What does getting the address of an array variable mean?

Question

Today I read a C snippet which really confused me:

#include 

int
main(void)
{
    int a[] = {0, 1, 2, 3};

    printf(\"%d\\n\", *(*(&a +         


        

Answer 1

Let's dissect it.

a has type int [4] (array of 4 int). It's size is 4 * sizeof(int).

&a has type int (*)[4] (pointer to array of 4 int).

(&a + 1) also has type int (*)[4]. It points to an array of 4 int that starts 1 * sizeof(a) bytes (or 4 * sizeof(int) bytes) after the start of a.

*(&a + 1) is of type int [4] (an array of 4 int). It's storage starts 1 * sizeof(a) bytes (or 4 * sizeof(int) bytes after the start of a.

*(&a + 1) - 1 is of type int * (pointer to int) because the array *(&a + 1) decays to a pointer to its first element in this expression. It will point to an int that starts 1 * sizeof(int) bytes before the start of *(&a + 1). This is the same pointer value as &a[3].

*(*(&a + 1) - 1) is of type int. Because *(&a + 1) - 1 is the same pointer value as &a[3], *(*(&a + 1) - 1) is equivalent to a[3], which has been initialized to 3, so that is the number printed by the printf.