Two Dimensional Array Implementation Using Double

问题:

Please consider the following code:

#include <stdio.h>
#include <stdlib.h>

#define NUM_ARRAYS     4
#define NUM_ELEMENTS   4
#define INVALID_VAL   -1

int main()
{
   int index            = INVALID_VAL;
   int array_index      = INVALID_VAL;
   int **ptr            = NULL;

   ptr = malloc(sizeof(int*)*NUM_ARRAYS);

   if (!ptr)
   {
      printf ("\nMemory Allocation Failure !\n\n");
      exit (EXIT_FAILURE);
   }

   for (index=0; index<NUM_ARRAYS; index++)
   {
      *(ptr+index) = malloc(sizeof(int)*NUM_ELEMENTS); 

      if (!*(ptr+index))
      {
         printf ("\nMemory Allocation Failure !\n");
         exit (EXIT_FAILURE);
      }
   }

   /* Fill Elements Into This 2-D Array */
   for (index=0; index<NUM_ARRAYS; index++)
   {
      for (array_index = 0; array_index<NUM_ELEMENTS; array_index++)
      {
         *(*(ptr+index)+array_index) = (array_index+1)*(index+1);
      }
   }

   /* Print Array Elements */
   for (index = 0; index<NUM_ARRAYS; index++)
   {
      printf ("\nArray %d Elements:\n", index);
      for (array_index = 0; array_index<NUM_ELEMENTS; array_index++)
      {
         printf (" %d ", *(*(ptr+index)+array_index));
      }
      printf ("\n\n");
   }

   return 0;
}

There is no problem with my code. It works fine.

Output:

Array 0 Elements:
 1  2  3  4 


Array 1 Elements:
 2  4  6  8 


Array 2 Elements:
 3  6  9  12 


Array 3 Elements:
 4  8  12  16 

I have a question about pointer arithmetic:

*(ptr+0) = Pointer to COMPLETE BLOCK (First Array)
*(ptr+1) = Pointer to COMPLETE BLOCK (Second Array).

But what is: (*ptr+1)?

GDB Output:

(gdb) p *(*ptr+1)
$1 = 2
(gdb) p *(*ptr+2)
$2 = 3
(gdb) p *(*ptr+3)
$3 = 4
(gdb) p *(*ptr+4)
$4 = 0

I am getting confused on this. Please provide me some explanation to resolve this doubt.

回答1:

*(ptr+i) is equals to ptr[i] and *(ptr+1) is ptr[1].

You can think, a 2-D array as array of array.

• ptr points to complete 2-D array, so ptr+1 points to next 2-D array.

---

In figure below ptr is 2-D and number of columns are 3

Original figure made by Mr. Kerrek SB, here , you should also check!

+===============================+==============================+====
|+---------+----------+--------+|+----------+---------+--------+|
||ptr[0,0] | ptr[0,1] | ptr[0,2]|||ptr[1,0] |ptr[1,1] | ptr[1,2]|| ...
|+---------+----------+--------+++----------+---------+--------++ ...
|            ptr[0]             |           ptr[1]              |
+===============================+===============================+====
   ptr

*(*ptr+1) = *( ptr[0] + 1 ) = ptr[0][1]

Understand following:

ptr points to complete 2-D.

*ptr = *(ptr + 0) = ptr[0] that is first row.

*ptr + 1 = ptr[1] means second row

*(*ptr+1) = *(*(ptr + 0) + 1 ) = *(ptr[0] + 1) = ptr[0][1]

Array 0 Elements:
1  2  3  4 

And GDB Output:

(gdb) p *(*ptr+1)
$1 = 2  

that is correct 2 this can be read using ptr[0][1].

回答2:

                               (*ptr)   (*ptr+1)     (*ptr+2)
                                 |         |            |
             __________      ____v___ _____v_____ ______v____ __________
  ptr------>|   *ptr   |--->|*(*ptr) |*(*ptr+1)  |*(*ptr+2)  |          |
            |__________|    |________|___________|___________|__________|
 (ptr+1)--->| *(ptr+1) |     ____________ _____________ __________
            |__________|--->|*(*(ptr+1)) |*(*(ptr+1)+1)|          |
            |          |    |____________|_____________|__________|
            |__________|          ^             ^
                                  |             |
                              *(ptr+1)     *(ptr+1)+1

2D array with double pointers that means that you have a main array and the elements of the main array are pointers (or addresses) to a sub arrays. As indicated in above figure

so if you have defined a double pointer as a pointer of this 2D array let's say int **ptr

so ptr is ponting to the main array which will contains pointers to sub arrays. ptr is ponting to the main array that's means ptr is pointing to the first element of the main array so ptr + 1 is pointing to the second element of the main array.

*ptr this means the content of the first element which the ptr is pointing on. And it is a pointer to a subarray. so *ptr is a pointer to the first subarray (the subarray is an array of int). so *ptr is pointing to the first element in the first subarray. so *ptr + 1 is a pointer to the second element in the first subarray

回答3:

Simplest way for creating 2-dimensinal array using pointer,assigning values and accessing elements from the array.

#include<stdio.h>
#include<stdlib.h>

int main()
{
int i,j;
int row,col;
printf("Enter the values for row and col:\n");
scanf("%d%d",&row,&col);
int **arr=(int**)malloc(row*(sizeof(int*)));
for(i=0;i<row;i++)
{
    *(arr+i)=(int*)malloc(sizeof(int)*col);
            //You can use this also. Meaning of both is same.
            //arr[i]=(int*)malloc(sizeof(int)*col);
}
for(i=0;i<row;i++)
for(j=0;j<col;j++)
{
    arr[i][j]=0;
}
for(i=0;i<row;i++)
{
    for(j=0;j<col;j++)
    {
        printf("%d ",arr[i][j]);
    }
    printf("\n");
}
}

回答4:

Unless you mistypes, (*ptr + 1) is equivalent to *(ptr + 0) + 1 which is a pointer to the second element in the first block.