When overloading the = operator, should one make the contents of one object equal to the contents of the other object OR do you make the pointer of the object point to the same object?
Reading back on the question it seems that the contents should be copied and not the pointers. But I just can't figure it out, So I would be grateful if someone would explain what I should do, I know how to do both, I'm just not sure which one to choose.
class IntObject
{
private:
int *pi_One;
public:
IntObject(void);
IntObject::IntObject(int const &i_one);
~IntObject(void);
IntObject & operator=(const IntObject&);
};
IntObject::IntObject()
{
pi_One = new int(0);
}
IntObject::IntObject(int const &i_one)
{
pi_One = new int(i_one);
}
IntObject::~IntObject(void)
{
delete pi_One;
}
IntObject & IntObject::operator=(const IntObject& c) {
//This copies the pointer to the ints
this->pi_One = c.pi_One;
return *this;
}
It depends on what semantics you want to have in your type. If you want value semantics, then copy the contents (deep copy, as is the case in std::vector), if you want reference semantics (shallow copy, as in std::shared_ptr)
You should definitely copy the contents, not the pointers. Think about what you will do when one of the objects which both hold the same pointer is destroyed; you can't delete the pointer because the other object would be affected as well, but you can't not delete it either because you'd cause memory leaks. You'd have to use reference counting and everything would get a whole lot more complicated.
The contents should be copied (in fact, changing the pointer of the object shouldn't actually be possible - I can't imagine how you would do that - and even if it is somehow, you're not supposed to do it). You also have to take care of the differences between deep and shallow copies, especially if your class contains pointers (or containers with pointers in them).
Now that I think about it, I'm not even sure which pointer you could possibly want to reassign. Unless you are already working with a pointer - those already have an '=' operator that shouldn't be overloaded though.
The principle of least astonishment would say to copy the content. When using operator= on any other object, you wouldn't expect it to copy pointers.
If you keep your destructor as it is, then you should change assignment overload. It would be also wise that nobody is attempting assigning IntObject to itself:
IntObject & IntObject::operator=(const IntObject& c) {
if (this != &c)
{
//This copies the pointer to the ints
*this->pi_One = *c.pi_One;
}
return *this;
}
Otherwise, there will be attempt to free freed memory in IntObject's destructor
Related
can someone walk through exactly what happens with the memory in this operator overload function? I am confused on how exactly the object created inside the operator function gets deallocated in the main.
Object& operator+(const Object& other) {
Object o(*this); //create instance of o that deep copies first argument
...
//copy contents of other and add onto o
return o;
}
int main() {
Object b;
Object c;
Object a = b + c;
}
Edit: to be more specific, isn't it bad practice to create a local object in a function and then return it by reference? Wouldn't that cause a memory leak?
Edit 2: I am referencing my textbook Data abstraction & problem solving with c++ carrano which suggests an operator + overload for LinkedLists in this format: LinkedList<ItemType>& operator+(const LinkedList<ItemType>& rightHandSide) const;. They implemented the method in the way I described.
Edit 2.5: the full method pseudocode given by the book:
LinkedList<ItemType>& operator+(const LinkedList<ItemType>& rightHandSide) const {
concatList = a new, empty instance of LinkedList
concatList.itemCount = itemCount + rightHandSide.itemCount
leftChain = a copy of the chain of nodes in this list
rightChain = a copy of the chain of nodes in the list rightHandSide
concatList.headPtr = leftChain.headPtr
return concatList
}
Edit 3: Asked my professor about this. Will get to the bottom of this by tomorrow.
Edit 4: The book is wrong.
Returning a reference to a local object
As everyone else correctly states, returning a reference to a local object results in undefined behaviour. You will end up with a handle to a destroyed function-scope object.
Returning references in arithmetic operators
If you think about it, a+b should give you a result, but it shouldn't change a nor b. C++ however leaves it up to you to define how operators work on your own types so it's possible to implement the behaviour you need. This is why the operator+ usually has to create a new object and can't return a reference.
