Understanding (int *)&i and int increment(int * &i)

Q1:
#include "stdio.h"
void main()
{
int i[] = {2, 3};
int *j = i;
int **k = &j; // Double pointer, correct;
int *p = NULL;

// k = &i; // 'i' represents the array address, and '&i' also seems like a double pointer—why does this fail?
// Error: cannot convert from 'int ()[2]' to 'int **'
// Isn't 'int ()[2]' an array pointer? For example, int (*c)[2]; c = arry[3][2]. Here, 'c' is also used like a double pointer~

// i = (int *)&i; // Error: cannot convert from 'int *' to 'int [2]'
// k = (int *)&i; // Error: cannot convert from 'int *' to 'int **'
// k = (int *)i; // Error: cannot convert from 'int *' to 'int **'

p = (int *)i;
p = (int *)&i;
// Both of these casts work. What's the actual difference between &i and i? Is & meaningless here?
}

A1:
i = (int *)&i; // Error: cannot convert from 'int *' to 'int [2]'

This line is definitely wrong because 'i', as an array name, is a pointer—but a constant pointer—and cannot be assigned to.
k = (int *)&i; // Error: cannot convert from 'int *' to 'int **'

This is also wrong. &i is already of type int**, so adding an explicit (int**) cast does more harm than good.
k = (int *)i; // Error: cannot convert from 'int *' to 'int **'

This is incorrect as well. 'k' is of type int**, and you cannot convert directly from int*.

Q2:
**Define a function, for example: int increment(int * & i)
{
i++;
return 0;
}

How should the parameter: int * & i be understood?**  

A2:

1. int increment(int * & i)
   {
   i++;
   return 0;
   }

'increment' is a function. It's essentially the same as int increment(int * i); the '&' just indicates that the parameter is passed by reference.
2) &i is simply an alias for i; it denotes a reference to an integer pointer;

• for pointer, & for reference
  int * & i
  means a reference to a pointer to an integer