Initialising array with unknown size in C

Question

I have this piece of code:

#include <stdio.h>

int average(int array []);

int main () {

int num_array[];


int x = 0;

  while(num_array[x] != -1){

     printf("Enter a number\n");
     scanf("%d",&num_array[x]);

     x++;

  }

  printf("%d\n", average(num_array));

  return 0;
}



int average(int array[]){

  int i;
  int total_size = (int)sizeof(array);
  int sum = 0;

  for(i = 0; i < total_size; i++){

    sum = sum + array[i];

  }

  return sum/total_size;

}

But i get an error at compile time, because i'm not initialising the array. However i can't do it, since i don't know it's size.

How can i solve this ?

@ForceBru as they said, you need to use `malloc()` with a pointer to an array. `int* num_array;` then use that pointer in `malloc()` — Dylan Lawrence, Mar 05 '15 at 14:49
In order to avoid `realloc`, first ask for the user about the array size. In order to avoid `malloc`, set the array's maximum size and allocate it statically. — Eugene Sh., Mar 05 '15 at 14:49
`int* num_array; num_array = (int*) malloc(size+1);` You should probably check if `malloc()` returns null, but this is the basic setup. — Dylan Lawrence, Mar 05 '15 at 14:52
@Lundin The point is that it's much easier to google it rather than write a question on SO.. So laziness isn't an excuse — Eugene Sh., Mar 05 '15 at 14:54
@DylanLawrence http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc — Lundin, Mar 05 '15 at 14:56
@Lundin if you do not want to provide any help, just vote the question down and make pointless observations, please go somewhere else. — laker001, Mar 05 '15 at 16:31
@laker001 Oh but I helped by doing the research for you, by finding one of many duplicates to the question. Read away! — Lundin, Mar 06 '15 at 07:13

score 1 · Answer 1 · answered Mar 05 '15 at 14:48

1

You can't declare an array like

int num_array[];

in C except when it is a parameter of a function. You must have to specify the size.

answered Mar 05 '15 at 14:48

haccks

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score 1 · Answer 2 · answered Mar 05 '15 at 14:51

This is actually a surprisingly tricky problem in C unless you're using an array library. What you're actually talking about is a growable array. For what you're trying to do, you start with an initial size, say 10, using malloc() and as the inputs grow, you have to realloc() the array to a larger size. Generally, you double the size of the array every time the limit is hit. If you're a beginning C programmer, this is probably too much to take on. I would set a fixed limit, and enforce it by not accepting any more input after the limit is hit. Given that you can comfortably allocate 1000's of ints and a human won't want to type that much, you can easily make a practical limit ahead of time and declare the array as such:

int num_array[1000];

score 0 · Answer 3 · answered Mar 05 '15 at 14:51

You need to provide the size of the array here: int num_array[];.

If you are working with a structure of unknown size, you could:

Ask the user how many numbers are going to be entered (not really practical, but might suffice if your case is a trivial one).
Initialize it with a default size, say 10 and then if you need, create larger arrays and copying your current data to the new one.
Or else, you use a linked list like data structure and create new elements as you go along, thus allowing your list to grow as much as you need it to be.

Initialising array with unknown size in C

3 Answers3