Menu
In simple 1D array:
node *nodes = new node[MAX_NODES];
Deleting by:
delete [] nodes;
Deletes all the nodes allocated in the array.
But in this case:
float (*buildingArray)[3] = new float[10][3];
Does this statement make buildingArray a single dimension array of 3 float pointers? And this is the deallocation line:
delete[] buildingArray;
Does the above deallocation delete the array, but I am doubtful about whether it will delete its references?
456
In C, if you are declaring your array as static ex: int a[10], no need to delete. It will be freed when your function ends or program terminates.
To delete a dynamic array, the delete or delete[] operator is used. It deallocates the memory from heap. The delete[] keyword deletes the array pointed by the given pointer. Therefore, to delete a dynamically allocated array, we use the delete[] operator.
Here i and j are the size of the two dimensions, i.e i denotes the number of rows while j denotes the number of columns. Example: int A[10][20]; Here we declare a two-dimensional array in C, named A which has 10 rows and 20 columns.
Delete is an operator that is used to destroy array and non-array(pointer) objects which are created by new expression.
Does the above de-allocation delete the array?
Yes it does.
Simply follow the rule:
You need to call delete or delete [] as many times you called new or new [] respectively.
If you had an array of pointers where each index was allocated dynamic memory, you would need to explicitly loop through it and deallocate each array element explicitly.
On a side note you are much better off using a std::vector or std::array rather than dynamically allocated array.
97
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
