How to choose the adequate, delete or delete[]? [duplicate] - c++

What is the difference between delete and delete[] operators in C++?

The delete operator deallocates memory and calls the destructor for a single object created with new.
The delete [] operator deallocates memory and calls destructors for an array of objects created with new [].
Using delete on a pointer returned by new [] or delete [] on a pointer returned by new results in undefined behavior.

The delete[] operator is used to delete arrays. The delete operator is used to delete non-array objects. It calls operator delete[] and operator delete function respectively to delete the memory that the array or non-array object occupied after (eventually) calling the destructors for the array's elements or the non-array object.
The following shows the relations:
typedef int array_type[1];
// create and destroy a int[1]
array_type *a = new array_type;
delete [] a;
// create and destroy an int
int *b = new int;
delete b;
// create and destroy an int[1]
int *c = new int[1];
delete[] c;
// create and destroy an int[1][2]
int (*d)[2] = new int[1][2];
delete [] d;
For the new that creates an array (so, either the new type[] or new applied to an array type construct), the Standard looks for an operator new[] in the array's element type class or in the global scope, and passes the amount of memory requested. It may request more than N * sizeof(ElementType) if it wants (for instance to store the number of elements, so it later when deleting knows how many destructor calls to done). If the class declares an operator new[] that additional to the amount of memory accepts another size_t, that second parameter will receive the number of elements allocated - it may use this for any purpose it wants (debugging, etc...).
For the new that creates a non-array object, it will look for an operator new in the element's class or in the global scope. It passes the amount of memory requested (exactly sizeof(T) always).
For the delete[], it looks into the arrays' element class type and calls their destructors. The operator delete[] function used is the one in the element type's class, or if there is none then in the global scope.
For the delete, if the pointer passed is a base class of the actual object's type, the base class must have a virtual destructor (otherwise, behavior is undefined). If it is not a base class, then the destructor of that class is called, and an operator delete in that class or the global operator delete is used. If a base class was passed, then the actual object type's destructor is called, and the operator delete found in that class is used, or if there is none, a global operator delete is called. If the operator delete in the class has a second parameter of type size_t, it will receive the number of elements to deallocate.

This the basic usage of allocate/DE-allocate pattern in c++
malloc/free, new/delete, new[]/delete[]
We need to use them correspondingly. But I would like to add this particular understanding for the difference between delete and delete[]
1) delete is used to de-allocate memory allocated for single object
2) delete[] is used to de-allocate memory allocated for array of objects
class ABC{}
ABC *ptr = new ABC[100]
when we say new ABC[100], compiler can get the information about how many objects that needs to be allocated(here it is 100) and will call the constructor for each of the objects created
but correspondingly if we simply use delete ptr for this case, compiler will not know how many objects that ptr is pointing to and will end up calling of destructor and deleting memory for only 1 object(leaving the invocation of destructors and deallocation of remaining 99 objects). Hence there will be a memory leak.
so we need to use delete [] ptr in this case.

The operators delete and delete [] are used respectively to destroy the objects created with new and new[], returning to the allocated memory left available to the compiler's memory manager.
Objects created with new must necessarily be destroyed with delete, and that the arrays created with new[] should be deleted with delete[].

When I asked this question, my real question was, "is there a difference between the two? Doesn't the runtime have to keep information about the array size, and so will it not be able to tell which one we mean?" This question does not appear in "related questions", so just to help out those like me, here is the answer to that: "why do we even need the delete[] operator?"

C++ delete[] operator ensures that Destructor for all object allocated with new[] is called. The following example demonstrates the same. Also, delete[] must be preferred (if new[] used previously) when the class has a non-default destructor to release the acquired resources. Otherwise, it might result in memory leaks.
Common Code:-
#include <iostream>
using namespace std;
class memTest{
public:
static int num;
memTest(){
cout<<"Constructor from object " << num++ << endl;
}
~memTest(){
cout<<"Destructor from object " << --num << endl;
}
};
int memTest::num=0;
Example 1:- use of new[] and delete may result in undefined behavior.
int main() {
memTest* Test1=new memTest[3];
delete Test1; //<-----
return 0;
}
Output 1:-
Constructor from object 0
Constructor from object 1
Constructor from object 2
Destructor from object 2 //<-----
Example 2: The correct behavior is using new[] and delete[].
int main() {
memTest* Test1=new memTest[3];
delete[] Test1; //<-----
return 0;
}
Output 2:-
Constructor from object 0
Constructor from object 1
Constructor from object 2
Destructor from object 2
Destructor from object 1 //<-----
Destructor from object 0 //<-----

delete is used for one single pointer and delete[] is used for deleting an array through a pointer.
This might help you to understand better.

