How to deal with a bidimesional array passed by pointer? C

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How to deal with a bidimesional array passed by pointer? C

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#1 by (3 votes)

3

As the title says, I can treat a two-dimensional array passed by pointer, an example to clarify my idea:

main()
{
char array[10][10];
llenar_array(array);
}

void llenar_array(char *array[])
{
 char registro[10] = "Cadena";
 //Sin punteros sería algo así array[0] = registro;
 *array[0] = registro;
}

What I want to do is pass a two-dimensional char array to a function but treat it with pointers. Any answer about the management of two-dimensional arrays using pointers will be appreciated. Thanks.

c

asked by Juanjoo Tocino 29.08.2017 в 03:41

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1 answer

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score 3 · Answer 1

This represents an array on the stack or stack :

char array[3][3];

Its fundamental characteristic is that its positions are contiguous in memory:

0x100 | 0x101 | 0x102 | 0x103 | 0x104 | 0x105 | 0x106 | 0x107 | 0x108 | 0x109
    0 |     1 |     2 |     3 |     4 |     5 |     6 |     7 |     8 |     9

However this is an array of pointers:

char *array[];

The main characteristic of this type is that the first thing you find in memory is a pointer that redirects you to the position where each row of data is:

0x100 | 0x101 | 0x102 | ... | 0x160 | 0x161 | 0x162 | ... | 0x240 | 0x241 | 0x242 | ...
0x160 | 0x240 | 0x300 | ... |     1 |     2 |     3 | ... |     4 |     5 |     6 | ...

So no, those statements are not equivalent. What you could do, if you need it, is to create an array of pointers that encapsulate the stack array, but it's a process that you have to program by hand:

#include <stdio.h>

void func(int* array[])
{
  for( int i=0; i<3; i++ )
    for( int j=0; j<3; j++ )
      array[i][j] = i*3 + j + 1;
}

int main()
{
  int array[3][3];
  int* wrapper[3];
  for( int i=0; i<3; i++ )
    wrapper[i] = array[i];

  func(wrapper);

  for( int i=0; i<3; i++ )
  {
    for( int j=0; j<3; j++ )
      printf("%d ",array[i][j]);
    printf("\n");
  }
}

Another thing you can do is encapsulate the creation of the array of pointers with a macro. You can get more elaborate examples but here is an example:

#define PREPARE_ARRAY(dest,source,SIZE1) \
{ \
  for( int i=0; i<SIZE1; i++ ) \
    dest[i] = source[i]; \
}

int main()
{
  int array[3][3];
  int* wrapper[3];

  // SIZE1 no puede ser inferido por el compilador... mala suerte        
  PREPARE_ARRAY(wrapper,array,3);

  func(wrapper);

  for( int i=0; i<3; i++ )
  {
    for( int j=0; j<3; j++ )
      printf("%d ",array[i][j]);
    printf("\n");
  }
}