Thread: function reveives a 2-d array as parameter

my problem solved, thanks to all.
according to matsp and Salem's method, i write two different version codes,
(they are 3d arrays)and they work fine for me , post it, hoping helpful
in the first version , i've a question about free();
version1:

Code:

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

void fun(int ***arr);
int ***send(void);

int main(void)
{ 
    int ***ptr;
    ptr = send();   
    fun(ptr);
    getchar();
return 0;
}

 void fun(int ***arr)
 {
   int i;
   for(i=0; i<10; i++)
   {
     printf("%d\n", arr[0][0][i]); // print just 10 elements
   }
  free(arr); // should i use free() 3 times as i used 3malloc for arr?
                  //  or 1 free() is enough? }
  
int ***send(void)
{
    int i;
    int j;
    int k;
    int ***twodarray;
    
    for(i=0; i<10; i++)
      twodarray = (int***)malloc(10*sizeof(int));
   
    for(i=0; i<10; i++)
      twodarray[i] = (int**)malloc(10*sizeof(int));
    
    for(i=0; i<10; i++)
      for(j=0; j<10; j++)
          twodarray[i][j] = (int*)malloc(10*sizeof(int));
          
    for(i=0; i<10; i++)
      for(j=0; j<10; j++)
        for(k=0; k<10; k++)
          twodarray[i][j][k] = 0;
          
    return twodarray;  
}

version2:

Code:

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

void fun(int (*arr)[10][10]);
int (*send(void))[10][10];

int main(void)
{
    int (*ptr)[10][10]; //"this is a pointer to an array of ten elements"
    ptr = send();       // ptr[10][10] "10 pointers to integers"
    fun(ptr);
    getchar();
return 0;
}

void fun(int (*arr)[10][10])
 {
   int i;
   for(i=0; i<10; i++)
   {
     printf("%d\n", arr[0][0][i]); // print just 10 elements
   }
 }
  
int (*send(void))[10][10]
{
    int static twodarray[10][10][10] = {0};
    
    return twodarray;  
}

I made a little change...thanks

Code:

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

void fun(int ***arr);
int ***send(void);

int main(void)
{ 
    int ***ptr;
    ptr = send();   
    fun(ptr);
    getchar();
return 0;
}

 void fun(int ***arr)
 {
   int i;
   int j;
   int k;

   for(i=0; i<10; i++)
     printf("&#37;d\n", arr[0][0][i]); // print just 10 elements
   
   for(i=0; i<10; i++)
      for(j=0; j<10; j++)
          free(arr[i][j]);

   for(i=0; i<10; i++)
      free(arr[i]); 

   free(arr);
 }
  
int ***send(void)
{
    int i;
    int j;
    int k;
    int ***twodarray;
    
    twodarray = (int***)malloc(10*sizeof(***twodarray)); // i cast malloc cause i edit the file in c++ enviorment.
   
    for(i=0; i<10; i++)
      twodarray[i] = (int**)malloc(10*sizeof(**twodarray));
    
    for(i=0; i<10; i++)
      for(j=0; j<10; j++)
          twodarray[i][j] = (int*)malloc(10*sizeof(*twodarray));
          
    for(i=0; i<10; i++)
      for(j=0; j<10; j++)
        for(k=0; k<10; k++)
          twodarray[i][j][k] = 0;
          
    return twodarray;  
}

> In general, you should use this form
> p = malloc ( numRequired * sizeof *p );

In case numRequired is a general expression, it's a little safer to put the sizeof on the left, like this:

Code:

p = malloc(sizeof(*p) * numRequired);

since the argument to malloc() should be calculated as type size_t and this is more likely to happen without explicit casts if sizeof(*p) is on the left. For example, on a 64-bit platform where size_t is 64 bits and int is 32 bits,

Code:

p = malloc(sizeof(*p) * 0x10000U * 0x10000U);

works properly, but

Code:

p = malloc(0x10000U * 0x10000U * sizeof(*p));

doesn't. It doesn't always eliminate the need for a cast, though, since for example if numRequired was a sum instead of a product, one would have to write something like

Code:

p = malloc(sizeof(*p) * ((size_t)num1 + num2));

but it eliminates some of them. This will matter more in 5 or 10 years when everyone is on a 64-bit platform and it's commonplace to allocate arrays > 2^32 elements.