allocation */ }
Bar(Bar const& b) = delete;
/* Other member functions */
private:
/* data members */
};
class Bar_Ptr {
public:
Bar_Ptr(Bar* ptr) {
if (ptr != nullptr) { ptr_ = unique_ptr<Bar> (ptr); }
} /* The user can still pass the address of an already existing Bar ... */
/* Other member functions */
private:
unique_ptr<Bar> ptr_;
};
You can't prevent programmers from doing stupid things. Both std::unique_ptr and std::shared_ptr contain the option to create an instance with an existing ptr. I've even seen cases where a custom deleter is passed in order to prevent deletion. (Shared ptr is more elegant for those cases)
So if you have a pointer, you have to know the ownership of it. This is why I prefer to use std::unique_ptr, std::shared_ptr and std::weak_ptr for the 'owning' pointers, while the raw pointers represent non-owning pointers. If you propagate this to the location where the object is created, most static analyzers can tell you that you have made a mistake.
Therefore, I would rewrite the class Bar_ptr to something like:
class Bar_ptr {
public:
explicit Bar_ptr(std::unique_ptr<Bar> &&bar)
: ptr(std::move(bar)) {}
// ...
}
With this, the API of your class enforces the ownership transfer and it is up to the caller to provide a valid unique_ptr. In other words, you shouldn't worry about passing a pointer which isn't allocated.
No one prevents the caller from writing:
Bar bar{};
Bar_ptr barPtr{std::unique_ptr<Bar>{&bar}};
Though if you have a decent static analyzer or even just a code review I would expect this code from being rejected.
No you can't. You can't stop people from doing stupid stuff. Declare a templated function that returns a new object based on the templated parameter.
I've seen something similar before.
The trick is that you create a function (let's call it make_unique) that takes the object (not pointer, the object, so maybe with an implicit constructor, it can "take" the class constructor arguments) and this function will create and return the unique_ptr. Something like this:
template <class T> std::unique_ptr<T> make_unique(T b);
By the way, you can recommend people to use this function, but no one will force them doing what you recommend...
You cannot stop people from doing the wrong thing. But you can encourage them to do the right thing. Or at least, if they do the wrong thing, make it more obvious.
For example, with Bar, don't let the constructor take naked pointers. Make it take unique_ptrs, either by value or by &&. That way, you force the caller to create those unique_ptrs. You're just moving them into your member variables.
That way, if the caller does the wrong thing, the error is in the caller's code, not yours.

