I have a struct like this:
class Items
{
private:
struct item
{
unsigned int a, b, c;
};
item* items[MAX_ITEMS];
}
Say I wanted to 'delete' an item, like so:
items[5] = NULL;
And I created a new item on that same spot later:
items[5] = new item;
Would I still need to call delete[] to clean this up? Or won't this be needed since bounds of array items[] are known before compiling?
Is setting that pointer to NULL valid or should I be calling delete there?
You need to call delete before setting it to NULL. (Setting it to NULL isn't required, it just helps reduce bugs if you accidentally try to dereference the pointer after deleting it.)
Remember that every time you use new, you will need to use delete later on the same pointer. Never use one without the other.
Also, new [] and delete [] go together in the same way, but you should never mix new [] with delete or new with delete []. In your example, since you created the object with new (rather than new [] which would create an array of objects) you must delete the object with delete (rather than delete []).
As Kluge pointed out, you'd leak the object at index 5 like that. But this one really sounds like you shouldn't do this manually but use a container class inside Item. If you don't actually need to store these item objects as pointers, use std::vector<item> instead of that array of MAX_ITEMS pointers. You can always insert or erase vector elements in the middle as well if you need to.
In case you need to store the objects as pointers (usually if struct item is actually polymorphic, unlike in your example), you can use boost::ptr_vector<item> from Boost.PtrContainer instead.
Example:
class Items {
private:
struct item {
unsigned int a, b, c;
};
std::vector<item> items;
}
if (items.size() > 5) // (just to ensure there is an element at that position)
items.erase(items.begin() + 5); // no need to use operator delete at all
To delete an item use:
delete items[5];
after deleting the item it is advisable to set the deleted pointer to NULL, so you won't have an error if you later delete it again by mistake.
items[5] = NULL
Say I wanted to 'delete' an item, like so:
items[5] = NULL;
I know little Visual Basic, but that smells like a Visual Basic programming idiom, since "Set a = None" (or Null, I'm not sure) would delete the object pointed by a (or rather decrement its reference count, for COM objects).
As somebody else noted, you should use either:
delete items[5];
items[5] = newContent;
or:
delete items[5];
items[5] = NULL;
After delete[5], the only possible use of the pointer stored in items[5] is causing you trouble. What's worse is that it might happen to work at the beginning, and start failing only when you allocate something else over the space previously used by *items[5]. Those are the causes which make C/C++ programming "interesting", i.e. really annoying (even for who likes C like me).
Writing just delete items[5]; saves what can be an useless write, but that's a premature optimization.
Just to be clear: you refer to calling "delete[]". I think you mean delete.
I mention this because C++ has two separate operators, operator delete and operator delete[]. The latter is used for deleting arrays of objects allocated with operator new[], and does not apply in this case. You have an array of pointers to objects, which you must have initialised with repeated calls to operator new rather than a single call to operator new[].
All I'm really trying to say is: your use of delete[] is confusing and ambiguous; change it to delete.
There are a few, related, questions here:
According to the code you posted, the array itself is not allocated on the heap unless the struct is, so you don't need to delete[] the array. If you created the array with new[] you would have to delete[] it.
The code posted doesn't say how the objects being pointed to from the array are allocated. If you allocate those objects on the stack you must not delete them (then again, this is highly unlikely because your pointers will become invalid when the objects they point to fall out of scope). If you allocated them on the heap (with new) then you must delete them when they fall out of scope.
As others have already suggested, life is much easier if you use a container -- especially an STL container -- and smart pointers -- which for now means pointers out of Boost.
C++ isn't my strong suit, but I'm pretty sure you'd be leaking the memory if you set the pointer to NULL.
EDIT: The memory being leaked would be the memory being pointed to by the pointer in the array.
Setting items[5] to NULL doesn't delete the memory associated with the item, it simply sets the pointer to that item to NULL, therefore the memory is leaked.
You can delete the item by calling:
delete items[5];
Since C++ has not automatic garbage collection, you need to delete any memory you no longer need.
Related
int* func()
{
int* i=new int[1];
//do something
return i;
}
void funcc()
{
int* tmp=func();
//delete allocated memory after use
delete tmp;
}
should delete working as described in the second function be a correct use ?
I think I didn't allocate memory to it by new ? new was used in the first, to be sure.
It should be delete [] tmp; because you're doing array new, but otherwise, it's correct.
As others have stated, you should use delete[] to delete arrays, not delete.
But, if you're asking whether it's okay to delete tmp because the memory it points to wasn't allocated with new, then your premise is incorrect.