Compound assignments (+=, -=, etc) on the other hand do change the object itself so a += b is changing a. This is why it's usually being implemented by returning a reference (not to a local object, but to the instance itself):
Object& Object::operator+=(const Object& rhs)
{
// do internal arithmetics to add 'rhs' to this instance
return *this; // here we return the reference, but this isn't a local object!
}
It wouldn't cause a memory leak, but o gets destroyed when it goes out of scope, when the function returns. So the reference the caller has is junk. It might appear to work fine for a short time until the memory is overwritten later.
It is simply undefined behavior.
In terms of what happens to memory, the memory will not be reserved to the object after the function returns (because the object is now out of scope).
So it can contain ANYTHING, including the same object by way of coincidence.
Since C++11, when using the move assignment operator, should I std::swap all my data, including POD types? I guess it doesn't make a difference for the example below, but I'd like to know what the generally accepted best practice is.
Example code:
class a
{
double* m_d;
unsigned int n;
public:
/// Another question: Should this be a const reference return?
const a& operator=(a&& other)
{
std::swap(m_d, other.m_d); /// correct
std::swap(n, other.n); /// correct ?
/// or
// n = other.n;
// other.n = 0;
}
}
You might like to consider a constructor of the form: - ie: there are always "meaningful" or defined values stores in n or m_d.
a() : m_d(nullptr), n(0)
{
}
I think this should be rewriten this way.
class a
{
public:
a& operator=(a&& other)
{
delete this->m_d; // avoid leaking
this->m_d = other.m_d;
other.m_d = nullptr;
this->n = other.n;
other.n = 0; // n may represents array size
return *this;
}
private:
double* m_d;
unsigned int n;
};
should I std::swap all my data
Not generally. Move semantics are there to make things faster, and swapping data that's stored directly in the objects will normally be slower than copying it, and possibly assigning some value to some of the moved-from data members.
For your specific scenario...
class a
{
double* m_d;
unsigned int n;
...it's not enough to consider just the data members to know what makes sense. For example, if you use your postulated combination of swap for non-POD members and assignment otherwise...
std::swap(m_d, other.m_d);
n = other.n;
other.n = 0;
...in the move constructor or assignment operator, then it might still leave your program state invalid if say the destructor skipped deleting m_d when n was 0, or if it checked n == 0 before overwriting m_d with a pointer to newly allocated memory, old memory may be leaked. You have to decide on the class invariants: the valid relationships of m_d and n, to make sure your move constructor and/or assignment operator leave the state valid for future operations. (Most often, the moved-from object's destructor may be the only thing left to run, but it's valid for a program to reuse the moved-from object - e.g. assigning it a new value and working on it in the next iteration of a loop....)
Separately, if your invariants allow a non-nullptr m_d while n == 0, then swapping m_ds is appealing as it gives the moved-from object ongoing control of any buffer the moved-to object may have had: that may save time allocating a buffer later; counter-balancing that pro, if the buffer's not needed later you've kept it allocated longer than necessary, and if it's not big enough you'll end up deleting and newing a larger buffer, but at least you're being lazy about it which tends to help performance (but profile if you have to care).
No, if efficiency is any concern, don't swap PODs. There is just no benefit compared to normal assignment, it just results in unnecessary copies. Also consider if setting the moved from POD to 0 is even required at all.
I wouldn't even swap the pointer. If this is an owning relationship, use unique_ptr and move from it, otherwise treat it just like a POD (copy it and set it to nullptr afterwards or whatever your program logic requires).
If you don't have to set your PODs to zero and you use smart pointers, you don't even have to implement your move operator at all.
Concerning the second part of your question:
As Mateusz already stated, the assignment operator should always return a normal (non-const) reference.
I'm new to the C++ world, but I have some experience with C and read some tutorials about C++.
Now, creating objects in C++ seems quite easy and works well for me as long as the class has only attributes that are values (not pointers).
Now, when I try to create objects which allocate memory in the constructor for some of their attributes, I figure out how exactly such objects are passed between functions.
