Allocation

The previous examples used either pointers to existing data structures or data structures with an underlying pointer nature. In many instances though one wishes to work with pointers directly. In such cases one needs another layer of working with memory allocation.

Can you guess what will happen with the following program?

#include <stdio.h>
int main(void) {
  int *count;
  *count = 44;
  printf("count=%d\n", *count);
  return 0;
}

Actually, what does happen depends on the compiler; in some cases it will compile, but with a warning, and in other cases, it may compile but give a segmentation fault when running. The reason for this is that, when dealing with pointers directly, it is up to the programmer to allocate sufficient memory to store the relevant values.

This memory allocation can be done through the use of the malloc function:

#include <stdio.h>
int main(void) {
  int *count;
  count = (int *) malloc(sizeof(int));
  *count = 44;
  printf("count=%d\n", *count);
  return 0;
}

The argument to malloc is the number of bytes to allocate (in this case, the size of a single integer), and the return value is of type void *, which subsequently must be cast into the desired type.

One important uses of dynamically allocating memory in this way is creating arrays of a specified size during the running of a program:

#include <stdio.h>
int main(void) {
  int *scores, n, i;
  printf("How many scores? ");
  scanf("%d", &n);
  scores = (int *) malloc(n * sizeof(int));
  for(i=0; i<n; i++) {
    scores[i] = i;
    printf("scores[%d] = %d\n", i, scores[i]);
  }
  free(scores);
  return 0;
}

Note the use of free here to free the memory allocated by malloc - this memory will be freed when the program terminates, but freeing the memory associated with variables when they will no longer be needed is a good practice to get into.