Pointer to an array of pointers

How to declare simply:
- pointer array? A pointer to a pointer does not mean anything. It is always a pointer to something (an object )
- pointer to array of pointers?
- table of pointers to arrays?
- Difference between pointer to an array and array of pointers?

Use pointer to pointer integers :

Code:

---

#include <iostream.h>

int main()
{
// int * (arr *) [10]; this is not correct
int **aa; // use this trick
int k;
aa=new int*[3];
for(j=0;j<3;j++)
aa[j]=new int[4];

return 0;
}

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Pointer to an array of pointers :

Code:

---

# include <stdio.h>

 # define N 3
 int main (void)
 (
  / * Array of pointers to char * /
  char * t [N] = ( "RAM", "the", "C");

  / * P is a pointer to an array of pointers to char and
 initialized at t above * /
  char * (* p) [N] = & t;

 
  int i;

  for (i = 0; i <N; i + +)
    printf ( "% s", (* p) [i]);
  printf ( "\ n");

  return 0;
 )

---

Pointer to an array :

Code:

---

# include <stdio.h>

 # define N 3
 int main (void)
 (
 int t [2] [N] = ((14, 19, 15), (21, 11, 2009));
 int (* p) [N] = & t [1];

 printf ( "% d \ n", (* p)  [2]);

 return 0;
 )

---

Array of pointers to arrays :

Code:

---

# include <stdio.h>

 # define N 3
 int main (void)
 (
  int t0 [5] = (55, 33, 75, 21, 29);
  int t1 [5] = (8, 23, 45, 54, 69);

  / * P is an array of 2 pointers to tables 5 int * /
  int (* p [2]) [5];

  / * Test * /
  p [0] = & t0;
  p [1] = & t1;

  printf ( "% d% d \ n", (* p [0]) [3], (* p [1]) [4]);

  return 0;
 )

---