Related

C++ What is the difference between delete and delete [ ]? [duplicate]

What is the difference between delete and delete[] operators in C++?
The delete operator deallocates memory and calls the destructor for a single object created with new.
The delete [] operator deallocates memory and calls destructors for an array of objects created with new [].
Using delete on a pointer returned by new [] or delete [] on a pointer returned by new results in undefined behavior.
The delete[] operator is used to delete arrays. The delete operator is used to delete non-array objects. It calls operator delete[] and operator delete function respectively to delete the memory that the array or non-array object occupied after (eventually) calling the destructors for the array's elements or the non-array object.
The following shows the relations:
typedef int array_type[1];
// create and destroy a int[1]
array_type *a = new array_type;
delete [] a;
// create and destroy an int
int *b = new int;
delete b;
// create and destroy an int[1]
int *c = new int[1];
delete[] c;
// create and destroy an int[1][2]
int (*d)[2] = new int[1][2];
delete [] d;
For the new that creates an array (so, either the new type[] or new applied to an array type construct), the Standard looks for an operator new[] in the array's element type class or in the global scope, and passes the amount of memory requested. It may request more than N * sizeof(ElementType) if it wants (for instance to store the number of elements, so it later when deleting knows how many destructor calls to done). If the class declares an operator new[] that additional to the amount of memory accepts another size_t, that second parameter will receive the number of elements allocated - it may use this for any purpose it wants (debugging, etc...).
For the new that creates a non-array object, it will look for an operator new in the element's class or in the global scope. It passes the amount of memory requested (exactly sizeof(T) always).
For the delete[], it looks into the arrays' element class type and calls their destructors. The operator delete[] function used is the one in the element type's class, or if there is none then in the global scope.
For the delete, if the pointer passed is a base class of the actual object's type, the base class must have a virtual destructor (otherwise, behavior is undefined). If it is not a base class, then the destructor of that class is called, and an operator delete in that class or the global operator delete is used. If a base class was passed, then the actual object type's destructor is called, and the operator delete found in that class is used, or if there is none, a global operator delete is called. If the operator delete in the class has a second parameter of type size_t, it will receive the number of elements to deallocate.
This the basic usage of allocate/DE-allocate pattern in c++
malloc/free, new/delete, new[]/delete[]
We need to use them correspondingly. But I would like to add this particular understanding for the difference between delete and delete[]
1) delete is used to de-allocate memory allocated for single object
2) delete[] is used to de-allocate memory allocated for array of objects
class ABC{}
ABC *ptr = new ABC[100]
when we say new ABC[100], compiler can get the information about how many objects that needs to be allocated(here it is 100) and will call the constructor for each of the objects created
but correspondingly if we simply use delete ptr for this case, compiler will not know how many objects that ptr is pointing to and will end up calling of destructor and deleting memory for only 1 object(leaving the invocation of destructors and deallocation of remaining 99 objects). Hence there will be a memory leak.
so we need to use delete [] ptr in this case.
The operators delete and delete [] are used respectively to destroy the objects created with new and new[], returning to the allocated memory left available to the compiler's memory manager.
Objects created with new must necessarily be destroyed with delete, and that the arrays created with new[] should be deleted with delete[].
When I asked this question, my real question was, "is there a difference between the two? Doesn't the runtime have to keep information about the array size, and so will it not be able to tell which one we mean?" This question does not appear in "related questions", so just to help out those like me, here is the answer to that: "why do we even need the delete[] operator?"
C++ delete[] operator ensures that Destructor for all object allocated with new[] is called. The following example demonstrates the same. Also, delete[] must be preferred (if new[] used previously) when the class has a non-default destructor to release the acquired resources. Otherwise, it might result in memory leaks.
Common Code:-
#include <iostream>
using namespace std;
class memTest{
public:
static int num;
memTest(){
cout<<"Constructor from object " << num++ << endl;
}
~memTest(){
cout<<"Destructor from object " << --num << endl;
}
};
int memTest::num=0;
Example 1:- use of new[] and delete may result in undefined behavior.
int main() {
memTest* Test1=new memTest[3];
delete Test1; //<-----
return 0;
}
Output 1:-
Constructor from object 0
Constructor from object 1
Constructor from object 2
Destructor from object 2 //<-----
Example 2: The correct behavior is using new[] and delete[].
int main() {
memTest* Test1=new memTest[3];
delete[] Test1; //<-----
return 0;
}
Output 2:-
Constructor from object 0
Constructor from object 1
Constructor from object 2
Destructor from object 2
Destructor from object 1 //<-----
Destructor from object 0 //<-----
delete is used for one single pointer and delete[] is used for deleting an array through a pointer.
This might help you to understand better.

