Imagine some array
uint8_t var[5] = {1,2,3,4,5};
so var will be pointer to the first element of this array, and
uint 8_t* a=var;
b=a[3]
and
b=var[3]
will give the same result.
But will
a = &var[2];
b = a[1];
and
b=var[3];
be same?
After this assignment
a = &var[2];
that is the same as
a = var + 2;
due to the implicit conversion of the array designator to a pointer to its first element the pointer a points to the element var[2].
So a[0] yields var[2] and a[1] yields var[3].
Pay attention to that the subscript operator a[i] is evaluated like *( a + i ).
So you have a[1] is equivalent to *( a + 1 ) that is in turn equivalent to *( var + 2 + 1 ) that is to *( var + 3 ).