How to delete 1D array - c++

I create an array like as below
int size = 5;
double my_arr[m_size];
for (int i = 0; i < size; i++)
{
my_arr[i] = (rand()%10+1) + ((double) rand() / (RAND_MAX));;
}
after doing some calculation on array I want to delete the array. So I do this
for (int i = 0; i < size; i++)
{
delete my_arr[i];
}
and I get this error
error: type ‘double’ argument given to ‘delete’, expected pointer
I searched internet and all solutions are related to pointer array. But I am not using any pointer. So how can I delete this array?

The array will be automatically deleted when leaving the scope in which the variable has been declared.
If you really need to free memory fast you can try put your code between embraces:
{ //create new scope
int size = 5;
double my_arr[m_size];
for (int i = 0; i < size; i++)
{
my_arr[i] = (rand()%10+1) + ((double) rand() / (RAND_MAX));;
}
//some stuff
} //all non-pointer objects (or arrays) will be deleted
Or you can use pointers :
double *pMyarr = new double[m_size] ;

First of all, I think you should consider using a memory leak detector in your programms, so you can know by yourself if your code is leaking and if you have to do something about it. In your code you would have seen that there is no need to delete anything. :)
As far as I know, you should worry about memory in only two cases:
You have allocated a C-style array with malloc or calloc. In this case, you need to use the function free to deallocate the array. But generally in C++, you don't want this. You prefer using a container like std::array or std::vector instead.
You have created a new instance of an class with new and you have got a pointer to this instance. In this case, you have to use delete on the pointer when you don't need this instance anymore. But generally in C++11 (or further), you don't want this. If you really have to use a pointer, you prefer creating a smart pointer like std::unique_ptr or std::shared_ptr which will handle the memory for you.
When you define a variable without using new, malloc or calloc, (for example int a = 1 ;), it will be automatically deleted when it goes out of scope. So you don't need to worry.
Here is a simple example of a variable going out of scope:
int a = 1 ;
{
int b = 1 ;
// b is implictly deleted here, just before the bracket.
}
a++; // It works, because a exists in this scope
b++; // It doesn't work, because b is out of scope.
// a is implicitly deleted here, assuming we are at the end of a function

Related

How to delete memory of a pointer to pointer in C++

Using Valgrind, I see that I have a problem while deleting the memory in the following function:
Obj1 Obj1::operator*(const Obj1& param) const {
int n = param.GetSize(2);
Obj2** s = new Obj2*[n];
for( int i = 0; i < n; ++i) {
s[i] = new Obj2(*this*param.GetColumn(i+1));
}
Obj1 res = foo(s,n);
for(int i=n-1;i>-1;i--) {
s[i]->~Obj2();
}
delete[] s;
return res;
Valgrind tells me that the leak comes from the line
s[i] = new Obj2(*this*param.GetColumn(i+1));
I'm not pretty sure if the problem is when I try to free the memory. Can anyone tell me how to fix this problem?
Here:
s[i] = new Obj2(*this*param.GetColumn(i+1));
you create a dynamic object and assign s[i]to point to it.
In order to delete it, you do this:
delete s[i];
Unless you do that, the allocation will leak.
You must repeat that in a loop for every i just like you repeated the allocations. You of course have to do this before you delete s itself.
s[i]->~Obj2();
Don't do that. Calling the destructor is not appropriate here. delete will call the destructor.
P.S. Don't use raw owning pointers. Use containers or smart pointers instead. std::vector is a standard containers for dynamic arrays.
P.P.S. You should avoid unnecessary dynamic allocation. Your example doesn't demonstrate any need to allocate the pointed objects dynamically. So, in this case you should probably use std::vector<Obj2>.

Free dynamic array of static arrays?

