Is Freeing Memory Used By Dynamic Arrays Required

[Delerna]

Been playing around with DirectX 10 in C++ and I have been using arrays to store info for vertex buffers. To simplify my question I will drop DX10 from the equation. I am an experienced programmer but have only dabbled in C++.

To The Point.
If i declare a variable ready for use as a dynamic array
int* ArrayName;

and initialize it so it has a size of 1
ArrayName=new int [1];


and then in a function in my code I want to expand the array to 2

so I create another array with the new dimensions
int* NewArray;
NewArray= new int[2];

and copy the data from ArrayName to NewArray
(I wont show the code as its not my question)

and then point ArrayName to the NewArray
ArrayName= &NewArray;

My question is this
What happens to the memory that ArrayName was pointing at before I pointed it at NewArray ??
Do I need to release it befor I repoint it??
How ??

[gpraghuram]

memory of ArrayName leaks.
Since you overwrite the memory with new value you dont have a chance to free the memeory of ArrayName variable

Raghuram

[weaknessforcats]

and copy the data from ArrayName to NewArray
(I wont show the code as its not my question)

and then point ArrayName to the NewArray
ArrayName= &NewArray;

Further, this does not copy the NewArray to the ArrayName array. It just copies the address.

Read this for a quick refresher on C++ arrays:

First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:

int array[5];
 

That is an array of 5 elements each of which is an int.

int array[];
 

won't compile. You need to declare the number of elements.

Second, this array:

int array[5][10];
 

is still an array of 5 elements. Each element is an array of 10 int.

int array[5][10][15];
 

is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.

int array[][10];
 

won't compile. You need to declare the number of elements.

Third, the name of an array is the address of element 0

int array[5];
 

Here array is the address of array[0]. Since array[0] is an int, array is the address of an int. You can assign the name array to an int*.

int array[5][10];
 

Here array is the address of array[0]. Since array[0] is an array of 10 int, array is the address of an array of 10 int. You can assign the name array to a pointer to an array of 10 int:

int array[5][10];
 
int (*ptr)[10] = array;
 

Fourth, when the number of elements is not known at compile time, you create the array dynamically:

int* array = new int[value];
int (*ptr)[10] = new int[value][10];
int (*ptr)[10][15] = new int[value][10][15];
 

In each case value is the number of elements. Any other brackets only describe the elements.

Using an int** for an array of arrays is incorrect and produces wrong answers using pointer arithmetic. The compiler knows this so it won't compile this code:

int** ptr = new int[value][10];    //ERROR
 

new returns the address of an array of 10 int and that isn't the same as an int**.

Likewise:

int*** ptr = new int[value][10][15];    //ERROR
 

new returns the address of an array of 10 elements where each element is an array of 15 int and that isn't the same as an int***.

With the above in mind this array:

int array[10] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Wheras this array:

int array[5][2] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Kinda the same, right?

So if your disc file contains

0 1 2 3 4 5 6 7 8 9

Does it make a difference wheher you read into a one-dimensional array or a two-dimensional array? No.

Therefore, when you do your read use the address of array[0][0] and read as though you have a
one-dimensional array and the values will be in the correct locations.

[Delerna]

memory of ArrayName leaks.
Since you overwrite the memory with new value you dont have a chance to free the memeory of ArrayName variable

Raghuram

Thanks Raghuram
I thought it would leak.
So how do I release it before repointing to NewArray

[Delerna]

[quote=weaknessforcats]Further, this does not copy the NewArray to the ArrayName array. It just copies the address.
QUOTE]

Thanks weaknessforcats
I realise that I need to copy the the contents of ArrayName over to NewArray
Thats why I stated that I wouldn't show the code because it wasn't my question :-)

I appreciate your time and effort and although I am familiar with the principles of your refresher it has been presented well, Short and concise.
I will be copying the refresher over to my help documents, hope you don't mind.

Thanks for the comments

[weaknessforcats]

I will be copying the refresher over to my help documents, hope you don't mind.

Not at all. I'm glad you found it useful.