A simple example of such class would be:
class A {
int *a;
public:
A(int value) {
this->a = new int;
*(this->a) = value;
}
~A() {
delete this->a;
}
int getValue() const { return this->a; }
}
I want to use the class and pass it by value to other functions, etc. At least these examples must work without creating memory leaks or double free errors.
A f1() {
// some function that returns A
A value(5);
// ...
return value;
}
void f2(A a) {
// takes A as a parameter
// ...
}
A a = f1();
A b = a;
f2(a);
f2(f1());
The class A is incomplete because I should override operator= and A(A& oldValue) to solve some of these problems.
As I understand it, the default implementation of these methods just copy the value of the members which is causing the destructor to be called twice on the same pointer values.
Am I right and what else am I missing?
In addition, do you know any good tutorial that explains this issue?
Use containers and smart pointers.
E.g. std::vector for dynamic length array, or boost::shared_ptr for dynamically allocated single object.
Don't deal directly with object lifetime management.
Cheers & hth.,
When you pass an object like that, you will create a copy of the object. To avoid doing that, you should pass a const reference...
void f2(A const & a)
{
}
This does mean that you are not allowed to change 'a' in your function - but, to be honest, you shouldn't be doing that anyways, as any changes won't be reflected back to the original parameter that was passed in. So, here the compiler is helping you out, but not compiling when you would have made a hard to find error.
Specifically, you must implement a copy constructor that properly copies the memory pointer for the a variable. Any default constructor would simply copy the memory location for the a variable, which would obviously be subject to a double-delete.
Even doing this:
A value(5);
// ...
return value;
won't work because when A falls out of scope (at the end of the section) the delete operator for A will be called, thus deleting the a sub-variable and making the memory invalid.
so I have a structure like
struct GetResultStructure
{
int length;
char* ptr;
};
I need a way to make a full copy of it meaning I need a copy to have a structure with new ptr poinnting on to copy of data I had in original structure. Is It any how possible? I mean any structure I have which contains ptrs will have some fields with its lengths I need a function that would copy my structure coping all ptrs and data they point to by given array of lengthes... Any cool boost function for it? Or any way how to create such function?
For the specific scenario you describe, use a std::vector or some other sequence container. If you do so, then simply copying objects of type GetResultStructure will make copies of the pointed-to data as well:
struct GetResultStructure {
std::vector<char> data;
};
GetResultStructure a = GetData();
GetResultStructure b = a; // a and b have distinct, equivalent data vectors
In general, when you do need to implement this yourself, you do so by implementing a copy constructor and a copy assignment operator. The easiest way to do that is to use the copy-and-swap idiom, covered in great detail in What is the copy-and-swap idiom?
It's pretty much up to you to implement that. Normally you want to do it as a copy constructor so you only have to do it in one place. Unfortunately, there's no real magic to avoid telling the computer about how to copy your structure.
Of course, that only applies if your structure really is substantially different from something that's already written. The one you've given looks a lot like a string or (possibly) vector. Unless you really need to implement something new, you're probably better off just using one of those that's already provided.
Both a copy constructor and assignment operator should be implemented (in the way stated above). A technique which may aid in this process, however, is using a dereference operator (*) when copying pointer data. This will copy the pointer data rather than the memory locations. If you do ptr1 = ptr2 it simply sets the memory location of ptr1 to ptr2 which is why we dereference.
For instance, I'll just show a quick example for a copy constructor:
GetResultStructure(const GetResultStructure& other)
: length(other.length), ptr(new char[length]) // <--- VERY _important_ - initialization of pointer
{
// Alternatively, put your initialization here like so:
// ptr = new char[length];
for(int i=0;i<length;++i)
{
ptr[i] = new char;
*ptr[i] = *other.ptr[i]; // Copy the values - not the memory locations
}
}
And then obviously be sure to clean up in your destructor to prevent memory leaks.
Regards,
Dennis M.