I have code similar to this
EDIT this code is supposed to be on interface so only POD is allowed. No vectors, no containers.
int (*neco)[2];
int ahoj = 2;
neco = new int[ahoj+2][2];
int iter = 1;
for (size_t i = 0; i < 4; i++)
{
for (size_t j = 0; j < 2; j++)
{
neco[i][j] =iter;
iter++;
}
}
for (size_t i = 0; i < ahoj + 2; i++)
{
delete[] neco[i];
}
This does not work. The delete deletes memory not allocated.
Not even
delete[] neco;
Or any other delete. Everyone of them deletes memory BEFORE the allocated block.
Googling and consulting with peers in the office had no results. :)
How do I allocate dynamic array of "points" (two coordinates) and then free them?
I am able to solve the problem via using different structure.
But apperently I am able to allocate the memory. Write in the allocated memory…
How do I correctly free this memory? (Just accademic question.)
Plain pointers to dynamically allocated memory get you into troubles.
Use a std::unique_ptr:
#include <memory>
// ...
std::unique_ptr<int[][2]> neco(new int[ahoj+2][2]);
And remove that loop with delete[].
Alternatively, use std::vector<int[2]> neco(ahoj+2) - it does memory management for you and can be resized.
The general rule is that every new expression must be matched by a single corresponding delete expression.
Your new expression is the right hand side of the statement
neco = new int[ahoj+2][2];
so the corresponding delete expression is given in the statement
delete [] neco;
If that doesn't work (which is what you claim) it means the problem is in some other code exhibiting undefined behaviour.
Your loop
for (size_t i = 0; i < ahoj + 2; i++)
{
delete[] neco[i];
}
is incorrect because none of the neco[i] are the result of a new expression. So delete [] neco[i] has undefined behaviourin every loop iteration.
Given that your code sample(s) and description are incomplete, I doubt anyone else can give more useful advice.
Also, you are mistaken in your belief that you can't use a standard container. You can.
#include <vector>
int main()
{
std::vector<int[2]> avec(ahoj + 2);
int (*neco)[2] = &avec[0];
for (size_t i = 0; i < ahoj + 2; ++i)
{
for (size_t j = 0; j < 2; ++j)
{
neco[i][j] = iter;
++iter;
}
}
}
The only difference is that avec does the dynamic memory allocation and deallocation for you. neco is still a pointer to an array as required (according to your description) by your API. The two constraints on this approach are;
don't resize theavec after initialising neco (or reinitialise neco whenever avec is resized)
don't use neco after avec ceases to exist (since the behaviour will then be undefined).
Also, none of the arrays are actually static. So your question is mistitled.
If I can recall properly, an array of arrays (being dynamic or static), is basicly a matrix.
So, the academic answer:
The same way you allocate memory (first allocate the memory of the array of arrays, then with a foryou allocate the memory of the arrays inside the array), to free it, you need to delete in a forthe memory of the arrays, and then, in a simple delete, free the memory of the dynamic array.
So:
arrayOfArrays[[array1][array2][array3][array4]]
means: allocate memory of arrayOfArrays, then in a loop, allocate memory of array(number)
Same goes the other way around.
Free memory of arrayOfArraysin a single sentence.

Using functions from an objet inside an array of pointers

How would it be possible to use tabEmployes[0].Function ???
CEmploye **tabEmployes[NB_EMPLOYES] = new CEmploye[NB_EMPLOYES];
int main()
{
for (int i = 0; i < NB_EMPLOYES - 1; i++)
{
int j = 0;
string sNom = *LES_EMPLOYES[i, j];
int iNum = atol(*LES_EMPLOYES[i, j + 1]);
int iNiv = atol(*LES_EMPLOYES[i, j + 2]);
CEmploye* unEmploye = new CEmploye(sNom, iNum, iNiv);
tabEmployes[i] = &unEmploye;
}
tabEmployees is an array of pointers to pointer. This means that tabEmployees[0] is a pointer to a pointer. To access the class methods you would have to dereference the pointer. You could use
(*(tabEmployees[0]))->methodName();
However, your current code points tabEmployees[0] to the memory address of a local variable, which will go out of scope at the end of the loop. This is undefined behavior and will most likely lead to undesired results (such as a crash, or all your pointers being the same).
I think you don't really want tabEmployees to be an array of pointers to pointers, especially as your code shouldn't compile right now.
This line:
CEmploye **tabEmployes[NB_EMPLOYES] = new CEmploye[NB_EMPLOYES];
should not compile, as you are assigning an array of pointers to an array of pointers to pointers.
As you are creating your pointers during the loop, it seems to me you don't need to assign any value to this array. If you change it to:
CEmploye **tabEmployes[NB_EMPLOYES];
Your code will now compile.
However, as I said above, you still have the problem that you are pointing to a local variable that will be going out of scope. I would suggest that your tabEmployees should be defined as:
CEmploye *tabEmployes[NB_EMPLOYES];
and assigned later as
tabEmployes[i] = unEmploye;
(note the removed &).
Now your array contains pointers to the newly allocated objects, rather than to the variable that pointed to that object. You would then access methods as follows
tabEmployes[i]->methodName();

