Why does a+1 == *(a+1) in this example?

Question

#include <iostream>

int main()
{
    int a[3][3] = {{22, 33, 44}, {55, 66, 77}, {88, 99, 100}};
    std::cout << a[1] << '\n' << a + 1 << '\n' << *(a + 1);
}

0x0013FF68
0x0013FF68
0x0013FF68

Why does a+1 == *(a+1)?

Answer 1

a + 1 is the address of the second element in a and could also be written as &a[1] (which is equivalent to &*(a + 1) by definition).

*(a + 1) is an lvalue referring to the second array. It's equivalent to a[1] by definition.
Just like with any other array to pointer decay, this lvalue decays to a pointer to the first element of the array it refers to, i.e. it decays to &a[1][0]. But that is equivalent to the address of that array object itself. So the value is the same as that of &a[1] ... which is precisely how we defined the value of the expression a + 1 above.

Note that the array is decayed to a pointer because the best match for the second insertion is operator<<(void const*). Consider

int (*p1)[3] = a + 1;

int (&p2)[3] = *(a + 1); // We could also have written *p1

int* p3 = p2; // The array-to-pointer decay

assert( static_cast<void*>(p1) == static_cast<void*>(p3) );