7

For example, I have in the main file 

1) char ** array[NUMBER];
2) array = build_array();

and in an imported file

char ** build_array()
{
  char ** array[NUMBER];
  strings[0] = "A";
  strings[1] = "B";
  return (char *) strings;
}

However, at line 2 in the main file, I get the error: "incompatible types when assigning to type 'char **[(unsighed int)NUMBER]' from type 'char **'

What am I doing wrong? Any suggestions or advice would be appreciated. Thank you in advance.

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6
  • 4
    Post the actual code? Where is strings defined? Why do you take a pointer to a literal string? Etc... Commented Oct 3, 2011 at 4:36
  • 1
    Is array supposed to be strings? Are you trying to return a pointer to a stack allocated object? Commented Oct 3, 2011 at 4:40
  • I just want to call the build_array() function and return the array of strings (pointers to char arrays) to use in main. Commented Oct 3, 2011 at 4:42
  • strings aren't pointers to char arrays, they are just pointer to chars. Commented Oct 3, 2011 at 4:44
  • Right, my mistake. How would I do this? Commented Oct 3, 2011 at 4:45

3 Answers 3

Reset to default
17

There seem to be some confusion about what a string is in C. In C, a null terminated sequence of chars is considered a string. It is usually represented by char*.

I just want to call the build_array() function and return the array of strings

You pretty much can't return an array, neither a pointer to a local array. You could however pass the array to build_array as an argument, as well as its size, and fill that instead.

void build_array( char* strings[], size_t size )
{
  // make sure size >= 2 here, based on your actual logic
  strings[0] = "A";
  strings[1] = "B";
}
...later called as:...
char *array[NUMBER];
build_array(array, NUMBER);

The alternatives are to return a pointer to a global or static allocated array, which would make your function non-reentrant. You probably don't care about this now, but is bad practice so I would recommend you avoid going that route.

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6 Comments

Thank you so much. It now compiles, however, when I try to print the contents, nothing prints or gibberish prints. Is there an issue with assigning values to the array?
@LearningPython: We have seen so little of your code, that I wouldn't know. How is NUMBER defined? How do you attempt to print the contents?
@LearningPython: Well I'm intrigued... what was it?
Using int for an array size? Really? ;)
@Chris Lutz: If in C++ I would have gone with std::size_t which is unsigned, is that your point?
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9

As littleadv pointed out, there are several problems with your code:

  • Mismatch between char ** and char **[ ]

  • Returning a pointer to a local variable

  • Etc.

This example might help:

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

#define NUMBER 2
#define MAX_STRING 80

char **
build_array ()
{
  int i = 0;
  char **array = malloc (sizeof (char *) * NUMBER);
  if (!array)
    return NULL;
  for (i = 0; i < NUMBER; i++) {
    array[i] = malloc (MAX_STRING + 1);
    if (!array[i]) {
      free (array);
      return NULL;
    }
  }
  strncpy (array[0], "ABC", MAX_STRING);
  strncpy (array[1], "123", MAX_STRING);
  return array;
}

int
main (int argc, char *argv[])
{
  char **my_array = build_array ();
  if (!my_array) {
    printf ("ERROR: Unable to allocate my_array!\n");
    return 1;
  }
  else {
    printf ("my_array[0]=%s, my_array[1]=%s.\n",
      my_array[0], my_array[1]);
  }
  return 0;
}
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2 Comments

In this great example NUMBER is known in advance. Could we get the length of the returned array from the caller function if that number is decided inside build_array() ?
Sure. One way is to pass #/elements back to the caller as an "out" parameter. EXAMPLE: char ** build_array(int * nelms), then inside build_array(): *nelms = number;
1

Your return type is char**, while you're assigning it to char**[], that's incompatible.

Other than that you should post the actual code that you have problem with, the code you posted doesn't compile and doesn't make much sense.

In order to fix your code, the function should be returning char **[NUMBER]. Note also, that you're casting the return value to char* instead of char** that you declared (or char **[NUMBER] that it should be, and in fact - is).

Oh, and returning a pointer to a local variable, as you do in your case, is a perfect recipe for crashes and undefined behavior.

What you probably meant was:

char *array[NUMBER];
int ret = build_array(array, NUMBER);
// do something with return value or ignore it

and in an imported file

int build_array(char **arr, int size)
{
  // check that the size is large enough, and that the
  // arr pointer is not null, use the return value to
  // signal errors
  arr[0] = "A";
  arr[1] = "B";
  return 0; // asume 0 is OK, use enums or defines for that
}
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1 Comment

What should the return type be to return an array of strings?