Placement forms of the operator delete functions - c++

In his new book TC++PL4, Stroustrup casts a slightly different light on a once usual practice regarding user-controlled memory allocation and placement new—or, more specifically, regarding the enigmatical "placement delete." In the book's sect. 11.2.4, Stroustrup writes:
The "placement delete" operators do nothing except possibly inform a garbage collector that the deleted pointer is no longer safely derived.
This implies that sound programming practice will follow an explicit call to a destructor by a call to placement delete.
Fair enough. However, is there no better syntax to call placement delete than the obscure
::operator delete(p);
The reason I ask is that Stroustrup's sect. 11.2.4 mentions no such odd syntax. Indeed, Stroustrup does not dwell on the matter; he mentions no syntax at all. I vaguely dislike the look of ::operator, which interjects the matter of namespace resolution into something that properly has nothing especially to do with namespaces. Does no more elegant syntax exist?
For reference, here is Stroustrup's quote in fuller context:
By default, operator new creates its object on the free store. What
if we wanted the object allocated elsewhere?... We can place objects
anywhere by providing an allocator function with extra arguments and
then supplying such extra arguments when using new:
void* operator new(size_t, void* p) { return p; }
void buf = reinterpret_cast<void*>(0xF00F);
X* p2 = new(buf) X;
Because of this usage, the new(buf) X syntax for supplying extra
arguments to operator new() is known as the placement syntax.
Note that every operator new() takes a size as its first argument
and that the size of the object allocated is implicitly supplied.
The operator new() used by the new operator is chosen by the
usual argument-matching rules; every operator new() has
a size_t as its first argument.
The "placement" operator new() is the simplest such allocator. It
is defined in the standard header <new>:
void* operator new (size_t, void* p) noexcept;
void* operator new[](size_t, void* p) noexcept;
void* operator delete (void* p, void*) noexcept; // if (p) make *p invalid
void* operator delete[](void* p, void*) noexcept;
The "placement delete" operators do nothing except possibly inform a
garbage collector that the deleted pointer is no longer safely
derived.
Stroustrup then continues to discuss the use of placement new with arenas. He does not seem to mention placement delete again.

If you don't want to use ::, you don't really have to. In fact, you generally shouldn't (don't want to).
You can provide replacements for ::operator new and ::operator delete (and the array variants, though you should never use them).
You can also, however, overload operator new and operator delete for a class (and yes, again, you can do the array variants, but still shouldn't ever use them).
Using something like void *x = ::operator new(some_size); forces the allocation to go directly to the global operator new instead of using a class specific one (if it exists). Generally, of course, you want to use the class specific one if it exists (and the global one if it doesn't). That's exactly what you get from using void *x = operator new(some_size); (i.e., no scope resolution operator).
As always, you need to ensure that your news and deletes match, so you should only use ::operator delete to delete the memory when/if you used ::operator new to allocate it. Most of the time you shouldn't use :: on either one.
The primary exception to that is when/if you're actually writing an operator new and operator delete for some class. These will typically call ::operator new to get a big chunk of memory, then divvy that up into object-sized pieces. To allocate that big chunk of memory, it typically (always?) has to explicitly specify ::operator new because otherwise it would end up calling itself to allocate it. Obviously, if it specifies ::operator new when it allocates the data, it also needs to specify ::operator delete to match.

First of all: No there isn't.
But what is the type of memory? Exactly, it doesn't have one. So why not just use the following:
typedef unsigned char byte;
byte *buffer = new byte[SIZE];
Object *obj1 = new (buffer) Object;
Object *obj2 = new (buffer + sizeof(Object)) Object;
...
obj1->~Object();
obj2->~Object();
delete[] buffer;
This way you don't have to worry about placement delete at all. Just wrap the whole thing in a class called Buffer and there you go.
EDIT
I thought about your question and tried a lot of things out but I found no occasion for what you call placement delete. When you take a look into the <new> header you'll see this function is empty. I'd say it's just there for the sake of completeness. Even when using templates you're able to call the destructor manually, you know?
class Buffer
{
private:
size_t size, pos;
byte *memory;
public:
Buffer(size_t size) : size(size), pos(0), memory(new byte[size]) {}
~Buffer()
{
delete[] memory;
}
template<class T>
T* create()
{
if(pos + sizeof(T) > size) return NULL;
T *obj = new (memory + pos) T;
pos += sizeof(T);
return obj;
}
template<class T>
void destroy(T *obj)
{
if(obj) obj->~T(); //no need for placement delete here
}
};
int main()
{
Buffer buffer(1024 * 1024);
HeavyA *aObj = buffer.create<HeavyA>();
HeavyB *bObj = buffer.create<HeavyB>();
if(aObj && bObj)
{
...
}
buffer.destroy(aObj);
buffer.destroy(bObj);
}
This class is just an arena (what Stroustrup calls it). You can use it when you have to allocate many objects and don't want the overhead of calling new everytime. IMHO this is the only use case for a placement new/delete.