Pros and Cons of usage of reference in case of PIMPL idiom

As mentioned here you can use reference (d-reference) instead of pointer (d-pointer) in case of PIMPL idiom.
I'm trying to understand if there are any serious issues with this implementation and what are the pros and cons.
Pros:
Shorter syntax because of usage of "." instead of "->".
...
Cons:
What if the new ObjectPivate() fails and new doesn't throw (e.g.: new(std::nothrow) or custom new) and returns nullptr instead? You need to implement additional stuff to check if the referance is valid. In case of pointer you just use:
if (m_Private)
m_Private->Foo();
In rare case of multiple constructors for the Object with complex initialisation logic the solution could be not applicable. [© JamesKanze]
It fills more natural to use pointer for memory management. [© JamesKanze]
Some additional implementation details needs to be considered (use of swap()) to ensure the exception-safety (e.g. implementation of assignment operator) [© Matt Yang]
...
Here the sample code for illustration:
// Header file
class ObjectPrivate;
class Object
{
public:
Object();
virtual ~Object();
virtual void Foo();
private:
ObjectPrivate& m_Private;
};
// Cpp file
class ObjectPrivate
{
public:
void Boo() { std::cout << "boo" << std::endl; }
};
Object::Object() :
m_Private(* new ObjectPrivate())
{
}
Object::~Object()
{
delete &m_Private;
}
void Object::Foo()
{
m_Private.Boo();
}
It's really just a matter of style. I tend to not use
references in classes to begin with, so using a pointer in the
compilation firewall just seems more natural. But there's
usually no real advantage one way or the other: the new can
only fail by means of an exception.
The one case where you might favor the pointer is when the
object has a lot of different constructors, some of which need
preliminary calculations before calling the new. In this
case, you can initialize the pointer with NULL, and then call
a common initialization routine. I think such cases are rare,
however. (I've encountered it once, that I can recall.)
EDIT:
Just another style consideration: a lot of people don't like something like delete &something;, which is needed if you use references rather than pointers. Again, it just seems more natural (to me, at least), that objects managing memory use pointers.
It's not convenient to write exception-safe code I think.
The first version of Object::operator=(Object const&) might be:
Object& operator=(Object const& other)
{
ObjectPrivate *p = &m_Private;
m_Private = other.m_Private; // Dangerous sometimes
delete *p;
}
It's dangerous if ObjectPrivate::operator=(ObjectPrivate const&) throws exception. Then what about using a temporary variable? Aha, no way. operator=() has to be invoked if you want change m_Private.
So, void ObjectPrivate::swap(ObjectPrivate&) noexcept can act as our savior.
Object& operator=(Object const& other)
{
ObjectPrivate *tmp = new ObjectPrivate(other.m_Private);
m_Private.swap(*tmp); // Well, no exception.
delete tmp;
}
Then consider the implementation of void ObjectPrivate::swap(ObjectPrivate&) noexcept. Let's assume that ObjectPrivate might contain a class instance without swap() noexcept or operator=() noexcept. I think it's hard.
Alright then, this assumption is too strict and not correct sometimes. Even so, it's not necessary for ObjectPrivate to provide swap() noexcept in most cases, because it's usually a helper structure to centralize data.
By contrast, pointer can save a lot of brain cells.
Object& operator=(Object const& other)
{
ObjectPrivate *tmp = new ObjectPrivate(*other.p_Private);
delete p_Private;
p_Private = tmp; // noexcept ensured
}
It's much more elegant if smart pointers are used.
Object& operator=(Object const& other)
{
p_Private.reset(new ObjectPrivate(*other.p_Private));
}
Some quick and obvious additions:
Pro
The reference must not be 0.
The reference may not be assigned another instance.
Class responsibilities/implementation are simpler due to fewer variables.
The compiler could make some optimizations.
Con
The reference may not be assigned another instance.
The reference will be too restrictive for some cases.

C++ Return by reference

Say, i have a function which returns a reference and i want to make sure that the caller only gets it as a reference and should not receive it as a copy.
Is this possible in C++?
In order to be more clear. I have a class like this.
class A
{
private:
std::vector<int> m_value;
A(A& a){ m_value = a.m_value; }
public:
A() {}
std::vector<int>& get_value() { return m_value; }
};
int main()
{
A a;
std::vector<int> x = a.get_value();
x.push_back(-1);
std::vector<int>& y = a.get_value();
std::cout << y.size();
return 0;
}
Thanks,
Gokul.
You can do what you want for your own classes by making the class non copyable.
You can make an class non copyable by putting the copy constructor and operator= as private or protected members.
class C
{
private:
C(const C& other);
const C& operator=(const C&);
};
There is a good example of making a NonCopyable class here that you can derive from for your own types.
If you are using boost you can also use boost::noncopyable.
Alt solution:
Another solution is to have a void return type and make the caller pass their variable by reference. That way no copy will be made as you're getting a reference to the caller's object.
If your function returns a reference to an object that shouldn't have been copied, then your function already has done what it could do to prevent copying. If someone else calls your function and copies the return value, then either
it's an error the caller made, because the object should never be copied (in which case the return type probably shouldn't have been copyable in the first place), or
it's irrelevant for the caller because the function is only called once in a week (in which case you must not try to cripple your callers' code), or
it's a pretty dumb oversight on the side of the caller (in which case the error will be found by profiling).
For #1, either you return have your own type or you can wrap whatever your return in your own type. Note that the only difference between #2 and #3 is the relevance - and if it's relevant, profiling will find it.
IMO you should not cripple your code by returning a pointer when what you need is a reference. Experienced programmers, seeing the pointer, will immediately ask whether they need to check for a NULL return value, whether the object is allocated dynamically and, if so, who is responsible for cleaning it up.
You should also not blindly forbid copying of whatever you return, if you cannot eliminate the possibility that copying is needed.
In the end it's the old motto, which C++ inherited from C: Trust your users to know what they are doing.
It "depends". Yes, you can hide the copy-constructor (and assignment operator), and your object becomes noncopyable:
struct foo
{
private:
foo(const foo&); // dont define
foo& operator=(const foo&); // dont define
}:
But if you're wondering about one specific function (i.e., normally copyable, but not for this function), no. In fact, what can you do about the caller anyway?
const foo& f = get_foo(); // okay, by reference, but...
foo f2 = foo(foo(foo(foo(foo(foo(f)))))); // :[
If your caller wants to do something, there isn't much you can do to stop it.
In C++11, you can prevent the copy constructor from being called by deleting it:
class A{
public:
A(const A&) = delete;
}
Are you trying to prevent a common typo that causes large objects to accidentally be copied? If so, you could return by pointer instead. Leaving off an & is pretty easy, but it takes a little bit of effort to copy an object from a pointer. OTOH, the resulting code will be uglier, so it's up to you whether it's worth it.