Delete a pointer array without deleting the pointed objects in memory?

I would like to know if there is a way to delete a pointer array without touching the pointed objects in memory.
I'm writing a restriction routine for a HashSet I implemented a couple of days ago, so when the hash table is full it gets replaced by another double sized table. I'm representing the hash table using an array of pointers to an object (User), and the array itself is declared dynamically in my HashSet class, so it can be deleted after copying all its content to the new table using a hash function.
So basically I need to:
Declare another table with a size that equals the double of the original array size.
Copy every pointer to User objects from my original array to the new one applying my hash function (it gets the User object from memory and it calculates the index using a string that represents the user's name).
After inserting all the pointers from the original array to the new one, I will have to free the allocated memory for the original array and replace the pointer in my HashSet class (member private userContainer) with the location of the new one (array).
The problem is that if I use delete[] userContainer to free the allocated memory for it, it will also delete every object in memory so the newly created replacement array will point to freed positions in memory!
What you describe does not sound right.
Let's say you have a class A and you create an array of As with:
A** array1 = new A*[32];
Then fill it:
for(int i = 0; i < 32; ++i)
array1[i] = new A();
Doing a delete[] array1 does not free the elements of array1.
So this is safe:
A** array1 = new A*[32];
for(int i = 0; i < 32; ++i)
array1[i] = new A();
A** arary2 = new A*[64];
for(i = 0; i < 32; ++i)
array2[i] = array1[i];
delete [] array1;
for(i = 0; i < 32; ++i)
// do something with array2[i]
In general, when you delete an array of pointers, whatever objects the pointers pointed to remain in existence. In fact, this is a potential source of large memory leaks.
But in some sort of reference-counted environment (eg, Objective-C or Qt), when you delete an array OBJECT (vs a simple [] array) then the reference counts are decremented and the objects will be deleted if the count goes to zero.
But if you're restructuring a hash table you'd better have somehow saved the pointer values before you delete the array, or else all the addressed objects will be lost. As you save them you can increment their reference counts (if you do it right).
(It would help to know what language you're dealing with, and what you mean by "array".)
I don't think your problem exists. Here's a baby example to show that there's nothing to worry about:
Foo * brr[10];
{
Foo * arr[10];
// This is not touching the objects!
for (Foo * it = arr; it != arr + 10; ++it) *it = new Foo;
std::copy(arr, arr + 10, brr);
} // no more arr
for (Foo * it = brr; it != brr + 10; ++it) delete *it; // fine
You can copy the pointers around freely as much as you like. Just remember to delete the object to which the pointers point when they're no longer needed.
A perhaps trivial reminder: Pointers don't have destructors; in particular, when a pointer goes out of scope, nothing happens.
Do you know the difference between malloc/free, new/delete and new[]/delete[]?
I figure that you might want to not use new[]/delete[] in your situation, as you don't want destructors to be called I guess?