This implies that sound programming practice will follow an explicit call to a destructor by a call to placement delete.
No it doesn't. IIUC Stroustrup does not mean placement delete is necessary to inform the garbage collector that memory is no longer in use, he means it doesn't do anything apart from that. All deallocation functions can tell a garbage colector memory is no longer used, but when using placement new to manage memory yourself, why would you want a garbage collector to fiddle with that memory anyway?
I vaguely dislike the look of ::operator, which interjects the matter of namespace resolution into something that properly has nothing especially to do with namespaces.
"Properly" it does have to do with namespaces, qualifying it to refer to the "global operator new" distinguishes it from any overloaded operator new for class types.
Does no more elegant syntax exist?
You probably don't ever want to call it. A placement delete operator will be called by the compiler if you use placement new and the constructor throws an exception. Since there is no memory to deallocate (because the pacement new didn't allocate any) all it does it potentially mark the memory as unused.

Related

Who defines the new operator?

Is this program supposed to compile?
int main(){
double* p = new double[4];
delete[] p;
return 0;
}
(It does compile with GCC 7.1.1)
The reason I ask this is because, I am not sure who is providing the definition of the operator new.
Is it the language? is the compiler by default #includeing <new>?
To make this question more clear I can actually define (overwrite?) operator new.
#include<cstdlib> // malloc and size_t
#include<cassert> // assert
void* operator new[](unsigned long int sz){ // what if this is not defined? who provides `new`?
assert(0); // just to check I am calling this function
return std::malloc(sz);
}
int main(){
double* p = new double[4]; // calls the new function above
delete[] p;
return 0;
}
What am I doing here?
overriding new? overloading new? Is the definition of new special (magic) for the compiler? (e.g. if not defined use a language provided one).
What is the role of #include<new> in all this?
Here you have a new expression, which invokes indeed operator new[]
void* operator new[]( std::size_t count );
The compiler is implicitly declaring the basic operator news in each translation unit (this is specified by the standard), see the cppreference documentation.
The versions (1-4) are implicitly declared in each translation unit even if the < new> header is not included.
In your second code snippet, you are overloading (technically you are actually replacing) operator new (not overriding, that's only for virtual functions), although you should overload operator new[] instead (note that if you don't, then operator new[] will fall back on operator new, so technically I believe your code is OK, but I'd just overload the array version for clarity).
There can be other overloads of operator new. You have scalar and array versions both provided by the compiler, possibly by the standard library, and possibly by user-code.
There are additional overloads of new (and delete) if you write them yourself (if they don't have the signature of the built-in versions) or if you #include new. If you provide operators that match the builtin signatures, then they will replace the builtin version. Beware. :)
The main reasons people include are for
1) constructing objects in a user-provided memory address
2) non-throwing versions of new
There are, of course, operator delete overloads as well, since new/delete must be overloaded pairwise to work together, or nearly certain destruction will follow. If your operator new matches the built-in signature, it will replace rather than overload the built-in version.
Remember, a "new expression" as simple as:
std::string myString = new std::string();
really works in 2 parts: 1) Allocating memory (via operator new), and then, 2) constructing an object in that memory. If successful, it returns the pointer to that object, and if failure, cleans up what was constructed, deallocates whatever was allocated, and then throws.
When you overload operator new, you are only dealing with the memory allocation, not the constructor call, and the object will be constructed in whatever address this operator returns. For the normal placement new, you might see code like this:
char myBuffer[1024]; // assume aligned; nobody needs more than 1024 bytes
std::string *ptr = new (myBuffer) std::string();
assert (ptr == &myBuffer);
The extra parameter to new is the myBuffer address, which is immediately returned by operator new and becomes the place where the object is constructed. The assertion should pass, and shows that the string was created in the bytes of myBuffer.
The no-throw versions of new are also available after #including new, which also uses an extra argument to the operator:
char * buf = new (std::nothrow) char[MAX_INT];
if (buf) {
std::cout << "WHOA, you have a lot of memory!!!\n";
}
Now instead of failing, it'll return a null pointer, so you have to check it.

