I have a function which grows an array when trying to add an element if it is full. Which of the execution blocks is better or faster?
I think my second block (commented out) may be wrong, because after doubling my array I then go back and point to the original.
When creating arrays does the compiler look for a contiguous block in memory which it entirely fits into? (On the stack/heap? I don't fully understand which, though it is important for me to learn it is irrelevant to the actual question.)
If so, would this mean using the second block could potentially overwrite other information by overwriting adjacent memory? (Since the original would use 20 adjacent blocks of memory, and the latter 40.)
Or would it just mean the location of elements in my array would be split, causing poor performance?
void Grow()
{
length *= 2; // double the size of our stack
// create temp pointer to this double sized array
int* tempStack = new int[length];
// loop the same number of times as original size
for(int i = 0; i < (length / 2); i++)
{
// copy the elements from the original array to the temp one
tempStack[i] = myStack[i];
}
delete[] myStack; //delete the original pointer and free the memory
myStack = tempStack; //make the original point to the new stack
//Could do the following - but may not get contiguous memory block, causing
// overwritten >data
#if 0
int* tempStack = myStack; //create temp pointer to our current stack
delete[] myStack; //delete the original pointer and free memory
myStack = new int[length *= 2]; //delete not required due to new?
myStack = tempStack;
#endif
}
The second block wouldn't accomplish what you want at all.
When you do
myStack = new int[length *= 2];
then the system will return a pointer to wherever it happens to allocate the new, larger array.
You then reassign myStack to the old location (which you've already de-allocated!), which means you're pointing at memory that's not allocated (bad!) and you've lost the pointer to the new memory you just allocated (also bad!).
Edit: To clarify, your array will be allocated on the heap. Additionally, the (new) pointer returned by your larger array allocation (new int[foo]) will be a contiguous block of memory, like the old one, just probably in a different location. Unless you go out of bounds, don't worry about "overwriting" memory.
Your second block is incorrect because of this sequence:
int* tempStack = myStack; //create temp pointer to our current stack
delete[] myStack; //delete the original pointer and free memory
tempStack and myStack and both simply pointers to the same block of memory. When you delete[] the pointer in the second line, you no longer have access to that memory via either pointer.
Using C++ memory management, if you want to grow an array, you need to create a new array before you delete the old one and copy over the values yourself.
That said, since you are working with POD, you could use C style memory management which supports directly growing an array via realloc. That can be a bit more efficient if the memory manager realizes it can grow the buffer without moving it (although if it can't grow the buffer in place, it will fall back on the way you grow your array in your first block).
C style memory management is only okay for arrays of POD. For non-POD, you must do the create new array/copy/delete old array technique.
This doesn't exactly answer your question, but you shouldn't be doing either one. Generally new[] or delete[] should be avoided in favor of using std::vector. new[] is hard to use because it requires explicit memory management (if an exception is thrown as the elements are being copied, you will need to catch the exception and delete the array to avoid a memory leak). std::vector takes care of this for you, automatically grows itself, and is likely to have an efficient implementation tuned by the vendor.
One argument for using explicit arrays is to have a contiguous block of memory that can be passed to C functions, but that also can be done with std::vector for any non-pathological implementation (and the next version of the C++ standard will require all conforming implementations to support that). (For reference, see http://www.gotw.ca/publications/mill10.htm by Herb Sutter, former convener of the ISO C++ standards committee.)
Another argument against std::vector is the weirdness with std::vector<bool>, but if you need that you can simply use std::vector<char> or std::vector<int> instead. (See: http://www.gotw.ca/publications/mill09.htm)
Related
This question is similar to Problem with delete[], how to partially delete the memory?
I understand that deleting an array after incrementing its pointer is not possible as it loses the track of how many bytes to clean. But, I am not able to understand why one-by-one delete/deallocation of a dynamic array doesn't work either.
int main()
{
int n = 5;
int *p = new int[n];
for(int i=0;i<n;++i){
delete &p[i];
}
}
I believe this should work, but in clang 12.0 it fails with the invalid pointer error. Can anyone explain why?
An array is a contiguous object in memory of a specific size. It is one object where you can place your data in and therefore you can only free/delete it as one object.
You are thinking that an array is a list of multiple objects, but that's not true. That would be true for something like a linked list, where you allocate individual objects and link them together.
You allocated one object of the type int[n] (one extent of memory for an array) using the operator new
int *p = new int[n];
Elements of the array were not allocated dynamically separately.
