#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;
}

我的代码没有问题.它工作正常.

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

我有一个关于指针算术的问题：

*(ptr 0)=指向COMPLETE BLOCK的指针(第一个数组)
*(ptr 1)=指向COMPLETE BLOCK(第二个数组)的指针.

但是：(* ptr 1)？

GDB输出：

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

我对此感到困惑.请给我一些解释来解决这个疑问.