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/
Related
Suppose this is a struct/class in c++
struct MyClass{
list<int> ls;
};
Now maybe due to some reason, it is not possible to initialize pointer of 'MyClass', and the memory is being allocated manually
MyClass *pointer = calloc(1, sizeof(MyClass));
pointer->ls.push_back(123); // <----- this line causes core dump
Although it works fine with pure C++ method like this:-
MyClass *pointer = new MyClass;
pointer->ls.push_back(123);// All good, no problem
So is there any function available in std::list to resolve this issue?
I tried doing the same with std::vector, by there was no problem in that when I used calloc()
struct MyClass{
vector<int> ls;
};
MyClass *pointer = calloc(1, sizeof(MyClass));//using calloc
pointer->ls.push_back(123); // All Good
But on doing this using malloc:-
struct MyClass{
vector<int> ls;
};
MyClass *pointer = malloc(1, sizeof(MyClass));//using malloc
pointer->ls.push_back(123); // Core dump :-(
C++ differentiates between memory and objects in that memory. You are allocating memory with malloc or calloc, but you are not creating any object in the memory you allocated.
You can manually create an object in allocated storage using the placement-new expression:
new(pointer) MyClass /*initializer*/;
This creates a new MyClass object at the storage location that pointer points to. /*initializer*/ is the optional initializer for the object and has the same meaning as in a variable declaration
MyClass my_class /*initializer*/;
You may need to #include<new> to use the placement-new form I used above and you need to make sure that the allocated memory block is large enough and sufficiently aligned for the MyClass object type.
Using a pointer as if it points to an object even though none was created at the storage location causes undefined behavior, which you are observing. Note that this is true for any (even fundamental) types technically. Probably that will change in the future to allow at least trivial types (or some similar category) to be created implicitly when used, but for non-trivial class types such as the standard containers, this certainly will not change.
Also note that malloc and calloc return void*, which in C++ cannot be implicitly cast to a different pointer type. You need to cast it explicitly:
MyClass *pointer = static_cast<MyClass*>(malloc(sizeof(MyClass)));
or rather than doing that, save the pointer as void* without any cast to hint at you that it is not actually pointing to an object, but just memory.
You can still pass that void* pointer to the placement-new and the placement-new will actually return you a pointer of the correct type pointing to the new object (there are some specific cases where you are actually required to use this pointer):
void *mem_ptr = malloc(sizeof(MyClass));
auto obj_ptr = new(mem_ptr) MyClass /*initializer*/;
You should also avoid using malloc and calloc and the like. C++ has its own memory allocation function, called (confusingly) operator new. It is used like malloc (initializing the memory to zero like calloc does is not useful, because the initializer in the placement-new can and/or will do that):
void* pointer = operator new(sizeof(MyClass));
and its free analogue is:
operator delete(pointer);
It still only allocates memory and does not create object as the placement new does.
The non-placement new expression new MyClass; allocates memory (by call to operator new) and creates an object as if by the placement-new form mentioned above.
You do not need to bother with all of this though, because you can just use a smart pointer and initialize it later, if you need that:
std::unique_ptr<MyClass> ptr; // No memory allocated or object created
ptr = std::make_unique<MyClass>(/*constructor arguments*/); // Allocates memory and creates object
// Object is automatically destroyed and memory freed once no `std::unique_ptr` references the object anymore.
You should not use raw pointers returned from new, malloc or the like as owning pointers in the first place. std::unique_ptr has the correct ownership semantics by default and requires no further actions by you.
For example, as simplified as possible:
class thing{
public:
char* arr;
thing();
}
thing::thing{
arr = new char[1000];
}
If I create a dynamically allocated instance of thing, then deallocate it with delete, will the memory that was dynamically allocated by the constructor also be deallocated, or do I have to deallocate arr first?
Not as that class is written. You will need to declare a destructor that does delete [] arr; - and you should follow the rule or three (or rule of five if you add move semantics).
No, you will leak the allocated array. A char* doesn't know it's supposed to delete itself when it goes out of scope — it's simply pointing to that memory.
If you use the right tool for the right job though, i.e. change arr to be a unique_ptr then it will automatically know to free the memory you assigned to it once the thing is deallocated.
If I declare a struct such as this one:
struct MyStruct
{
int* nums;
};
MyStruct ms;
ms.nums = new int[4];
Do I need to call delete ms.nums; before exiting my program or will the member variable nums automatically get deallocated because the MyStruct instance ms wasn't declared on the heap?
Yes, you have to delete that.. struct's default destructor won't do that. It will just delete the pointer variable which holds the address of the object and object will be left alone forever.
