Raw pointers can point to objects allocated on the stack or on the heap.
Heap allocation example:
// heap allocation
int* rawPtr = new int(100);
std::cout << *rawPtr << std::endl; // 100
Stack allocation example:
int i = 100;
int* rawPtr = &i;
std::cout << *rawPtr << std::endl; // 100
Heap allocation using auto_ptr example:
int* rawPtr = new int(100);
std::unique_ptr<int> uPtr(rawPtr);
std::cout << *uPtr << std::endl; // 100
Stack allocation using auto_ptr example:
int i = 100;
int* rawPtr = &i;
std::unique_ptr<int> uPtr(rawPtr); // runtime error
Are 'smart pointers' intended to be used to point to dynamically created objects on the heap? For C++11, are we supposed to continue using raw pointers for pointing to stack allocated objects? Thank you.
Smart pointers are usually used to point to objects allocated with new and deleted with delete. They don't have to be used this way, but that would seem to be the intent, if we want to guess the intended use of the language constructs.
The reason your code crashes in the last example is because of the "deleted with delete" part. When it goes out of scope, the unique_ptr will try to delete the object it has a pointer to. Since it was allocated on the stack, this fails. Just as if you had written, delete rawPtr;
Since one usually uses smart pointers with heap objects, there is a function to allocate on the heap and convert to a smart pointer all in one go. std::unique_ptr<int> uPtr = make_unique<int>(100); will perform the actions of the first two lines of your third example. There is also a matching make_shared for shared pointers.
It is possible to use smart pointers with stack objects. What you do is specify the deleter used by the smart pointer, providing one that does not call delete. Since it's a stack variable and nothing need be done to delete it, the deleter could do nothing. Which makes one ask, what's the point of the smart pointer then, if all it does is call a function that does nothing? Which is why you don't commonly see smart pointers used with stack objects. But here's an example that shows some usefulness.
{
char buf[32];
auto erase_buf = [](char *p) { memset(p, 0, sizeof(buf)); };
std::unique_ptr<char, decltype(erase_buf)> passwd(buf, erase_buf);
get_password(passwd.get());
check_password(passwd.get());
}
// The deleter will get called since passwd has gone out of scope.
// This will erase the memory in buf so that the password doesn't live
// on the stack any longer than it needs to. This also works for
// exceptions! Placing memset() at the end wouldn't catch that.
The runtime error is due to the fact that delete was called on a memory location that was never allocated with new.
If an object has already been created with dynamic storage duration (typically implemented as creation on a 'heap') then a 'smart pointer' will not behave correctly as demonstrated by the runtime error.
Are 'smart pointers' intended to be used to point to dynamically
created objects on the heap? For C++11, are we supposed to continue
using raw pointers for pointing to stack allocated objects?
As for what one is supposed to do, well, it helps to think of the storage duration and specifically how the object was created.
If the object has automatic storage duration (stack) then avoid taking the address and use references. The ownership does not belong with the pointer and a reference makes the ownership clearer.
If the object has dynamic storage duration (heap) then a smart pointer is the way to go as it can then manage the ownership.
So for the last example, the following would be better (pointer owns the int):
auto uPtr = std::make_unique<int>(100);
The uPtr will have automatic storage duration and will call the destructor when it goes out of scope. The int will have dynamic storage duration (heap) and will be deleteed by the smart pointer.
One could generally avoid using new and delete and avoid using raw pointers. With make_unique and make_shared, new isn't required.
Are 'smart pointers' intended to be used to point to dynamically created objects on the heap?
They are intended for heap-allocated objects to prevent leaks.
The guideline for C++ is to use plain pointers to refer to a single object (but not own it). The owner of the object holds it by value, in a container or via a smart pointer.
Are 'smart pointers' intended to be used to point to dynamically created objects on the heap?
Yes, but that's just the default. Notice that std::unique_ptr has a constructor (no. (3)/(4) on that page) which takes a pointer that you have obtained somehow, and a "deleter" that you provide. In this case the unique pointer will not do anything with the heap (unless your deleter does so).
For C++11, are we supposed to continue using raw pointers for pointing to stack allocated objects? Thank you.
