1

I would like to allocate memory in a function and then use that space in the main().
All is fine in the function, but I never can access the data in the main().
There are two memory allocation instructions for simulating a two dimensionnal array of chars

Here's my code:

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

int getNetworks(char **networks) {

  networks = malloc(5 * sizeof(char*));
  printf("networks in function:%p\n", networks);

  networks[0] = (char*) malloc(4);
  strcpy(networks[0], "aaa");
  networks[1] = (char*) malloc(3);
  strcpy(networks[1], "bb");
  networks[2] = (char*) malloc(5);
  strcpy(networks[2], "cccc");
  networks[3] = (char*) malloc(6);
  strcpy(networks[3], "ddddd");
  networks[4] = (char*) malloc(2);
  strcpy(networks[4], "e");
  return 5;
}

int main ()
{
  char **networks = NULL;
  int nbNetworks;

  printf("networks before call:%p\n", networks);
  nbNetworks = getNetworks(&*networks);
  printf("networks after call:%p\n", networks);
  for (int i=0; i<=nbNetworks-1; i++) {
    printf ("%s\n", networks[i]);
  }
  return 0;
}

Output is

networks before call:0x0
networks in function:0x7feb65c02af0
networks after call:0x0
Segmentation fault: 11
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3
  • 1
    That should be *networks = malloc(5 * sizeof(char*)); because you want to modify the parent's variable, not the value parameter on the stack. Commented Mar 16, 2019 at 11:36
  • 1
    getNetworks(&*networks); smells bad. Commented Mar 16, 2019 at 11:37
  • In the same way it should be (*networks)[0] = (char*) malloc(4); as you are working with the parent's array. Commented Mar 16, 2019 at 11:38

2 Answers 2

Reset to default
4

The following needs to be fixed:

int getNetworks(char ***networks)

as it is a pointer to an array of pointers to chars. (You can also write this as
char **networks[]).

In the function you must first dereference the pointer to work on the parent's variable:

    *networks = malloc(5 * sizeof(char*));
    (*networks)[0] = (char*) malloc(4);

In main, your declaration is correct, however, you must call as:

    getNetworks(&networks);

so the function can work on the parent's variable.

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1 Comment

This worked right away thanks!
1

There are many problems in your code, I will not comment them, but propose you this solution:

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

char** getNetworks(int *n)
{
  int count = 5;
  char ** networks = malloc(count * sizeof(char*));
  printf("networks in function:%p\n", networks);

  networks[0] = (char*) malloc(4);
  strcpy(networks[0], "aaa");
  networks[1] = (char*) malloc(3);
  strcpy(networks[1], "bb");
  networks[2] = (char*) malloc(5);
  strcpy(networks[2], "cccc");
  networks[3] = (char*) malloc(6);
  strcpy(networks[3], "ddddd");
  networks[4] = (char*) malloc(2);
  strcpy(networks[4], "e");
  *n = count;
  return networks;
}

int main ()
{
  char **networks = NULL;
  int nbNetworks;

  printf("networks before call:%p\n", networks);
  networks = getNetworks(&nbNetworks);
  printf("networks after call:%p\n", networks);
  for (int i=0; i < nbNetworks; i++) {
    printf ("%s\n", networks[i]);
  }
  return 0;
}
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4 Comments

warning : malloc(*n * sizeof(char*)) while *n (in fact nbNetworks) is non initialized, move *n = 5; before (I am not the DV ;-) )
@bruno Thanks, corrected. I did not execute it, it was a mistake.
it works. Just a semicolon missing after "int count = 5". Nice solution.
@Noury ok, happy. I wrote the code directly in editor, so some bugs were present.

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