delete operator with chained dynamic allocation

Let's assume I have two classes named classA and classB. I create a pointer to classA, and then dynamically allocate a single object using the new operator as follows:
classA *ptrA = new classA
classA has a member of type pointer to classB, and in its default constructor it allocates an array of objects of type classB on the heap as follows:
memberOfA = new classB[10]
and to complicate things further, classB has a member variable of type int*, and in its default constructor it allcoates an array of integer on the heap as follows:
memberOfB = new int[100]
Now, if I call a delete on ptrA using delete ptrA, how will the compiler go about deallocating the memory allocated by memberOfA and memberOfB.
Now, if I call a delete on ptrA using delete ptrA, how will the compiler go about deallocating the memory allocated by memberOfA and memberOfB.
When you call delete ptrA, the compiler will call the destructor of classA. If that destructor doesn't delete [] memberOfA then you're leaking both memberOfA and consequently memberOfB.
If instead of manual memory management (new and delete) you used automatic memory management (SBRM/RAII), then you wouldn't have to write any destructors and the resources you allocated would be freed as expected, for example:
std::unique_ptr<classA> ptrA(new classA);
std::unique_ptr<classB[]> memberOfA(new classB[10]);
std::unique_ptr<int[]> memberOfB(new int[100]);
This is in fact the modern C++ way to do it. The code becomes simpler to understand as the ownership semantics of the pointers are written in their type and there are no destructors lying around. An even better way would be to use std::make_unique(...) in place of new but I left it out for simplicity.
memberOfA and memberOfB is only deallocated if you put a delete[] in the destrutors of classA and classB. So if you had written the destructor of classA so that it freed the memory pointed to by memberOfA like this:
classA::~classA() {
delete [] memberOfA;
}
In this case the destructor would free the array calling the destructors of the elements pointed by the entries in memberOfA. The same can be said for the destructor of classB:
classB::~classB() {
delete [] memberOfB;
}
Remember using new must be paired with a subsequent delete at some point, if not you are look at a leak. Perhaps you should consider one of the smart pointer classes: unique_ptr, and shared_ptr.
If your design allows for classA and classB to use a unique_ptr or shared_ptr to the array, much of the dangers of a leak would have been overcome:
std::unique_ptr<classB[]> memberOfA(new classB[10]);
If the order of the deletion is of interest to you then, invoking the delete ptrA will result in the destructor of classA being called, and if we assume that it is implemented as hinted above (using delete []) then the delete [] operator will call the destructors of all the members in decreasing address order, quote from the Standard, draft N3690, ยง 5.3.5 / 6:
If the value of the operand of the delete-expression is not a null pointer value, the delete-expression will invoke the destructor (if any) for the object or the elements of the array being deleted. In the case of an array, the elements will be destroyed in order of decreasing address (that is, in reverse order of the completion of their constructor; see 12.6.2).
This will of course lead to the destructor of classB being called for each element, which will again utilize the delete [] memberOfB which will release the array pointing to the integers.
It will not ... you should do that by implementing the destructor! You can do that by using the delete [] operator.
classA::~classA() {
delete [] memberOfA;
}
and
classB::~classB() {
delete [] memberOfB;
}
and then you do delete ptrA the order of execution will be:
call classA::~classA()
call classB::~classB() on each of the elements of memberOfA
deallocate memory for the array of ints pointed by memberofB
deallocate memory for the array of memberofA
deallocate memory for classA
Deallocation of memory can actually happen at later stages in some cases.
Alternatively, if you don't really need to work with pointer use arrays or vectors to hold your elements.
class classB
{
...
std::array<int,100> memberOfB;
...
};
class classA
{
...
std::array<classB,10> memberOfA;
...
};
std::unique_ptr<A> ptrA(new classA)

