pass string array out of function in c

Question

I have a function that I'm using to build a string array. Inside the function i can print the array no worries however the program returns from the called function, the array (arr) remains NULL. How do I get the data out of the array ?

Thanks.

main(){
  char **arr = NULL;
  funct(arr);
  printf("%s\n", arr[2];
}

funct(char **arr){
  arr = malloc(10*sizeof(char *));
  arr[1...n] = calloc(SIZE, sizeof(char));
  // Add data to array
}

if you're wanting to build a char*[] in the function, including both the pointer array and its content, you're missing a level of indirection on your input parameter, though honestly I don't see why you don't just return a char** and call it good. — WhozCraig
– WhozCraig, Commented Mar 23, 2014 at 10:47
I might do that... Is the only difference that it looks a little cleaner to do it that way? (I'm still learning). So that should fix the issue? — DangerousDave23
– DangerousDave23, Commented Mar 23, 2014 at 23:46
No, thats not the only difference. C is a by-value function-parameter language. If you want to modify something on the caller's side, the "value" you're passing needs to be an address, and the formal parameter needs to be a pointer-to type, dereferenced to write to the caller's var. If you have a pointer p and you're not dereferencing it at-assignment, like p = <<something>> you're changing nothing on the caller side. In your case, arr = malloc(... allocates memory assigning the result to simply arr, and promptly leaks it on return. — WhozCraig
– WhozCraig, Commented Mar 24, 2014 at 0:02

Manuel Selva · Accepted Answer · 2014-03-23 15:25:32Z

2

I can see several problemens in your code:

in the line printf("%s\n", arr[2]; you forgot a closing )
Your arr variable local to the main function is never initialized. In C, parameters are passed by value meaning that you are passing the NULL pointer to your function and inside this function, the local pointer arr is allocated but not the one of the main function.

The solution is to allocate the array in the main function and pass a pointer to the array and the size to the function that fills it:

main(){
  char **arr = malloc(10*sizeof(char *));
  funct(arr, 10);
  printf("%s\n", arr[2]);
}

funct(char **arr, int size){
  // Add data to array
  arr[0] = "first data";
  ....
  arr[size -1] = "last data";
}

edited Mar 23, 2014 at 15:25

answered Mar 23, 2014 at 10:35

Manuel Selva

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2 Comments

DangerousDave23 Over a year ago

Manuel thanks for your reply. My question is, if it's a pointer why am I not able to allocate memory to the pointer inside the function? What if inside the function I need to realoc?

WhozCraig Over a year ago

@DangerousDave23 then either return the reallocated array as the function result or pass the pointer-to-pointer by address (yes, another level of indirection, a pointer-to pointer-to-pointer) and (shuddering to say this) join the three-star club. And yes, that name normally means you're probably doing something wrong (and thats a separate debate), but if you know how to use it, it works.

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