So to delete it you just need to write
delete []p;
If for example you allocated an array of pointers like
int **p = new int *[n];
and then for each pointer of the array you allocated an object of the type int like
for ( int i = 0;i < n;++i )
{
p[i] = new int( i );
}
then to delete all the allocated objects you need to write
for ( int i = 0; i < n; ++i )
{
delete p[i];
}
delete []p;
That is the number of calling of the operator delete or delete [] one to one corresponds to the number of calling operator new or new [].
One new always goes with one delete. Just as that.
In detail, when we request an array using new, what we actually do is to get a pointer that controls a contiguous & fixed block on the memory. Whatever we do with that array, we do it through that pointer and this pointer associates strictly with the array itself.
Furthermore, let's assume that you were able to delete an elemnent in the middle of that array. After the deletion, that array would fall apart and they are not contiguous anymore! By then, an array would not really be an array!
Because of that, we can not 'chop off' an array into separate pieces. We must always treat an array as one thing, not distinctive elements scattered around the memory.
Greatly simplyfyinh: in most systems memory is allocated in logical blocks which are described by the starting pointer of the allocated block.
So if you allocate an array:
int* array = new int[100];
OS stores the information of that allocation as a pair (simplifying) (block_begin, size) -> (value of array ptr, 100)
Thus when you deallocate the memory you don't need to specify how much memory you allocated i.e:
// you use
delete[] array; // won't go into detail why you do delete[] instead of delete - mostly it is due to C++ way of handling destruction of objects
// instead of
delete[100] array;
In fact in bare C you would do this with:
int* array = malloc(100 * sizeof(int))
[...]
free(array)
So in most OS'es it is not possible due to the way they are implemented.
However theoretically allocating large chunk of memory in fact allocate many smaller blocks which could be deallocated this way, but still it would deallocate smaller blocks at a time not one-by-one.
All of new or new[] and even C's malloc do exactly the same in respect to memory: requesting a fix block of memory from the operating system.
You cannot split up this block of memory and return it partially to the operating system, that's simply not supported, thus you cannot delete a single element from the array either. Only all or none…
If you need to remove an element from an array all you can do is copy the subsequent elements one position towards front, overwriting the element to delete and additionally remember how many elements actually are valid – the elements at the end of the array stay alive!
If these need to be destructed immediately you might call the destructor explicitly – and then assure that it isn't called again on an already destructed element when delete[]ing the array (otherwise undefined behaviour!) – ending in not calling new[] and delete[] at all but instead malloc, placement new for each element, std::launder any pointer to any element created that way and finally explicitly calling the constructor when needed.
Sounds like much of a hassle, doesn't it? Well, there's std::vector doing all this stuff for you! You should this one it instead…
Side note: You could get similar behaviour if you use an array of pointers; you then can – and need to – maintain (i.e. control its lifetime) each object individually. Further disadvantages are an additional level of pointer indirection whenever you access the array members and the array members indeed being scattered around the memory (though this can turn into an advantage if you need to move objects around your array and copying/moving objects is expensive – still you would to prefer a std::vector, of pointers this time, though; insertions, deletions and managing the pointer array itself, among others, get much safer and much less complicated).
I'm currently trying to create vector-like container. It uses memory allocated by new[] as it's base. the problem arises when I need to expand the array. I allocate a bigger chunk of memory with new[], then memcpy old memory into there and delete[] the old memory. Thing is, trying to store any pointer or any pointer-containing object inside result in memory corruption. So I need a way to free the memory used without destroying objects inside
Edit: Some code to understand the problem:
template<typename T>
class myvector
{
private:
T* _data;
size_t _size, _capacity;
static constexpr float multiplier = 1.5;
public:
void expand()
{
size_t e_size = sizeof(T);
size_t old_capacity = this->_capacity;
this->_capacity = (unsigned long)(float(this->_capacity) * myvector::multiplier);
T *tmp = new T[this->_capacity];
memcpy(tmp, this->_data, e_size * (old_capacity));
// this will destroy all the objects inside the container
// which will result in destruction of any allocated memory
delete[] this->_data;
// so now we have an array of invalid pointers. fun time
this->_data = tmp;
}
}
How to free memory allocated by new[]?
Using delete[]. This must be done before the pointer value is lost, and it must be done after the last use of the pointer value. And it must be done exactly once. And it must not be done for any other pointer than one that was returned by array new.
Thing is, trying to store any pointer or any pointer-containing object inside result in memory corruption.
Then there is a bug in how you use the objects. Storing such objects in your template should by itself not be a problem.
So I need a way to free the memory used without destroying objects inside
That is simply not possible. An object cannot exist without storage (except for special cases that don't apply here).
delete[] this->_data;
// so now we have an array of invalid pointers. fun time
Why would there be an array of invalid pointers? Do the pointers in the array point to this->_data?