You should use raw pointers in code that does not "own" the pointer - does not need to concern itself with allocation or deallocation; and that is regardless of whether you're pointing into the heap or the stack or elsewhere.
Another place to use it is when you're implementing some class that has a complex ownership pattern, for protected/private members.
PS: Please, forget about std::auto_ptr... pretend it never existed :-)
Related
I have a question regarding the thrust library when using CUDA.
I am using a thrust function, i.e. exclusive_scan, and I want to use raw pointers. I am using raw (device) pointers because I want to have full control of when the memory is allocated and deallocated.
After the function call, I will hand over the pointer to another data structure and then free the memory in either the destructor of this data structure, or in the next function call, when I recompute my (device) pointers. I came across for example this problem here now, which recommends to wrap the data structure in a device_vector. But then I run into the problem that the memory is freed once my device_vector goes out of scope, which I do not want. Having the device pointer globally is also not an option, since I am hacking code, i.e. it is used as a buffer and I would have to rewrite a lot if I wanted to do something like that.
Does anyone have a good workaround regarding this? The only chance I do see right now is to rewrite the thrust-function on my own, only using raw device-pointers.
EDIT: I misread, I can wrap it in a device_ptr instead of a device_vector.
Asking further though, how could I solve this if there wasn't the option of using a device_ptr?
There is no problem using plain pointers in thrust methods.
For data on the device do:
....
struct DoSomething {
__device__ int operator()(int item) { return 1; }
};
int* IntData;
cudaMalloc(&IntData, sizeof(int) * count);
auto dev_data = device_pointer_cast(IntData);
thrust::generate(dev_data, dev_data + count, DoSomething());
thrust::sort(dev_data, dev_data + count);
....
cudaFree(IntData);
For data on the host use plain malloc/free and raw_pointer_cast instead of device_pointer_cast.
See: thrust: Memory management
I wrote some code like this:
shared_ptr<int> r = make_shared<int>();
int *ar = r.get();
delete ar; // report double free or corruption
// still some code
When the code ran up to delete ar;, the program crashed, and reported "double free or corruption", I'm confused why double free? The "r" is still in the scope, and not popped-off from stack. Do the delete operator do something magic?? Does it know the raw pointer is handled by a smart pointer currently? and then counter in "r" be decremented to zero automatically?
I know the operations is not recommended， but I want to know why?
You are deleting a pointer that didn't come from new, so you have undefined behavior (anything can happen).
From cppreference on delete:
For the first (non-array) form, expression must be a pointer to an object type or a class type contextually implicitly convertible to such pointer, and its value must be either null or pointer to a non-array object created by a new-expression, or a pointer to a base subobject of a non-array object created by a new-expression. If expression is anything else, including if it is a pointer obtained by the array form of new-expression, the behavior is undefined.
If the allocation is done by new, we can be sure that the pointer we have is something we can use delete on. But in the case of shared_ptr.get(), we cannot be sure if we can use delete because it might not be the actual pointer returned by new.
shared_ptr<int> r = make_shared<int>();
There is no guarantee that this will call new int (which isn't strictly observable by the user anyway) or more generally new T (which is observable with a user defined, class specific operator new); in practice, it won't (there is no guarantee that it won't).
The discussion that follows isn't just about shared_ptr, but about "smart pointers" with ownership semantics. For any owning smart pointer smart_owning:
The primary motivation for make_owning instead of smart_owning<T>(new T) is to avoid having a memory allocation without owner at any time; that was essential in C++ when order of evaluation of expressions didn't provide the guarantee that evaluation of the sub-expressions in the argument list was immediately before call of that function; historically in C++:
f (smart_owning<T>(new T), smart_owning<U>(new U));
could be evaluated as:
T *temp1 = new T;
U *temp2 = new U;
auto &&temp3 = smart_owning<T>(temp1);
auto &&temp4 = smart_owning<U>(temp2);
This way temp1 and temp2 are not managed by any owning object for a non trivial time:
obviously new U can throw an exception
constructing an owning smart pointer usually requires the allocation of (small) ressources and can throw
So either temp1 or temp2 could be leaked (but not both) if an exception was thrown, which was the exact problem we were trying to avoid in the first place. This means composite expressions involving construction of owning smart pointers was a bad idea; this is fine:
auto &&temp_t = smart_owning<T>(new T);
auto &&temp_u = smart_owning<U>(new U);
f (temp_t, temp_u);
Usually expression involving as many sub-expression with function calls as f (smart_owning<T>(new T), smart_owning<U>(new U)) are considered reasonable (it's a pretty simple expression in term of number of sub-expressions). Disallowing such expressions is quite annoying and very difficult to justify.