How safe is overloading new array operator?

Assuming I'm using a struct to be allocated on heap and used as
new [] and delete[]
shared_ptr(new [])
then can I simply overload its new array operator and don't touch any delete [] operator
struct alignas(256) MyStruct
{
Item i1,i2;
void * operator new[](unsigned long int size)
{
return aligned_alloc(256,size);
}
void * operator new (unsigned long int size)
{
return aligned_alloc(256,size);
}
};
and consider it done without any leaks?
GCC 6.3 and c++0x.
The truth is, there is no guarantee on how your standard library implements those operator new() functions, whatsoever. Your standard implementation might just call through to malloc() and free(), but it's not required to do so. It may just as well use the sbreak() syscall itself to back the memory objects it manages. Or it could use the mmap() syscall.
Either of these three is perfectly possible, and incompatible with the other two implementations. And, just as bad, your standard operator delete() implementation may consult some hidden data fields in front of the pointer that you pass into it to perform its internal bookkeeping. If the pointer that is passed into your standard operator delete() is not actually a pointer that was returned by the matching operator new(), you have undefined behavior. And the probability of something really bad happening is very high.
There is really no way around it: If you supply operator new(), you must also supply operator delete(). Otherwise, all hell may break loose.
If you overload new on a object, you should overload delete as well. Likewise, if you overload new[], you should overload delete[]. This is because the object will use the default delete, which may lead to a crash, depending on how you've messed with new.
Depending on your compiler, delete may call free() anyway, but you should get into the habit of overloading both new and delete, especially if you change how your object gets allocated.

overriding delete with parameters

I can override global operator new with different parameters, so for example I can have:
void* operator new (std::size_t size) throw (std::bad_alloc);
void* operator new (std::size_t size, int num) throw (std::bad_alloc);
which can be called separately as
int* p1 = new int; // calls new(size_t)
int* p2 = new(5) int; // calls new(size_t, int)
since each of these can potentially use some different allocation scheme, I would need a separate delete() function for each. However, delete(void*) cannot be overloaded in the same way! delete(void*) is the only valid signature. So how can the above case be handled?
P.S. I am not suggesting this is a good idea. This kind of thing happened to me and so I discovered this "flaw" (at least in my opinion) in c++. If the language allows the new overrides, it must allow the delete overrides, or it becomes useless. And so I was wondering if there is a way around this, not if this a good idea.
I can override global operator new with different parameters
Those are called placement allocation functions.
delete(void*) is the only valid signature.
No.
First, some terminology: A delete expression such as delete p is not just a function call, it invokes the destructor then calls a deallocation function, which is some overload of operator delete that is chosen by overload resolution.
You can override operator delete with signatures to match your placement allocation function, but that overload will only be used if the constructor called by the placement new expression throws an exception e.g.
struct E {
E() { throw 1; }
};
void* operator new(std::size_t n, int) throw(std::bad_alloc) { return new char[n]; }
void operator delete(void* p, int) { std::puts("hello!"); delete[] (char*)p; }
int main()
{
try {
new (1) E;
} catch (...) {
puts("caught");
}
}
The placement deallocation function which matches the form of placement new expression used (in this case it has an int parameter) is found by overload resolution and called to deallocate the storage.
So you can provide "placement delete" functions, you just can't call them explicitly. It's up to you to remember how you allocated an object and ensure you use the corresponding deallocation.
If you keep track of the different memory regions you allocate with your different new overloads, you can tag them with the version of new that was called.
Then at delete time you can look the address up to find which new was called, and do something different in each case.
This way you can guarantee that the correct logic is automatically associated with each different new overload.
As pointed out by baruch in the comments below, there is a performance overhead associated with the maintenance of the data you use for tracking, and this logic will also only work as long as the overloaded delete is not passed anything allocated using the default delete.
As far as tracking overhead, it seems to me that the minimum overhead method of tracking the type of the allocation is to allocate the amount requested, plus a small amount of additional space at the start of the allocated region in which to tag the request type (sized according to conservative alignment requirements). You can then look at this tag region on delete to determine which logic to follow.
There is a placement delete which matches the placement operator new. And you can't call it directly. This is because the placement delete is only used to deallocate memory when the constructor of the new'ed object throws.
void *operator new(size_t, Area&); // placement new
void operator delete(void*, Area&); // matching placement delete
...
Area &area;
SomeType *t=new(area) SomeType();
// when SomeType() throws then `delete(t,area)` from above is called
// but you can't do this:
delete (area) t;
A common way to overcome this, is to use write an overloaded "destroy" function, which accepts all kinds of parameters.
template<class T> void destroy(Area &a, T* &pt) //<-you can't do this with 'delete'
{
if (pt) {
pt->~T(); // run the destructor
a.freeMem(pt); // deallocate the object
pt=NULL; // nulls the pointer on the caller side.
}
}
The simple answer is: Do not do this. All forms of non-placement new are redundant in C++11 and horrifically unsafe, for example, return raw pointer. If you want to allocate objects in a custom place, then use a class with an allocate function if stateful or a free function if not. The best treatment for new and indeed, delete, is to excise them from your program with prejudice, with the possible exception of placement new.
Edit: The reason why it's useless for you is because you're trying to use it for a purpose which it was not intended for. All you can use the extra params for is stuff like logging or other behaviour control. You can't really change the fundamental semantics of new and delete. If you need stateful allocation, you must use a class.
You're wrong. It is possible to provide a placement delete.