Indeed, your data structure does not have stable addresses for its elements. Expansion will invalidate any references to the elements. If you need such stability, then you must use a node based data structure such as a linked list.
Your template does have a limitation that it is well defined only for trivially copyable classes. Perhaps you've overlooked this limitation. It is easy to get rid of this limitation: Simply use std::copy (or perhaps std::move from <algorithm>, depending on exception safety guarantees that you need) instead of std::memcpy.
Is there a way to expand a dynamic memory array? like this:
int *a = new int[5];
*a = new int[2];
Is this legal?
You cannot expand this type of a dynamic memory array. You can use malloc and realloc though if you need this facility but I would advice against that and suggest including <vector> and using std::vector instead. It has a resize method.
Also, what you described won't compile. The following will:
1: int *a = new int[5];
2: a = new int[2];
The above will allocate two memory blocks, neither of which will be destroyed. Second line will simply assign a new array to the same int *a pointer. When an allocated memory stops being referenced by any pointer, this is called a memory leak. The above code loses any reference to new int[5] and there is no way to free this memory to the operating system.
Although this is not a very practical example, there are multiple ways to resize an array/vector.
As it is usually practical to increase the array size, I will do just this:
{ // C++ vector on the stack (although internally vector uses memory from the heap)
std::vector<int> a(1024);
// do smth
a.resize(4096); // note: this does not always reallocate
// do smth else
}
{ // C++ everything on the heap
std::vector<int> *a = new std::vector<int>(1024);
// do smth
a->resize(4096); // note: this does not always reallocate
// do smth else
delete a;
}
{ // C style
int *a = (int*)malloc(1024*sizeof(int));
// do smth
a = realloc(a, 4096*sizeof(int));
// do smth else
free(a);
}
It is worth to note that realloc does not do anything smart. All it does is:
Allocate new memory block malloc
Copy data from old memory block to new memory block memcpy
Free old memory block free
Return new memory block
You can certainly expand an array, but you need to take care of copying the contents and of freeing the old array (your code, apart from being incorrect syntax, shrinks the array, btw.).
Which is exactly how std::vector works, just you don't have to care.
So basically, having int *a already allocated, what needs to happen is something like:
{
std::unique_ptr<int[]> d(a);
a = new int[desired_new_size];
for(unsigned int i = 0; i < min_old_size_and_new_size; ++i)
a[i] = d[i];
}
Note that strictly speaking "expanding" never really expands the array, but replaces it with another bigger one (that is true for any containers offering the same functionality too). But this is transparent to any code using the pointer later, nobody will know.
You should never use realloc (or any other C memory allocation functions) in combination with memory allocated or freed by operator new and delete (or new[] and delete[]) as pointed out above.
This may work (and usually will), but it's conceptually wrong, and it's pure luck (unknown implementation detail) if it does not crash.
I come from a java background and there's something I could do in Java that I need to do in C++, but I'm not sure how to do it.
I need to declare an array, but at the moment I don't know the size. Once I know the size, then I set the size of the array. I java I would just do something like:
int [] array;
then
array = new int[someSize];
How do I do this in C++?
you want to use std::vector in most cases.
std::vector<int> array;
array.resize(someSize);
But if you insist on using new, then you have do to a bit more work than you do in Java.
int *array;
array = new int[someSize];
// then, later when you're done with array
delete [] array;
No c++ runtimes come with garbage collection by default, so the delete[] is required to avoid leaking memory. You can get the best of both worlds using a smart pointer type, but really, just use std::vector.
In C++ you can do:
int *array; // declare a pointer of type int.
array = new int[someSize]; // dynamically allocate memory using new
and once you are done using the memory..de-allocate it using delete as:
delete[]array;
Best way would be for you to use a std::vector. It does all you want and is easy to use and learn. Also, since this is C++, you should use a vector instead of an array. Here is an excellent reason as to why you should use a container class (a vector) instead of an array.
Vectors are dynamic in size and grow as you need them - just what you want.
The exact answer:
char * array = new char[64]; // 64-byte array
// New array
delete[] array;
array = new char[64];
std::vector is a much better choice in most cases, however. It does what you need without the manual delete and new commands.
As others have mentioned, std::vector is generally the way to go. The reason is that vector is very well understood, it's standardized across compilers and platforms, and above all it shields the programmer from the difficulties of manually managing memory. Moreover, vector elements are required to be allocated sequentially (i.e., vector elements A, B, C will appear in continuous memory in the same order as they were pushed into the vector). This should make the vector as cache-friendly as a regular dynamically allocated array.