It was allocated with new. The address of the memory that you allocate within the function for i is passed back to be stored in tmp, so it does point to memory that was allocated by new.
You're correct that i itself is out of scope at that point but memory allocated by new survives the scope change on exit from the function. And, since you're storing its location into tmp, you still have access to it.
Hence the deletion (once you make it an array deletion with []) is quite valid.
This is Undefined Behaviour™. You can only use delete if you used new. If you used new[], you MUST delete[] it. More than that, this is hideously unsafe code- you need a smart pointer.
No. new T[] should match delete []t. And new T should match delete t. Otherwise, your code would invoke undefined bevavior.
And it doesn't matter if you do delete []tmp outside the function. Its okay to do so. All that you need to keep in mind that the form of delete.
My spidey-senses tell me that you're wondering whether the dynamically-allocated int that you create in func is the same one that you attempt to delete in funcc.
The answer is: yes. Although you don't use strictly the same pointer variable, both pointers point to the same object.
However, please remember to use delete[] (with the []) when you used new[] (with the []). Your code is broken until you fix this.
Also, try to avoid newing in one place and deleteing in another. Perhaps consider a std::vector instead, which is far less error-prone.
Hi there i have some questions about pointers and arrays in c++:
when i want to pass an array of pointers in a function i must write:
foo(test *ptest[])
right?
when i want to delete a array of pointers, does
delete ptest;
all for me?
when i pass a pointer to another class and the first class make a delete, the pointer is deletete in all classes right?
must i always create a array of a constant size?
First of all, forget about arrays and pointers and tell us what you really want to achieve. You are probably better off using std::vector<T> and pass that around by (const) reference, saving you all the headaches of manual resource management and crazy array-to-pointer-decay rules inherited from C.
Now to answer your questions:
when i want to pass an array of pointers in a function i must write:
foo(test *ptest[])
right?
Yes, but remember that array parameters are always rewritten by the compiler to pointers. Your code is equivalent to:
foo(test **ptest)
This means that all information about the size of the array has been lost. Moving on:
when i want to delete a array of pointers, does
delete ptest;
all for me?
No. If you acquire via new[], you must release via delete[], otherwise you get undefined behavior. Note that deleting an array of pointers only gets rid of the array itself, not any objects pointed to by the pointers stored inside the array.
Again: Forget about arrays and pointers and use the facilities provided by the C++ standard library.
when i want to delete a array of pointers, does delete ptest; all for me?
Nope. It tries to delete what's been allocated to the address ptest. If ptest is not something which was allocated dynamically, your program will crash. You can only call delete on the addresses which you have received from new.
when i pass a pointer to another class and the first class make a delete, the pointer is deletete in all classes right?
Nope, you need to have exactly one delete for every new. (yah, there are "smart pointers" which call delete for you, but the point still holds.)
must i always create a array of a constant size?
Nope. In C++ there's std::vector for having arrays of dynamic size (can be used for constant-size arrays too) and in C there's realloc for changing a size of an array.
To delete any array you need:
delete [] ptest; // note the [] part
Arrays can be any size you like (within various practical limits) and unless you make copies of an array then you should delete it exactly once (it doesn't matter where you do this, so long as you don't try to acess the data afterwards).
Chances are, without knowing what you're aiming at and any other contextual information, you should use std::vector<T> instead and forget about these worries.
That said, to delete any primitive array arr you must say
delete[] arr;
Of course, if the pointers in the array themselves are each the last pointing at heap memory then you should free them first. You can do that nicely with a reusable Delete functor:
struct Delete {
template<class P>
void operator()(P ptr) const { delete ptr; }
};
// ...
std::foreach(arr, arr+ARR_LENGTH, Delete());
delete[] arr;
You must of course make sure no two pointers point at the same address. To further simplify things beyond what a std::vector<T> can do for you you should consider using a boost::shared_ptr<T>; this will allow you to forget about the delicate case in which two pointers point at the same address.
Why is there a delete[]? From my understanding its to behave differently for arrays. However, why does it really exist? There's only free in C and no free_array. Also in syntax the only difference between delete var and delete []var is the [] which has no params (I'm not telling the length of the array).
So why does delete[] really exist? I know someone will say you can overload delete and delete[] (at least i think that is possible) but lets say we are not overloading it. Why does it exist?
Typically, for non-POD classes, a delete[] expression must call destructors on a variable number of class instances that cannot be determined at compile time. The compiler typically has to implement some run time "magic" that can be used to determine the correct number of objects to destroy.