Filling an array of pointers, deleting when exiting

In C++, Lets say I'm creating an array of pointers and each element should point to a data type MyType. I want to fill this array in a function fillArPtr(MyType *arPtr[]). Lets also say I can create MyType objects with a function createObject(int x). It works the following way:
MyType *arptr[10]; // Before there was a mistake, it was written: "int *arptr[10]"
void fillArPtr(MyType *arptr[])
{
for (int i = 0; i < 10; i++)
{
MyType myObject = createObject(i);
arptr[i] = new MyType(myobject);
}
}
Is it the best way to do it? In this program how should I use delete to delete objects created by "new" (or should I use delete at all?)
Since you asked "What is the best way", let me go out on a limb here and suggest a more C++-like alternative. Since your createObject is already returning objects by value, the following should work:
#include <vector>
std::vector<MyType> fillArray()
{
std::vector<MyType> res;
for (size_t i = 0; i != 10; ++i)
res.push_back(createObject(i));
return res;
}
Now you don't need to do any memory management at all, as allocation and clean-up is done by the vector class. Use it like this:
std::vector<MyType> myArr = fillArray();
someOtherFunction(myArr[2]); // etc.
someLegacyFunction(&myArr[4]); // suppose it's "void someLegacyFunction(MyType*)"
Do say if you have a genuine requirement for manual memory management and for pointers, though, but preferably with a usage example.
Your method places the array of pointers on the stack, which is fine. Just thought I'd point out that it's also possible to store your array of pointers on the heap like so. Youd do this if you want your array to persist beyond the current scope
MyType **arptr = new MyType[10];
void fillArPtr(MyType *arptr[])
{
for (int i = 0; i < 10; i++)
{
MyType myObject = createObject(i);
arptr[i] = new MyType(myobject);
}
}
If you do this, don't forget to delete the array itself from the heap
for ( int i = 0 ; i < 10 ; i++ ) {
delete arptr[i];
}
delete [] arptr;
If you're going to use vector, and you know the size of the array beforehand, you should pre-size the array. You'll get much better performance.
vector<MyType*> arr(10);
for (int i = 0; i < 10; i++)
{
delete arptr[i];
arptr[i] = 0;
}
I suggest you look into boost shared_ptr (also in TR1 library)
Much better already:
std::vector<MyType*> vec;
for (int i=0; i<10; i++)
vec.push_back(new MyType(createObject(i));
// do stuff
// cleanup:
while (!vec.empty())
{
delete (vec.back());
vec.pop_back();
}
Shooting for the stars:
typedef boost::shared_ptr<MyType> ptr_t;
std::vector<ptr_t> vec;
for (int i=0; i<10; i++)
vec.push_back(ptr_t(new MyType(createObject(i)));
You would basically go through each element of the array and call delete on it, then set the element to 0 or null.
for (int i = 0; i < 10; i++)
{
delete arptr[i];
arptr[i] = 0;
}
Another way to do this is with an std::vector.
Use an array of auto_ptrs if you don't have to return the array anywhere. As long as you don't make copies of the auto_ptrs, they won't change ownership and they will deallocate their resources upon exiting of the function since its RAII based. It's also part of the standard already, so don't need boost to use it :) They're not useful in most places but this sounds like a good one.
You can delete the allocated objects using delete objPtr. In your case,
for (int i = 0; i < 10; i++)
{
delete arptr[i];
arptr[i] = 0;
}
The rule of thumb to remember is, if you allocate an object using new, you should delete it. If you allocate an array of objects using new[N], then you must delete[] it.
Instead of sticking pointers into a raw array, have a look at std::array or std::vector. If you also use a smart pointer, like std::unique_ptr to hold the objects within an std::array you don't need to worry about deleting them.
typedef std::array<std::unique_ptr<MyType>, 10> MyTypeArray;
MyTypeArray arptr;
for( MyTypeArray::iterator it = arptr.begin(), int i = 0; it != arptr.end(); ++it ) {
it->reset( new MyType( createObject(i++) ) );
}
You don't need to worry about deleting those when you're done using them.
Is the createObject(int x) function using new to create objects and returning a pointer to this?. In that case, you need to delete that as well because in this statement
new MyType( createObject(i++) )
you're making a copy of the object returned by createObject, but the original is then leaked. If you change createObject also to return an std::unique_ptr<MyType> instead of a raw pointer, you can prevent the leak.
If createObject is creating objects on the stack and returning them by value, the above should work correctly.
If createObject is not using new to create objects, but is creating them on the stack and returning pointers to these, your program is not going to work as you want it to, because the stack object will be destroyed when createObject exits.