While the same end result could definitely be accomplished by declaring a pointer to int and manually managing the memory, that would mean extra work:
Every time you need more memory, you must manually allocate it
You must be very careful to delete any previously allocated memory before assigning a new value to the pointer, lest you'll be stuck with huge memory leaks
Unlike std::vector, this approach is not RAII-friendly. Consider the following example:
void function()
{
int* array = new int[32];
char* somethingElse = new char[10];
// Do something useful.... No returns here, just one code path.
delete[] array;
delete[] somethingElse;
}
It looks safe and sound. But it isn't. What if, upon attempting to allocate 10 bytes for "somethingElse", the system runs out of memory? An exception of type std::bad_alloc will be thrown, which will start unwinding the stack looking for an exception handler, skipping the delete statements at the end of the function. You have a memory leak. That is but one of many reasons to avoid manually managing memory in C++. To remedy this (if you really, really want to), the Boost library provides a bunch of nice RAII wrappers, such as scoped_array and scoped_ptr.
use std::array when size is known at compile time otherwise use std::vector
#include <array>
constexpr int someSize = 10;
std::array<int, someSize> array;
or
#include <vector>
std::vector<int> array; //size = 0
array.resize(someSize); //size = someSize
Declare a pointer:
int * array;
How does std::vector implement the management of the changing number of elements: Does it use realloc() function, or does it use a linked list?
Thanks.
It uses the allocator that was given to it as the second template parameter. Like this then. Say it is in push_back, let t be the object to be pushed:
...
if(_size == _capacity) { // size is never greater than capacity
// reallocate
T * _begin1 = alloc.allocate(_capacity * 2, 0);
size_type _capacity1 = _capacity * 2;
// copy construct items (copy over from old location).
for(size_type i=0; i<_size; i++)
alloc.construct(_begin1 + i, *(_begin + i));
alloc.construct(_begin1 + _size, t);
// destruct old ones. dtors are not allowed to throw here.
// if they do, behavior is undefined (17.4.3.6/2)
for(size_type i=0;i<_size; i++)
alloc.destroy(_begin + i);
alloc.deallocate(_begin, _capacity);
// set new stuff, after everything worked out nicely
_begin = _begin1;
_capacity = _capacity1;
} else { // size less than capacity
// tell the allocator to allocate an object at the right
// memory place previously allocated
alloc.construct(_begin + _size, t);
}
_size++; // now, we have one more item in us
...
Something like that. The allocator will care about allocating memory. It keeps the steps of allocating memory and constructing object into that memory apart, so it can preallocate memory, but not yet call constructors. During reallocate, the vector has to take care about exceptions being thrown by copy constructors, which complicates the matter somewhat. The above is just some pseudo code snippet - not real code and probably contains many bugs. If the size gets above the capacity, it asks the allocator to allocate a new greater block of memory, if not then it just constructs at the previously allocated space.
The exact semantics of this depend on the allocator. If it is the standard allocator, construct will do
new ((void*)(_start + n)) T(t); // known as "placement new"
And the allocate allocate will just get memory from ::operator new. destroy would call the destructor
(_start + n)->~T();
All that is abstracted behind the allocator and the vector just uses it. A stack or pooling allocator could work completely different. Some key points about vector that are important
After a call to reserve(N), you can have up to N items inserted into your vector without risking a reallocation. Until then, that is as long as size() <= capacity(), references and iterators to elements of it remain valid.
Vector's storage is contiguous. You can treat &v[0] as a buffer containing as many elements you have currently in your vector.
One of the hard-and-fast rules of vectors is that the data will be stored in one contiguous block of memory.
That way you know you can theoretically do this:
const Widget* pWidgetArrayBegin = &(vecWidget[0]);
You can then pass pWidgetArrayBegin into functions that want an array as a parameter.
The only exception to this is the std::vector<bool> specialisation. It actually isn't bools at all, but that's another story.
So the std::vector will reallocate the memory, and will not use a linked list.
This means you can shoot yourself in the foot by doing this:
Widget* pInteresting = &(vecWidget.back());
vecWidget.push_back(anotherWidget);
For all you know, the push_back call could have caused the vector to shift its contents to an entirely new block of memory, invalidating pInteresting.
The memory managed by std::vector is guaranteed to be continuous, such that you can treat &vec[0] as a pointer to the beginning of a dynamic array.
Given this, how it actually manages it's reallocations is implementation specific.
std::vector stored data in contiguous memory blocks.
Suppose we declare a vector as
std::vector intvect;
So initially a memory of x elements will be created . Here x is implementation depended.
If user is inserting more than x elements than a new memory block will be created of 2x (twice the size)elements and initial vector is copied into this memory block.
Thats why it is always recommended to reserve memory for vector by calling reserve
function.
intvect.reserve(100);
so as to avoid deletion and copying of vector data.