Question

I’m having some trouble understanding the difference between these two code segments:
I allocate space for an array of integers dynamically within my code with the following statement

int *arr = calloc(cnt, sizeof(int));

In another function, where I pass in arr, I would like to determine the size (number of elements) in arr.
When I call

int arr_sz = sizeof(arr)/sizeof(int);

it only returns 1, which is just the number of bytes in an int for both arguments I am assuming (4/4)=1.
I just assumed it would be the same as using an array

  int arr[8];
  int arr_sz = sizeof(arr)/sizeof(int);

which returns the actual number of elements in the array.

If anyone could clear this up, that would be great. Thanks!

Answer 1

int *arr;   ----> Pointer
int arr[8]; ----> Array

First up what you got there – int *arr is a pointer, pointing to some bytes of memory location, not an array.

The type of an Array and a Pointer is not the same.

In another function where I pass in arr, I would like to determine the size (number elements) in arr. When I call

int arr_sz = sizeof(arr)/sizeof(int);

it only returns 1, which is just the number of bytes in an int for both arguments I am assuming (4/4)=1. I just assumed it would be the same as using an array

Even if it is assumed to be an Array — that’s because Arrays get decayed into pointers when passed into functions. You need to explicitly pass the array size in functions as a separate argument.

Go through this:

Sizeof an array in the C programming language?

There is a difference between a static array and dynamic memory allocation.

The sizeof operator will not work on dynamic allocations.
AFAIK it works best with stack-based and predefined types.