C pointer notation compared to array notation: When passing to function

Question

My question is base on the following code:

int myfunct(int ary[], int arysize)   
int myfunct2(int *ary, int arysize)

 int main(void){
   int numary[10];
   myfunct(numary, 10)
   myfunct2(numary, 10)
   return;
 }

int myfunct(int ary[], int arysize) {   
      //Whatever work is done

  }

int myfunct2(int *ary, int arysize) {
     // Whatever work is done

  }

Is there a reason to use one of these over the other? To elaborate, when concerned with numeric arrays, is there any reason one would want to use pointer notation over array notation. If one uses pointer notation then within the function pointer arithmetic would be used etc.. AND if one uses the [] array notation, one could work with the array as usual. I'm new to programming and I currently do not see any benefit to using the pointer notation.

My precise question, is there any reason to pass a numeric array to a function using pointer notation and therefore using pointer manipulations within the function.

Answer 1

When you declare a function parameter as an array, the compiler automatically ignores the array size (if any) and converts it to a pointer. That is, this declaration:

int foo(char p[123]);

is 100% equivalent to:

int foo(char *p);

In fact, this isn't about notation but about the actual type:

typedef char array_t[42];
int foo(array_t p);  // still the same function

This has nothing to do with how you access p within the function. Furthermore, the [] operator is not "array notation". [] is a pointer operator:

a[b]

is 100% equivalent to:

*(a + b)