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Thread: Pointer to an array of pointers

  1. #1
    Join Date
    Aug 2006
    Posts
    173

    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?

  2. #2
    Join Date
    Apr 2008
    Posts
    2,010

    Re: Pointer to an 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;
    }

  3. #3
    Join Date
    May 2008
    Posts
    2,302

    Re: Pointer to an array of pointers

    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;
     )

  4. #4
    Join Date
    May 2008
    Posts
    2,015

    Re: Pointer to an array of pointers

    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;
     )

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