First a little reminder (or something new if you didn't know this before): For any array or pointer p and index i the expression p[i] is exactly the same as *(p + i).
Now to hopefully help you understand what's going on...
The array a in your program is stored somewhere in memory, exactly where doesn't really matter. To get the location of where a is stored, i.e. get a pointer to a, you use the address-of operator & like &a. The important thing to learn here is that a pointer by itself doesn't mean anything special, the important thing is the base type of the pointer. The type of a is int[4], i.e. a is an array of four int elements. The type of the expression &a is a pointer to an array of four int, or int (*)[4]. The parentheses are important, because the type int *[4] is an array of four pointers to int, which is quite a different thing.
Now to get back to the initial point, that p[i] is the same as *(p + i). Instead of p we have &a, so our expression *(&a + 1) is the same as (&a)[1].
Now that explains what *(&a + 1) means and what it does. Now let us think for a while about the memory layout of the array a. In memory it looks something like
+---+---+---+---+
| 0 | 1 | 2 | 3 |
+---+---+---+---+
^
|
&a
The expression (&a)[1] treats &a as it was an array of arrays, which it definitely isn't, and accessing the second element in this array, which will be out of bounds. This of course technically is undefined behavior. Let us run with it for a moment though, and consider how that would look like in memory:
+---+---+---+---+---+---+---+---+
| 0 | 1 | 2 | 3 | . | . | . | . |
+---+---+---+---+---+---+---+---+
^ ^
| |
(&a)[0] (&a)[1]
Now remember that the type of a (which is the same as (&a)[0] and therefore means that (&a)[1] must also be this type) is array of four int. Since arrays naturally decays to pointers to its first element, the expression (&a)[1] is the same as &(&a)[1][0], and its type is pointer to int. So when we use (&a)[1] in an expression what the compiler gives us is a pointer to the first element in the second (non-existing) array of &a. And once again we come to the p[i] equals *(p + i) equation: (&a)[1] is a pointer to int, it's p in the *(p + i) expression, so the full expression is *((&a)[1] - 1), and looking at the memory layout above subtracting one int from the pointer given by (&a)[1] gives us the element before (&a)[1] which is the last element in (&a)[0], i.e. it gives us (&a)[0][3] which is the same as a[3].
So the expression *(*(&a + 1) - 1) is the same as a[3].
It's long-winded, and passes through dangerous territory (what with the out-of-bounds indexing), but due to the power of pointer arithmetic it all works out in the end. I don't recommend you ever write code like this though, it needs people to be really know how these transformations work to be able to decipher it.
Best Answer
The name of an array usually evaluates to the address of the first element of the array, so
arrayand&arrayhave the same value (but different types, soarray+1and&array+1will not be equal if the array is more than 1 element long).There are two exceptions to this: when the array name is an operand of
sizeofor unary&(address-of), the name refers to the array object itself. Thussizeof arraygives you the size in bytes of the entire array, not the size of a pointer.For an array defined as
T array[size], it will have typeT *. When/if you increment it, you get to the next element in the array.&arrayevaluates to the same address, but given the same definition, it creates a pointer of the typeT(*)[size]-- i.e., it's a pointer to an array, not to a single element. If you increment this pointer, it'll add the size of the entire array, not the size of a single element. For example, with code like this:We can expect the second pointer to be 16 greater than the first (because it's an array of 16 char's). Since %p typically converts pointers in hexadecimal, it might look something like: