array size stays the same even after memory allocation

Question

i know this may be silly but i do not understand why after i use malloc to allocate memory for cars, cars size is 8. it does not matter what size i enter to carsAmount, please help. when i checked the array size before allocating it- it was 8. i tried to free it but got an error. i tried using realloc-didnt work.

this is my code snippet:

int* cars = NULL;
int carsAmount;

printf("Enter cars amount:\n");
scanf("%d",&carsAmount);
cars = realloc(cars, (carsAmount)*sizeof(int));

Answer 1

From realloc man page

void *realloc(void *ptr, size_t size);

If ptr is NULL, then the call is equivalent to malloc(size)

Which is equivalent to

cars = malloc(carsAmount*sizeof(int))

The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL.

To check if allocation succeeded, simply check if

if (cars != NULL)
{
    // allocation ok 
} else {
   // Allocation failed
}

---

sizeof(int*) / sizeof(int)
    |            |
    V            V
sizeof(cars)/sizeof(int)

Is actually size of pointer which is probably 8 and size of int is probably 4. This is why you get 2.

Answer 2

It's not clear from the question how you are trying to get the array size, but I am guessing you are using sizeof(cars) and always getting 8. That's because sizeof(cars) will give you the size of the cars pointer, and all pointers in a 64-bit architecture are 8 bytes long.

Malloc and realloc keep the size of the buffer internally but you as the user are supposed to keep track of it.