Placement new behaviour equivalent

I have a question regarding placement new syntax in C++. Are the following two code snippets functionally equivalent and can be used interchangeably (I am not implying that the second should be used, when the first one is suitable)?
#1
T* myObj = new T();
// Do something with myObj
delete myObj;
#2
char* mem = new char[sizeof(T)];
T* myObj = new (mem) T();
// Do something with myObj
myObj->~T();
delete[] mem;
Is there something I should be especially careful of, when I am using the placement new syntax like this?
They are not equivalent, because they have different behaviour if the constructor or the destructor of T throws.
new T() will free any memory that has been allocated before letting the exception propagate any further. char* mem = new char[sizeof(T)]; T* myObj = new (mem) T(); will not (and unless you explicitly do something to ensure that it gets freed you will have a leak). Similarly, delete myObj will always deallocate memory, regardless of whether ~T() throws.
An exact equivalent for T* myObj = new T();/*other code*/delete myObj; would be something like:
//When using new/delete, T::operator new/delete
//will be used if it exists.
//I don't know how do emulate this in
//a generic way, so this code just uses
//the global versions of operator new and delete.
void *mem = ::operator new(sizeof(T));
T* myObj;
try {
myObj = new (mem) T();
}
catch(...) {
::operator delete(mem);
throw;
}
/*other code*/
try {
myObj->~T();
::operator delete(mem);
}
catch(...) {
//yes there are a lot of duplicate ::operator deletes
//This is what I get for not using RAII ):
::operator delete(mem);
throw;
}
Since you're allocating raw memory, a closer equivalent would be:
void *mem = operator new(sizeof(T));
T *myobj = new(mem) T();
// ...
myobj->~T();
operator delete(mem);
Note that if you've overloaded ::operator new for a particular class, this will use that class' operator new, where yours using new char [] would ignore it.
Edit: though I should add that I'm ignoring the possibility of exceptions here. #Mankarse's answer seems (to me) to cover that part fairly well.
On a fundamental level, you're doing the same thing in both situations: i.e. you're instantiating a new object on the heap and you're releasing the memory, but in the second case you're (obviously) using the placement new operator.
In this case both of them would yield the same results, minus the differences if the constructor throws as Mankarse explained.
In addition, I wouldn't say that you can use them interchangeably, because the second case is not a realistic example of how placement new is used. It would probably make a lot more sense to use placement new in the context of a memory pool and if you're writing your own memory manager so you can place multiple T objects on the per-allocated memory (in my tests it tends to save about 25% CPU time). If you have a more realistic use case for placement new then there will be a lot more things that you should worry about.
Well, you are pre-allocating memory for your T object. And it should be fine. Nevertheless it makes sense if you will reuse allocated area one more time. Otherwise It will be slower.
Yes. Your example is too simple to demonstrate this, but the memory you allocated in advance, "mem," should manage the object stored within, "myObj."
Perhaps put a better way, scenario #1 allocates space on the heap, and constructs an object in that space. Scenario #2 allocates space on the heap, "mem," then constructs an object somewhere within that space.
Now, put a second object somewhere else within "mem." It gets complicated, right?
The construction and destruction of myObj are happening identically in both scenarios (except in the case of your constructor throwing an exception, as pointed out by Mankarse), but the allocator is taking care of your memory management for you in scenario #1, and not in scenario #2.
So, be careful of managing "mem" appropriately. One common approach is the following:
template<class T> void destroy(T* p, Arena& a)
{
if (p) {
p->~T(); // explicit destructor call
a.deallocate(p);
}
}