GetResultStructure doCopy(GetResultStructure const& copy) {
GetResultStructure newstruct;
newstruct.length = copy.length;
newstruct.ptr = new char[newstruct.length];
memcpy(newstruct.ptr, copy.ptr, newstruct.length*sizeof(char));
return newstruct;
}
Should be simple. Yes, the sizeof(char) isn't really necessary, but there to show what to do for other data types.
Since you tagged it as C++: Write a copy constructor and an assignment operator,
within which you implement your deep copy code:
struct GetResultStructure
{
GetResultStructure(const GetResultStructure& other)
{
// Deep copy code in here
}
GetResultStructure& operator=(const GetResultStructure& other)
{
if (this != &other) {
// Deep copy code in here
}
return *this
}
int length;
char* ptr;
};
EDIT: I know in this case, if it were an actual class i would be better off not putting the string on the heap. However, this is just a sample code to make sure i understand the theory. The actual code is going to be a red black tree, with all the nodes stored on the heap.
I want to make sure i have these basic ideas correct before moving on (I am coming from a Java/Python background). I have been searching the net, but haven't found a concrete answer to this question yet.
When you reassign a pointer to a new object, do you have to call delete on the old object first to avoid a memory leak? My intuition is telling me yes, but i want a concrete answer before moving on.
For example, let say you had a class that stored a pointer to a string
class MyClass
{
private:
std::string *str;
public:
MyClass (const std::string &_str)
{
str=new std::string(_str);
}
void ChangeString(const std::string &_str)
{
// I am wondering if this is correct?
delete str;
str = new std::string(_str)
/*
* or could you simply do it like:
* str = _str;
*/
}
....
In the ChangeString method, which would be correct?
I think i am getting hung up on if you dont use the new keyword for the second way, it will still compile and run like you expected. Does this just overwrite the data that this pointer points to? Or does it do something else?
Any advice would be greatly appricated :D
If you must deallocate the old instance and create another one, you should first make sure that creating the new object succeeds:
void reset(const std::string& str)
{
std::string* tmp = new std::string(str);
delete m_str;
m_str = tmp;
}
If you call delete first, and then creating a new one throws an exception, then the class instance will be left with a dangling pointer. E.g, your destructor might end up attempting to delete the pointer again (undefined behavior).
You could also avoid that by setting the pointer to NULL in-between, but the above way is still better: if resetting fails, the object will keep its original value.
As to the question in the code comment.
*str = _str;
This would be the correct thing to do. It is normal string assignment.
str = &_str;
This would be assigning pointers and completely wrong. You would leak the string instance previously pointed to by str. Even worse, it is quite likely that the string passed to the function isn't allocated with new in the first place (you shouldn't be mixing pointers to dynamically allocated and automatic objects). Furthermore, you might be storing the address of a string object whose lifetime ends with the function call (if the const reference is bound to a temporary).
Why do you think you need to store a pointer to a string in your class? Pointers to C++ collections such as string are actually very rarely necessary. Your class should almost certainly look like:
class MyClass
{
private:
std::string str;
public:
MyClass (const std::string & astr) : str( astr )
{
}
void ChangeString(const std::string & astr)
{
str = astr;
}
....
};
Just pinpointing here, but
str = _str;
would not compile (you're trying to assign _str, which is the value of a string passed by reference, to str, which is the address of a string). If you wanted to do that, you would write :
str = &_str;
(and you would have to change either _str or str so that the constnest matches).
But then, as your intuition told you, you would have leaked the memory of whatever string object was already pointed to by str.
As pointed earlier, when you add a variable to a class in C++, you must think of whether the variable is owned by the object, or by something else.
If it is owned by the object, than you're probably better off with storing it as a value, and copying stuff around (but then you need to make sure that copies don't happen in your back).
It is is not owned, then you can store it as a pointer, and you don't necessarily need to copy things all the time.
Other people will explain this better than me, because I am not really confortable with it.