Better to do the deletion inside the struct's destructor
struct MyStruct
{
int* nums;
public :
~MyStruct()
{
if(nums != NULL)
delete nums;
}
};
The member variable nums (which is a pointer) will get automatically deallocated when ms, the struct containing it, is popped off the stack.
However, the memory pointed to by nums will NOT. You should call delete [] ms.nums; before your program exits.
You always have to call delete to deallocate all the memory that you explicitly allocated using new.
When you allocate memory using new, memory is allocated in heap and the base address of the block is returned back.
Your struct is present in stack which contains a pointer pointing to that base address. When the struct object leaves the scope, the object is deallocated , i.e. the pointer containig the address of allocated heap memory block is gone, but the heap memory itself is still allocated which you can't deallocate now.
You need to call delete if memory allocated and pointer returned to num. Here num is a pointer inside a structure. Memory for num is allocated when you declare structure variable. But memory address which is allocated by new should be un-allocated by calling delete
Memory for the pointer itself will be deallocated when the object is destroyed, as it is for any class member. However, the array it points to won't be deleted - there is no way to know what the pointer points to, whether it was allocated by new or new[] or not dynamically allocated at all, or whether anything else still wants to use it. To avoid memory leaks, you should delete it, with the array form delete [] ms.nums;, once you've finished with it.
Since it's often hard to do this correctly, why not use a library class to manage the array correctly for you?
#include <vector>
struct MyStruct
{
std::vector<int> nums;
};
MyStruct ms;
ms.nums.resize(4);
Now all the memory will be released automatically when ms is destroyed; and the class is correctly copyable and (since C++11) efficiently movable.
The struct is allocated on the stack, which basically is a pointer to an Array. the array is allocated on the heap. Every time you write "new" that means that the allocation is on the heap therefore you need to "delete" it. On the other hand the struct will be remove as soon as you exit the function.
When program exit, OS will free all memory allocated by your program.
On the other hand, if a memory allocation occurs multiple times during the life of the program, then surely it should be released to prevent dangling pointers.
There is such code:
#include <iostream>
int main()
{
int a;
int* p = new (&a) int(2);
std::cout << a << std::endl;
// delete p; error BLOCK TYPE IS INVALID
std::cin.get();
return 0;
}
The output is:
2
Why is it possible to dynamically allocate memory on stack? (I thought that heap is the right place to do this). And, why does delete operator return error in this case, but new operator work?
This is using the placement new syntax. Placement new does not allocate memory at all, rather, it is a way to construct an object at a particular location. In this example, the memory comes from the stack. It doesn't have to. delete has issues because you didn't new the memory.
There are ways to dynamically allocate memory from the stack (alloca) but that's not what is happening here.
int* p = new (&a) int(2);
This is called placement-new. It doesn't allocate memory. It constructs the object in the same memory of a. In placement new, it's the user who specifies the memory region where new operator constructs the object. In your code above, you specify the memory region by writing (&a) expression just after the new keyword. Since &a is not a dynamically allocated memory, you cannot delete it:
delete p; //runtime-error
It would give runtime error, it attempts to delete the memory where the variable a resides.
However, if you dynamically allocate the memory, then you can do delete it. Lets suppose, A is some class, then you should be doing this:
char *buffer = new char[sizeof(A)]; //allocate memory of sizeof(A);
///ASSUMPTION: the buffer is properly align as required by the type A
//use placement-new to construct an object at the specified memory region
A *pA = new (buffer) A(/*..parameters..*/);
//...
//incorrect way to delete the memory!
//delete pA; //incorrect
//before deleting the memory you should be calling the destructor explicitly as
pA->~A(); //call the destructor explicitly - must do it
//now deallocate the memory as
delete []buffer;
This is simplest example of placement-new which explains the syntax only. But the story doesn't end here; it is the beginning and to make it work properly, the memory pointed to by buffer has to be aligned properly for the object type, and in the above example, I simply assumed so. In the real code, you cannot make such dangerous assumption. Now read this FAQ:
What is "placement new" and why would I use it?
This is called placement new: http://www.parashift.com/c++-faq-lite/dtors.html#faq-11.10
You can optionally pass an address to new, and it will only call the object's constructor (if it has one). Because no memory was allocated, it would be an error to deallocate it with delete. Simply call the object's destructor (if it has one) and you're done.
C++ separates the notions of memory allocation and object lifetime. This is perhaps one of the most important "new" aspects of the language compared to C. In C there is no such distinction because variables are entirely determined by their memory, while in C++ objects have a more abstract notion of a "state" which is distinct from the underlying memory.
Let's look at memory first:
{
char buf[100]; // automatic allocation, scoped lifetime
}
{
void * p = std::malloc(100); // dynamic allocation, manually managed
void * q = ::operator new(100); // essentially identical
// ...