[This is one reason, and in my opinion the most compelling reason, why the non determinism of the order of evaluation was removed by the C++ standardisation committee so that such code is not safe. (This was an issue not just for memory allocated, but for any managed allocation, like file descriptors, database handles...)]
Because code frequently needed to do things such as smart_owning<T>(allocate_T()) in sub-expressions, and because telling programmers to decompose moderately complex expressions involving allocation in many simple lines wasn't appealing (more lines of code doesn't mean easier to read), the library writers provided a simple fix: a function to do the creation of an object with dynamic lifetime and the creation of its owning object together. That solved the order of evaluation problem (but was complicated at first because it needed perfect forwarding of the arguments of the constructor).
Giving two tasks to a function (allocate an instance of T and a instance of smart_owning) gives the freedom to do an interesting optimization: you can avoid one dynamic allocation by putting both the managed object and its owner next to each others.
But once again, that was not the primary purpose of functions like make_shared.
Because exclusive ownership smart pointers by definition don't need to keep a reference count, and by definition don't need to share the data needed for the deleter either between instances, and so can keep that data in the "smart pointer"(*), no additional allocation is needed for the construction of unique_ptr; yet a make_unique function template was added, to avoid the dangling pointer issue, not to optimize a non-thing (an allocation that isn't done in the fist place).
(*) which BTW means unique owner "smart pointers" do not have pointer semantic, as pointer semantic implies that you can makes copies of the "pointer", and you can't have two copies of a unique owner pointing to the same instance; "smart pointers" were never pointers anyway, the term is misleading.
Summary:
make_shared<T> does an optional optimization where there is no separate dynamic memory allocation for T: there is no operator new(sizeof (T)). There is obviously still the creation of an instance with dynamic lifetime with another operator new: placement new.
If you replace the explicit memory deallocation with an explicit destruction and add a pause immediately after that point:
class C {
public:
~C();
};
shared_ptr<C> r = make_shared<C>();
C *ar = r.get();
ar->~C();
pause(); // stops the program forever
The program will probably run fine; it is still illogical, indefensible, incorrect to explicitly destroy an object managed by a smart pointer. It isn't "your" resource. If pause() could exit with an exception, the owning smart pointer would try to destroy the managed object which doesn't even exist anymore.
It of course depends on how library implements make_shared, however most probable implementation is that:
std::make_shared allocates one block for two things:
shared pointer control block
contained object
std::make_shared() will invoke memory allocator once and then it will call placement new twice to initialize (call constructors) of mentioned two things.
| block requested from allocator |
| shared_ptr control block | X object |
#1 #2 #3
That means that memory allocator has provided one big block, which address is #1.
Shared pointer then uses it for control block (#1) and actual contained object (#2).
When you invoke delete with actual object kept by shred_ptr ( .get() ) you call delete(#2).
Because #2 is not known by allocator you get an corruption error.
See here. I quot:
std::shared_ptr is a smart pointer that retains shared ownership of an object through a pointer. Several shared_ptr objects may own the same object. The object is destroyed and its memory deallocated when either of the following happens:
the last remaining shared_ptr owning the object is destroyed;
the last remaining shared_ptr owning the object is assigned another pointer via operator= or reset().
The object is destroyed using delete-expression or a custom deleter that is supplied to shared_ptr during construction.
So the pointer is deleted by shared_ptr. You're not suppose to delete the stored pointer yourself
UPDATE:
I didn't realize that there were more statements and the pointer was not out of scope, I'm sorry.
I was reading more and the standard doesn't say much about the behavior of get() but here is a note, I quote:
A shared_ptr may share ownership of an object while storing a pointer to another object. get() returns the stored pointer, not the managed pointer.