What I end up doing a lot is writing code like this :
class Foo {
private :
Bar & dep_bar_;
Baz & dep_baz_;
Bing * p_bing_;
public:
Foo(Bar & dep_bar, Baz & dep_baz) : dep_bar_(dep_bar), dep_baz_(dep_baz) {
p_bing = new Bing(...);
}
~Foo() {
delete p_bing;
}
That is, if an object depends on something in the 'Java' / 'Ioc' sense (the objects exists elsewhere, you're not creating it, and you only wants to call method on it), I would store the dependency as a reference, using dep_xxxx.
If I create the object, I would use a pointer, with a p_ prefix.
This is just to make the code more "immediate". Not sure it helps.
Just my 2c.
Good luck with the memory mgt, you're right that it is the tricky part comming from Java ; don't write code until you're confortable, or you're going to spend hours chasing segaults.
Hoping this helps !
The general rule in C++ is that for every object created with "new" there must be a "delete". Making sure that always happens in the hard part ;) Modern C++ programmers avoid creating memory on the heap (i.e. with "new") like the plague and use stack objects instead. Really consider whether you need to be using "new" in your code. It's rarely needed.
If you're coming from a background with garbage collected languages and find yourself really needing to use heap memory, I suggest using the boost shared pointers. You use them like this:
#include <boost/shared_ptr.hpp>
...
boost::shared_ptr<MyClass> myPointer = boost::shared_ptr<MyClass>(new MyClass());
myPointer has pretty much the same language semantics as a regular pointer, but shared_ptr uses reference counting to determine when delete the object it's referencing. It's basically do it yourself garbage collection. The docs are here: http://www.boost.org/doc/libs/1_42_0/libs/smart_ptr/smart_ptr.htm
I'll just write a class for you.
class A
{
Foo * foo; // private by default
public:
A(Foo * foo_): foo(foo_) {}
A(): foo(0) {} // in case you need a no-arguments ("default") constructor
A(const A &a):foo(new Foo(a.foo)) {} // this is tricky; explanation below
A& operator=(const &A a) { foo = new Foo(a.foo); return *this; }
void setFoo(Foo * foo_) { delete foo; foo = foo_; }
~A() { delete foo; }
}
For classes that hold resources like this, the copy constructor, assignment operator, and destructor are all necessary. The tricky part of the copy constructor and assignment operator is that you need to delete each Foo precisely once. If the copy constructor initializer had said :foo(a.foo), then that particular Foo would be deleted once when the object being initialized was destroyed and once when the object being initialized from (a) was destroyed.
The class, the way I've written it, needs to be documented as taking ownership of the Foo pointer it's being passed, because Foo * f = new Foo(); A a(f); delete f; will also cause double deletion.
Another way to do that would be to use Boost's smart pointers (which were the core of the next standard's smart pointers) and have boost::shared_ptr<Foo> foo; instead of Foo * f; in the class definition. In that case, the copy constructor should be A(const A &a):foo(a.foo) {}, since the smart pointer will take care of deleting the Foo when all the copies of the shared pointer pointing at it are destroyed. (There's problems you can get into here, too, particularly if you mix shared_ptr<>s with any other form of pointer, but if you stick to shared_ptr<> throughout you should be OK.)
Note: I'm writing this without running it through a compiler. I'm aiming for accuracy and good style (such as the use of initializers in constructors). If somebody finds a problem, please comment.
Three comments:
You need a destructor as well.
~MyClass()
{
delete str;
}
You really don't need to use heap allocated memory in this case. You could do the following:
class MyClass {
private:
std::string str;
public:
MyClass (const std::string &_str) {
str= _str;
}
void ChangeString(const std::string &_str) {
str = _str;
};
You can't do the commented out version. That would be a memory leak. Java takes care of that because it has garbage collection. C++ does not have that feature.
When you reassign a pointer to a new object, do you have to call delete on the old object first to avoid a memory leak? My intuition is telling me yes, but i want a concrete answer before moving on.
Yes. If it's a raw pointer, you must delete the old object first.
There are smart pointer classes that will do this for you when you assign a new value.