::operator delete(q); // manual deallocation
std::free(p); // ditto
}
On the other hand, an object's lifetime is a separate topic:
{
Foo x; // automatic storage, scoped lifetime.
// Implies automatic memory allocation for sizeof(Foo) bytes.
}
{
Foo * px = ::new Foo; // dynamic storage, manual lifetime,
// implies dynamic allocation via ::operator new()
Foo * py = ::new (q) Foo; // dynamic storage and manual lifetime, uses memory at q
// ...
delete px; // destroy object _and_ deallocate memory
py->~Foo(); // destroy object. Memory was never our business to start with.
}
As you can see, the separation of memory and object lifetime adds a lot of flexibility: We can have dynamic objects living in automatic memory, or take care of allocation ourselves and reuse memory for repeated object constructions. The standard new and delete expressions combine allocation and construction, but this is only a shortcut for the most frequently used operations. In principle, you're entirely free to handle memory and object lifetime separately.
This idea underpins the notion of allocators, which are a core concept in the C++ standard library.
Can we call delete on the pointer which is allocated with the placement new? If no then why? Please explain in details.
I know that there is no placement delete. But I wonder why just delete opetator can not delete the memory without caring how that memory on which the pointer points is allocated?
delete is doing two things:
Calls destrucor
Frees memory
And I see no reaason for delete not to be able to call either of these two operations on the object which was created by placement new. Any idea about reasons?
You must only call delete on pointers that were created with operator new. If you use placement new with a memory location that was allocated by the normal operator new then you may safely use delete on it (provided you get the types and pointers right). However, you can use placement new on any memory, so you usually will manage that memory some other way and call the object's destructor manually.
For instance, in this convoluted and usually unnecessary scenario, it is safe to delete the memory you used placement new on, but only because you allocated it with new before:
char* mem = new char[sizeof(MyObject)];
MyObject* o = new (mem) MyObject;
// use o
o->~MyObject(); // with placement new you have to call the destructor by yourself
delete[] mem;
However, this is illegal:
char mem[16]; // create a buffer on the stack, assume sizeof(MyObject) == 16
MyObject* o = new (mem) MyObject; // use stack memory to hold a MyObject
// note that after placement new is done, o == mem
// pretend for this example that the point brought up by Martin in the comments didn't matter
delete o; // you just deleted memory in the stack! This is very bad
Another way to think of it is that delete only deallocates memory allocated previously by the normal new. With placement new, you do not have to use memory that was allocated by the normal new, so with the possibility of not having been allocated by normal new, delete cannot deal with it.
EDIT1: I know that there is no placement delete. But I wonder why just
delete opetator can not delete the memory without caring how that
memory on which the pointer points is allocated?
Because each flavour of memory allocation uses some implementation specific tracking of the memory (usually a header block that precedes the user address) and this make the allocation/deallocation to work only when paired up correctly:
new must pair with delete
new[] must pair with delete[] (most implementations though forgive mixing the newand new[])
malloc and frieds must pair with free
CoTaskMemAlloc pairs with CoTaskMemFree
alloca pairs with nothing (stack unwinding takes care of it)
MyCustomAllocator pairs with MyCustomFree
Attempting to call the wrong deallocator will result in unpredictable behavior( most likely seg fault now or later). Therefore calling delete on memory allocated by anything else other than new will result in bad things.
Furthermore the placement new may be called on any address, may not even be an allocated address. It can be called on an address located in the middle of some larger object, it may be called on a memory mapped region, it may be called on a raw virtual committed region, anything goes. delete woul attempt, in all these cases, to do what its implementation tell him to do: subtract the header size, interpret it as a new header, link it back into the heap. Kaboom.
The one that know how to release the memory of a placement new address is you, since you know exactly how was that memory allocated. delete will only do what it knows, and it may not be the right thing.
No, since delete not only calls the destructor but also frees the memory, but if you used placement new you must have allocated the memory yourself using malloc() or stack. You do, however, have to call the destructor yourself. Also see the C++ FAQ.
No. There is no placement-delete expression.
Typical scenario:
void * const addr = ::operator new(sizeof(T)); // get some memory
try {
T * const pT = new (addr) T(args...); // construct
/* ... */
p->~T(); // nap time
}
catch (...) {
}
::operator delete(addr); // deallocate
// this is _operator_-delete, not a delete _expression_
Note that the placement-new operator does have a corresponding delete operator which is mandated to be precisely void ::operator delete(void* [, size_t]) { }, a no-op; this is what gets called if the constructor of T throws an exception.
No, because a placement new doesn't allocate any memory. You use placement new on previously allocated raw memory. The only thing it does is call the constructor of the object.