So it looks that it is allowed that the pointer returned by get() is not necessarily the same pointer allocated by the shared_ptr (presumably using new). So delete that pointer is undefined behavior. I will be looking a little more into the details.
UPDATE 2:
The standard says at § 20.7.2.2.6 (about make_shared):
6 Remarks: Implementations are encouraged, but not required, to perform no more than one memory allocation. [ Note: This provides efficiency equivalent to an intrusive smart pointer. — end note ]
7 [ Note: These functions will typically allocate more memory than sizeof(T) to allow for internal bookkeeping structures such as the reference counts. — end note ]
So an specific implementation of make_shared could allocate a single chunk of memory (or more) and use part of that memory to initialize the stored pointer (but maybe not all the memory allocated). get() must return a pointer to the stored object, but there is no requirement by the standard, as previously said, that the pointer returned by get() has to be the one allocated by new. So delete that pointer is undefined behavior, you got a signal raised but anything can happen.
I've created a list of stacks.
using namespace std;
list<stack<int>> stacks;
stack<int> *st = new stack<int>(); //LINE0
stacks.push_back(*st);
st->push(10);
stack<int> *last = &stacks.back();
stacks.pop_back(); //LINE1
delete last; //LINE2
LINE2 will cause an unhandled exception. Will LINE1 automatically deallocate the memory allocated at LINE0? Is LINE2 not necessary?
To answer your question, stacks contains objects, not pointers, and none of them were created by you using new. So the line delete last; is an error because you are giving it an address that wasn't received from new.
The key here is that stacks.push_back(*st); stores in the list a copy of the object pointed to by st.
On the other hand, st does contain a pointer received from a new expression. You should do a delete st; otherwise you will leak that memory.
Manual resource management can be error prone and difficult to reason about so I suggest you simply stop using new unless and until it's really necessary.
Yes, containers delete stored object automatically.
C++ containers store exactly the type of data which is set as parameter. In this case list stores stack<int>, not pointer or reference to it. It means that container creates copy of object when push_back is running.
If you want store pointers in container, you should inform compiler about it:
list<stack<int>*>
That list won't delete stacks. It will delete pointers only.
By the way, I don't recommend to abuse allocating memory dynamically. It often leads to memory leaks.
D's docs saying that when you use scope for local variables, then they will be allocated on stack (even if you're allocating class instance). But what about auto keyword? Does it guarantee that the instance will be allocated on stack?
void foo() { auto instance = new MyClass();}
void foo() { scope instance = new MyClass();}
So can I suggest that this two statements are equal (in terms of allocation)?
No, auto only infers the type.
There's no point in using auto if you want it to be allocated on the stack; that's what scope is (was) for.
They've brilliantly (read: not so much) decided to remove scope, delete, etc. from the language, so it will probably allocate on the heap anyway. Your best bet is to use the function called scoped in one of the modules, to allocate on the stack.
To answer the second question: in D1 those two statements are not equal. First one allocates on the heap, second one is (supposed) to allocate on the stack.
I am developping on Windows with DevStudio, in C/C++ unmanaged.
I want to allocate some memory on the stack instead of the heap because I don't want to have to deal with releasing that memory manually (I know about smart pointers and all those things. I have a very specific case of memory allocation I need to deal with), similar to the use of A2W() and W2A() macros.
_alloca does that, but it is deprecated. It is suggested to use malloca instead. But _malloca documentation says that a call to ___freea is mandatory for each call to _malloca. It then defeats my purpose to use _malloca, I will use malloc or new instead.
Anybody knows if I can get away with not calling _freea without leaking and what the impacts are internally?
Otherwise, I will end-up just using deprecated _alloca function.
It is always important to call _freea after every call to _malloca.
_malloca is like _alloca, but adds some extra security checks and enhancements for your protection. As a result, it's possible for _malloca to allocate on the heap instead of the stack. If this happens, and you do not call _freea, you will get a memory leak.
In debug mode, _malloca ALWAYS allocates on the heap, so also should be freed.
Search for _ALLOCA_S_THRESHOLD for details on how the thresholds work, and why _malloca exists instead of _alloca, and it should make sense.
