Difference between passing array and array pointer into function in C

First, some standardese:

6.7.5.3 Function declarators (including prototypes)
...
7 A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. If the keyword static also appears within the [ and ] of the array type derivation, then for each call to the function, the value of the corresponding actual argument shall provide access to the first element of an array with at least as many elements as specified by the size expression.

So, in short, any function parameter declared as T a[] or T a[N] is treated as though it were declared T *a.

So, why are array parameters treated as though they were declared as pointers? Here's why:

6.3.2.1 Lvalues, arrays, and function designators
...
3 Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue. If the array object has register storage class, the behavior is undefined.

Given the following code:

int main(void)
{
  int arr[10];
  foo(arr);
  ...
}

In the call to foo, the array expression arr isn't an operand of either sizeof or &, so its type is implicitly converted from "10-element array of int" to "pointer to int" according to 6.2.3.1/3. Thus, foo will recei