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)