Allocating memory for array of pointers of int dynamically

Question

If I want to dynamically allocate memory for array of pointers of int, how can I achieve this requirement?

Suppose I declare an array of pointers of int like this:

int (* mat)[];

Is there a way I can allocate memory for K number of pointers dynamically and assign it to mat? If I do

mat = new int * [K];

It gives error : cannot convert 'int**' to 'int (*)[]' in assignment. I understand this memory allocation is implicitly got converted to int **. Is there any way to allocate memory for above scenario?

Even when I try to assignment of statically allocated array of pointers of int to array of pointers of int, like this:

int (*mat)[] = NULL;
int (* array_pointers)[26];
mat = array_pointers;

Compilation gives this error: cannot convert 'int (*)[26]' to 'int (*)[]' in assignment.

Can someone please explain to me why this is an error or why it should be an error?

is this C or C++?

AShelly
– AShelly

2016-10-25 05:05:41 +00:00
Commented Oct 25, 2016 at 5:05 — AShelly
– AShelly, Commented Oct 25, 2016 at 5:05
If C++, you may try to use std::vector<int*>

Mine
– Mine

2016-10-25 05:11:18 +00:00
Commented Oct 25, 2016 at 5:11 — Mine
– Mine, Commented Oct 25, 2016 at 5:11
In C you can simplify use int ** mat; for array of pointers

ObliteratedJillo
– ObliteratedJillo

2016-10-25 05:15:34 +00:00
Commented Oct 25, 2016 at 5:15 — ObliteratedJillo
– ObliteratedJillo, Commented Oct 25, 2016 at 5:15

M.M · Accepted Answer · 2016-10-25 05:10:11Z

In C, int (* mat)[]; is a pointer to array of int with unspecified size (not an array of pointers). In C++ it is an error, the dimension cannot be omitted in C++.

Your question says new int *[K] so I assume this is really a C++ question. The expression new T[n] evaluates to a T * already, there is no implicit conversion.

The code to allocate an array of null pointers using new is:

int **mat = new int *[10]();

Then mat points to the first one of those. Another option (less commonly used) is:

int * (*mat)[10] = new int *[1][10]();

where *mat designates the entire array of 10 pointers.

NB. This sort of code is not useful for anything except demonstration purposes perhaps, whatever you are trying to do has a better solution.

Ivan Rubinson · Accepted Answer · 2016-10-25 05:32:53Z

In C and C++, in some cases, arrays and pointers are interchabgable. int*[] decays in to int**.

To dynamically allocate an array, use:

int* mat;
// ...
mat = new int[size];
// ...
delete[] mat;

Note that you need to delete all memory you take with new, or else you'll leak memory. It's not always easy to do.

C++ has a thing called RAII to help. To use RAII (meaning all resources taken in a scope will be released when the scope ends), put the new in a classes constructor, and delete in the destructor.

The STL has some comtainers ready for you to use. Look in to std::vector.

If you want an array of pointers as opposed to a dynamic array, the syntax is more like

int** arr = new int[size];
arr[i] = &val;
// ...
delete[] arr;

Don't forget to initialize each pointer in the array!

If size is known at compile time, this can be rewritten as:

std::array<int*, size> arr;
arr[i] = &val;

No need to delete, and it checks overflows.

If size is unknown at compile time, you can do:

std::vector<int*> arr;
arr.resize(size);
arr[i] = &val;

This allocates an array of int. OP asked for an array of pointers to int.
Your second suggestion doesn't compile . The result of new is a prvalue and & cannot be applied to it.
ideone suppresses error messages sometimes. The code in your link is OK but different to what you have in your answer. Also, it can only be rewritten with std::array if size is known at compile-time (unclear from OP's post whether that is true or not)

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