Swap method with const members

I want to implement a Swap() method for my class (let's call it A) to make copy-and-swap operator=(). As far as I know, swap method should be implemented by swapping all members of the class, for example:
class A
{
public:
void swap(A& rhv)
{
std::swap(x, rhv.x);
std::swap(y, rhv.y);
std::swap(z, rhv.z);
}
private:
int x,y,z;
};
But what should I do if I have a const member? I can't call std::swap for it, so I can't code A::Swap().
EDIT: Actually my class is little bit more complicated. I want to Serialize and Deserialize it. Const member is a piece of data that won't change (its ID for example) within this object. So I was thinking of writing something like:
class A
{
public:
void Serialize(FILE* file) const
{
fwrite(&read_a, 1, sizeof(read_a), file);
}
void Deserialize(FILE* file) const
{
size_t read_a;
fread(&read_a, 1, sizeof(read_a), file);
A tmp(read_a);
this->Swap(tmp);
}
private:
const size_t a;
};
and call this code:
A a;
FILE* f = fopen(...);
a.Deserialize(f);
I'm sorry for such vague wording.
I think what you really want is to have an internal data structure that you can easily exchange between objects. For example:
class A
{
private:
struct A_Data {
int x;
int y;
const int z;
A_Data(int initial_z) : z(initial_z) {}
};
std::auto_ptr<A_Data> p_data;
public:
A(int initial_z) : p_data(new A_Data(initial_z)) {}
void swap(A& rhv) {
std::swap(p_data, rhv.p_data);
}
};
This keeps the z value constant within any instance of A object internal data, but you can swap the internal data of two A objects (including the constant z value) without violating const-correctness.
After a good nights sleep I think the best answer is to use a non-const pointer to a const value -- after all these are the semantics you are trying to capture.
f0b0s, a good design principle is to design your objects to be immutable. This means that the object can't change once created. To "change" the object, you must copy the object and make sure to change the elements you want.
That being said, in this case you should look at using a copy constructor instead to copy the objects you want to swap, and then actually swap the references to the object. I can understand it'd be tempting just to be able to change the elements of an object under the hood, but it'd be better to make a copy of the object and replace the references to that object with the NEW object instead. This gets you around any const nastiness.
Hope this helps.
I suggest you use pointers to the instances. The pointers can be swapped much easier than the data in the class or struct.
The only way to swap a constant value is to create another object, or clone the current object.
Given a struct:
struct My_Struct
{
const unsigned int ID;
std::string name;
My_Struct(unsigned int new_id)
: ID(new_id)
{ ; }
};
My understanding is that you want to swap instances of something like My_Struct above. You can copy the mutable (non-const) members but not the const member. The only method to alter the const member is to create a new instance with a new value for the const member.
Perhaps you need to rethink your design.
IMHO you must consider not to swap CONST members.
PD: I think you could consider to use reflection in your approach. so you don't have to maintain the function.
This is why const_cast was created. Just remember not to shoot your foot off.
Edit: OK, I concede - const_cast wasn't made for this problem at all. This might work with your compiler, but you can't count on it and if demons come flying out of your nostrils, please don't blame me.
tl;dr; : It's Undefined Behavior.
Reference/reason: CppCon 2017: Scott Schurr “Type Punning in C++17: Avoiding Pun-defined Behavior, #24m52s +- ”
My interpretation, by example:
Suppose you create an object of type T, which have some const members. You can pass this object as a non-const reference to a function f(&T) that manipulates it, but you'd expect the const members to remain unalterable after the call. swap can be called in non-const references, and it can happen inside the function f, breaking the premise of const members to the caller.
Every part of your code that uses swap would have to assert that the object of type T being swapped does not belong to any context where the const members are assumed constant. That is impossible to automatically verify*.
*I just assumed that this is impossible to verify because it seems like an extension of the undecidability of the halting problem.