Edit:
There have been comments suggesting that the person just allocate on the heap, and use smart pointers, etc.
There are advantages to stack allocations, which _malloca will provide you, so there are reasons for wanting to do this. _alloca will work the same way, but is much more likely to cause a stack overflow or other problem, and unfortunately does not provide nice exceptions, but rather tends to just tear down your process. _malloca is much safer in this regard, and protects you, but the cost is that you still need to free your memory with _freea since it's possible (but unlikely in release mode) that _malloca will choose to allocate on the heap instead of the stack.
If your only goal is to avoid having to free memory, I would recommend using a smart pointer that will handle the freeing of memory for you as the member goes out of scope. This would assign memory on the heap, but be safe, and prevent you from having to free the memory. This will only work in C++, though - if you're using plain ol' C, this approach will not work.
If you are trying to allocate on the stack for other reasons (typically performance, since stack allocations are very, very fast), I would recommend using _malloca and living with the fact that you'll need to call _freea on your values.
Another thing to consider is using an RAII class to manage the allocation - of course that's only useful if your macro (or whatever) can be restricted to C++.
If you want to avoid hitting the heap for performance reasons, take a look at the techniques used by Matthew Wilson's auto_buffer<> template class (http://www.stlsoft.org/doc-1.9/classstlsoft_1_1auto__buffer.html). This will allocate on the stack unless your runtime size request exceeds a size specified at compiler time - so you get the speed of no heap allocation for the majority of allocations (if you size the template right), but everything still works correctly if your exceed that size.
Since STLsoft has a whole lot of cruft to deal with portability issues, you may want to look at a simpler version of auto_buffer<> which is described in Wilson's book, "Imperfect C++".
I found it quite handy in an embedded project.
To allocate memory on the stack, simply declare a variable of the appropriate type and size.
I answered this before, but I'd missed something fundamental that meant that it only worked in debug mode. I moved the call to _malloca into the constructor of a class that would auto-free.
In debug this is fine, as it always allocates on the heap. However, in release, it allocates on the stack, and upon returning from the constructor, the stack pointer has been reset, and the memory lost.
I went back and took a different approach, resulting in a combination of using a macro (eurgh) to allocate the memory and instantiate an object that will automatically call _freea on that memory. As it's a macro, it's allocated in the same stack frame, and so will actually work in release mode. It's just as convenient as my class, but slightly less nice to use.
I did the following:
class EXPORT_LIB_CLASS CAutoMallocAFree
{
public:
CAutoMallocAFree( void *pMem ) : m_pMem( pMem ) {}
~CAutoMallocAFree() { _freea( m_pMem ); }
private:
void *m_pMem;
CAutoMallocAFree();
CAutoMallocAFree( const CAutoMallocAFree &rhs );
CAutoMallocAFree &operator=( const CAutoMallocAFree &rhs );
};
#define AUTO_MALLOCA( Var, Type, Length ) \
Type* Var = (Type *)( _malloca( ( Length ) * sizeof ( Type ) ) ); \
CAutoMallocAFree __MALLOCA_##Var( (void *) Var );
This way I can allocate using the following macro call, and it's released when the instantiated class goes out of scope:
AUTO_MALLOCA( pBuffer, BYTE, Len );
Ar.LoadRaw( pBuffer, Len );
My apologies for posting something that was plainly wrong!
If you're using _malloca() then you must call _freea() to prevent memory leak because _malloca() can do the allocation either on stack or heap. It resorts to allocate on heap if the given size exceeds_ALLOCA_S_THRESHOLD value. Thus, it's safer to call _freea() which won't do anything if allocation happened on stack.
If you're using _alloca() which seems to be deprecated as of today; there is no need to call _freea() as the allocation happens on stack.
If your concern is having to free temp memory, and you know all about things like smart-pointers then why not use a similar pattern where memory is freed when it goes out of scope?
template <class T>
class TempMem
{
TempMem(size_t size)
{
mAddress = new T[size];
}
~TempMem
{
delete [] mAddress;
}
T* mAddress;
}
void foo( void )
{
TempMem<int> buffer(1024);
// alternatively you could override the T* operator..
some_memory_stuff(buffer.mAddress);
// temp-mem auto-freed
}