placement delete inside or outside destructor

I found the following code while reading placement new operator.
#include <iostream>
using namespace std;
class MyClass {
public:
// Placement new operator
void* operator new (size_t sz, void* v) {
cout << "Placement new invoked" << endl;
return v;
}
~MyClass() {
// Cleanup
}
};
int main()
{
// Create a buffer to store the object
int buffer[16];
cout << "Starting address of my buffer = " << &buffer << endl;
// Create the object. Use placement new
MyClass* obj = new (buffer) MyClass();
cout << "Location of my object = " << obj << endl;
// Don't delete object created with placement delete
// Call the destructor explicitly
obj->~MyClass();
}
I have few questions related to delete object which was created using placement new:
what is the clean up code needs to write in destructor in order to
free memory which was occupied obj in buffer memory.
is it not needed to define placement delete, if yes whether it needs to be inside destructor or outside destructor. If it is outside destructor how it will be called ?
Regular new does two things:
allocate memory for your object
construct your object in that memory space.
Placement new means you manage one of those and the other is managed as before.
You allocate / provide the memory for your object
Constructor is called in that memory space.
The reverse is delete of which the regular delete does the following:
Invoke the destructor of the object to clean-up
Free the memory that was allocated for it.
Note that they are done in reverse order for obvious reasons. You cannot free the memory that contains information about what needs to be cleaned up, unti you have finished using that memory. Whereas in construction you need to get hold of the memory first.
In what you call placement delete but is really the reverse of placement new, you need to perform the first step of destruction but not the second. Thus you call the object's destructor, and then you can free up the memory it uses / use it for something else.
The most common example of usage of placement new is in the implementation of std::vector which requires a contiguous buffer for its data, and which lets you reserve ahead (and if you don't it will probably do it for you). That part allocates the memory but does not construct the objects in it. Thus when they are constructed later, placement new is used.
what is the clean up code needs to write in destructor in order to free memory which was occupied obj in buffer memory
The destructor should just do what any destructor does: clean up any resources managed by the object. In this case, it doesn't manage any resources, so there's no need to do anything.
It shouldn't do anything special according to how the storage for the object itself was allocated. Managing that storage is the job for custom new and delete operators, when required.
is it not needed to define placement delete
No. Placement-new is used to construct an object in storage that you're managing yourself, and it's your responsibility to free that storage yourself. In this case, the storage is automatic, so it's automatically freed when the function exits.
Just as the allocator for placement-new does nothing (just returning the provided pointer to pre-allocated storage), so the corresponding deallocator would do nothing; and so it doesn't exist. You just need to destroy the object by calling its destructor directly, before disposing of the storage yourself.
The first thing to ask is: what are you trying to do? If you
define a placement operator new in a class, then that is the
only operator new which will be found when you write new
MyClass; you must always specify the extra arguments. And in
almost every case where you define a class specific operator
new, you should also define a class specific operator delete;
otherwise the global operator delete function will be called
when you write delete p, and that generally won't work.
If your goal is to systematically require separation of
allocation and initialization, and that is why you are defining
a member operator new, then you can provide a no-op operator
delete; if the constructor of the class can throw, then you'll
also want to provide a placement operator delete, since this is
what will be called if the constructor of a newed object exits
via an exception. There is no other way to call it, however.
When providing placement operator new, you must provide
a default operator delete which does the right thing; and when
providing several operator new for the same type, you need to
memorize somehow in each allocation which one was called, in
order to dispatch in the non-placement operator delete.
And by the way, just allocating a buffer as a local variable
does not guarantee sufficient alignment for anything but the
declared buffer type.
EDIT:
Just an example of what would be needed for the operator
delete functions (which must be members):
void operator delete( void* p ) {}
void operator delete( void* p, void* ) {}