is the following new overload leaking memory?

I have encountered the following code:
class a {
public:
void * operator new(size_t l, int nb);
double values;
};
void *a::operator new (size_t l,int n)
{
return new char[l+ (n>1 ? n - 1 : 0)*sizeof(double)];
}
From what I get it is then used to have an array like structure that start at "values":
double* Val = &(p->a->values) + fColumnNumber;
My question is :
is there a memory leak? I am very new to overloading new operator, but I'm pretty sure that the memory allocated is not deallocated properly. Also does that mean I can never create a "a" class on the stack?
thanks
I believe it technically produces UB as it is, though it's a form of UB that will probably never cause a visible side effect (it's using new [], but I believe that'll get matched up with delete -- but for char, this usually won't cause a visible problem).
IMO, it's almost worse that it's using a new expression to allocate what should really be raw bytes instead of objects. If I were doing it, I'd write it like:
void *a::operator new (size_t l,int n)
{
return ::operator new(l+ (n>1 ? n - 1 : 0)*sizeof(double));
}
You'd match that up with:
void a::operator delete(void *block)
{
::operator delete(block);
}
I don't see why the default operator delete called on an a * wouldn't be able to correctly deallocate the memory allocated by this custom operator new. The best way to check would be to actually write some code and find out, although rather than rob05c's technique I'd probably run it in a profiler such as valgrind. I assume the questioner sees a memory leak happening and suspects this as the cause, so writing a test case around this operator seems like a worthwhile endeavour.
Obviously it will leak if nobody gets around to actually deleting it afterwards...
I'd question the necessity of overriding new for this kind of functionality, but I also assume this was somebody else's code.
It's fairly easy to find out. Write a loop that constructs and deconstructs lots of a's, and watch your memory usage. It'll go up pretty fast if it's leaking.
Its fine as it is, but you'd need to use delete[], not delete, from the code that uses this class as it allocates an array. Note that the user wouldn't get any hints that they need to do this - so overloading a delete operator for them would be a good idea.
You can definitely create Class "a" on the stack.
There are 4 (actually more but will stick with the basics) new & delete method signatures you should know.
void* operator new (std::size_t size) throw (std::bad_alloc);
void* operator new[] (std::size_t size) throw (std::bad_alloc);
void operator delete (void* ptr) throw ();
void operator delete[] (void* ptr) throw ();
You are allocating an array within the "operator new" method which should be done in the "operator new[]" method. This will get rid of your nasty check. Write both, "operator new" and "operator new[]"
Don't forget you want to give the caller an object of type "a" ( a myA = new a) so make sure you return "a" not char*, therefore you need to also be casting.
You need to write the corresponding delete[] and delete methods.
To answer your question, I believe it does leak memory. The new signature you've provided is called the "placement new". This allows you to allocate a new pointer without allocating memory but to give it a location to point to. Example: If you needed a pointer to a specific point in memory.
long z = 0x0F9877F80078;
a myA = new (z) a[5]; // 5 pointers that point to 0x0F9877F80078
By definition the placement-new operator is not supposed to allocate memory and since you have you make be leaking. Get rid of your second argument, which you can do now since you have 2 versions of operator new and you're good to go. Don't forget to return an object "a".
Check out IBM's Info Center:
http://publib.boulder.ibm.com/infocenter/comphelp/v8v101/index.jsp?topic=%2Fcom.ibm.xlcpp8a.doc%2Flanguage%2Fref%2Fcplr318.htm
And the reference or references, cpluplus.com:
http://www.cplusplus.com/reference/std/new