A delete expression doesn't have to worry about this, it simply has to destroy the one object that the supplied pointer is pointing to. Because of this, it can have a more efficient implementation.
By splitting up delete and delete[], delete can be implemented without the overhead needed to correctly implement delete[] as well.
If you delete an array, only first object's destructor will be called. delete[] calls destructors of all objects in array and frees array's memory.
Assume delete[] didn't exist, write the code for deleting the array vs deleting only the first element in the array.
delete array; // Deletes first element, oops
delete &array; // Deletes first element, oops
delete &array[0]; // Deletes first element
A pointer to an array being an alias for a pointer to the first element of the array is of course an old C "feature".
Consider:
int* a = new int[25];
int* b = a;
delete b; // only deletes the first element
The C++ compiler has no idea whether b points to an array or a single element. Calling delete on an array will only delete the first element.
I saw an example of using the function: delete in cpp and I didn't completely understand it.
the code is:
class Name {
const char* s;
//...
};
class Table {
Name* p;
size_t sz;
public:
Table(size_t s = 15){p = new Name[sz = s]; }
~Table { delete[] p; }
};
What is the exact action of the command: delete[] p;?
I think the aim was to delete all the pointers in the container Table.
The brackets in delete[] give me a clue that it deletes an array of pointers to Name but the size of the array is not specified, so how does the destructor "know" how many pointers to delete?
delete isn't a function, it's an operator.
A delete expression using [] destroys objects created with new ... [] and releases the associated memory. delete[] must be used for pointers returned by new ... []; non-array delete only on pointers returned by non-array new. Using the non-matching delete form is always incorrect.
The delete expression in ~Table() (missing () in your code) will destroy the dynamically created array of Name objects ensuring that the Name destructor is called for each member of the array.
It is up the the implementation to implement some mechanism of recording the number of elements in arrays allocated with new ... [] the programmer doesn't have to worry about this.
In many implementations, where the array elements have non-trivial destructors, a new[] expression will allocate extra space to record the element count before the space for all the array members. This hidden count is then looked up when delete[] is used to ensure the correct number of destructors are called. This is just an implementation detail, though, other implementations are possible.
In short, delete[] knows the size of the array it is deleting because it is required to.
Because the C++ language standard states that it must know.
So when you allocate an array, it is up to the system to store the size somewhere where delete[] can find it.
One option is to allocate a few bytes more than needed. Use the first bytes to specify the size, and then instead of returning a pointer to the first allocated byte, return a pointer to the first byte past the size field.
Then delete[] just has to subtract a few bytes from the pointer in order to find the size.
Another option could be to have a global map<void*, int>, where the key is the pointer to a memory allocation, and the value is the size of that allocation. There are plenty of other ways in which delete[] can find out the size. The C++ standard doesn't specify which one to use. It just says that delete[] must know the size, and leaves it up to the implementers to figure out how to pull it off.
I have a class with a vector of pointers to objects. I've introduced some elements on this vector, and on my main file I've managed to print them and add others with no problems. Now I'm trying to remove an element from that vector and check to see if it's not NULL but it is not working.
I'm filling it with on class Test:
Other *a = new Other(1,1);
Other *b = new Other(2,2);
Other *c = new Other(3,3);
v->push_back(a);
v->push_back(b);
v->push_back(c);
And on my main file I have:
Test t;
(...)
Other *pointer = t.vect->at(0);
delete t.vect->at(0);
t.vect->erase(t.vect->begin());
if (pointer == NULL) { cout << "Nothing here.."; } // Never enters here..
Deleting a pointer doesn't have to zero it, it just frees the memory assigned there. Using pointer is undefined after the delete though, as the memory is free to be used for other things. Technically the C++ spec intentionally leaves it up to implementations whether they want to zero a deleted pointer or not, but practically none that I know of do so
The deletion of memory pointed by a pointer, doesn't set the pointer to NULL.
You set pointer equal to the address of something, and never touched it again, so of course it won't be null. The fact that you then did something to the object is irrelevant.
Using simple pointers, there is no safe way for pointer to determine whether the object it once pointed to has been deleted. The simplest way to do what you seem to want to do is by leaving it to the containers: if you're interested in that object, search for pointer in the vector to see whether it's still there (and don't delete the object without erasing the corresponding element from the vector, or you'll have the same problem all over again).
To overcome the checking a deleted ptr problem you could use boost::shared_ptr.
Instead of delete use .reset() and to check if the ptr is still valid use .get()
(Actually you can just use if(p) where p is the shared_ptr because it has a conversion to bool)