how do you delete an object allocated with placement new

there are quite a few faces for the new operator in c++, but I'm interested in placement new.
Suppose you allocate memory at a specific memory location
int memoryPool[poolSize*sizeof(int)];
int* p = new (mem) int; //allocates memory inside the memoryPool buffer
delete p; //segmentation fault
How can I correctly deallocate memory in this case?
What if instead of built-in type int I would use some class called myClass?
myClass memoryPool[poolSize*sizeof(myClass )];
myClass * p = new (mem) myClass ; //allocates memory inside the memoryPool buffer
delete p; //segmentation fault
Thanks for your help.
In the first case, there's no point in using placement new, since int doesn't have a constructor.
In the second case, it's either pointless (if myClass is trivial) or wrong, since there are already objects in the array.
You use placement new to initialise an object in a block of memory, which must be suitably aligned, and mustn't already contain a (non-trivial) object.
char memory[enough_bytes]; // WARNING: may not be properly aligned.
myClass * c = new (memory) myClass;
Once you've finished with it, you need to destroy the object by calling its destructor:
c->~myClass();
This separates the object's lifetime from that of its memory. You might also have to release the memory at some point, depending on how you allocated it; in this case, it's an automatic array, so it's automatically released when it goes out of scope.
In your case there is no need to deallocate it, your int array will be deallocated once you return from your function. You should only call explicitly your destructor:
p->~myclass();
to keep you buffer correctly aligned use std::aligned_storage, look in here for example:
http://www.cplusplus.com/reference/type_traits/aligned_storage/

why destructor is not called implicitly in placement new"?

As referenced in this site...
http://www.parashift.com/c++-faq-lite/dtors.html#faq-11.10
But i did not find the reason, why we should explicitly call the desturctor?
You can think of it as a call to delete, but since you used placement new, you don't want to use delete, as that would attempt to free the memory. If you wanted it to be called automatically, you could use RAII:
// Could use a templated version, or find an existing impl somewhere:
void destroy_fred(Fred* f) {
f->~Fred();
}
void someCode()
{
char memory[sizeof(Fred)];
void* p = memory;
boost::shared_ptr<Fred> f(new(p) Fred(), destroy_fred);
// ...
// No need for an explicit destructor, cleaned up even during an exception
}
Normally, newed objects are destructed when delete is called. delete is never called for placement new, so you have to figure out when to destruct the object yourself, and do so explicitly.
You can't call ordinary delete because that would attempt free the memory, which was not allocated with ordinary new, so you call the destructor directly and then handle the underlying memory allocation depending on the architecture you are using.
Normally, an object is created using new, and destroyed when you call delete on the object. However, because 'delete' will also try to free the underlying memory back to the heap, you should never call delete on an object which which was instantiated using placement new.
Nevertheless, (no matter how it was allocated) every object ought to be destroyed (because its destructor might do something important).
Therefore instead of the object's being destroyed using delete, you have to destroy the object yourself, and do so explicitly, by invoking its destructor as shown in the FAQ you cited.
The example from your link:
void someCode()
{
char memory[sizeof(Fred)];
void* p = memory;
Fred* f = new(p) Fred();
...
f->~Fred(); // Explicitly call the destructor for the placed object
}
"placement new" return the memory offset that we provide. Right after new the compiler adds a call to the ctor ("placement new" is a also a trick to explicity call a ctor).
"memory" is allocated in the stack so we can not delete it.
string *ps = new string("Hello ");
The new in the above statement is called new operator which do two things
Call operator new(allocates
necessary memory like malloc) and
Call constructor
on raw memory created by operator
new to initialize
delete ps;
Again delete corresponds to
Destruction of object and
Deallocation of memory(i.e, free)
In c++ user is not allowed to call constructor directly,Placement new is used when you have raw memory allocated by some means and wanted to initialize the bits with constructor.Because the memory allocation process did not happen through operator new, delete cannot deallocate the memory.
void someCode()
{
char memory[sizeof(Fred)];//memory allocated on stack
void* p = memory;
Fred* f = new(p) Fred();//placement new
...
f->~Fred(); // Explicitly call the destructor for the placed object
}