An array of structures

问题:

How is an array of structures created in C without knowing the eventual amount of structures to be stored in the array?

I would like to loop in a for loop, create a tempStruct set its variables, add this to an array and then loop again, creating a new tempStruct and adding it to the array.

Im having some issues wrapping my head around how this is done in C while trying to relate from objective C.

回答1:

Dynamically allocated arrays (using malloc) can be reallocated (using realloc).

Therefore the solution will look something like this:

1. malloc initial array (arbitrary size)
2. while still space in array, add structures
3. when array full, realloc to bigger size
4. goto 2

回答2:

You could create a double linked list which points to parent and child

struct list{
 list* next;
 list* prev;
 special_data* data;
}

easy to do and flexible

回答3:

You can't create an array in C without knowing number of it's members up front. Your options for adding are:

• (Faster) Create new array with +1 element, copy entire array and add new element to the end
• (Better) Create your own implementation of linked list (Linked list) which will dynamically allocate memory for each new member.

回答4:

You can use malloc to create your structure.

Edit: The following demonstrates one way to do what you're asking by creating a linked list:

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

typedef struct {
  int data;
  void* next;
} tempStruct;

#define NUM_STRUCTS 4

int main(void) {
  tempStruct* cur_ptr;
  tempStruct* root_ptr;
  int i;

  root_ptr = malloc(sizeof(tempStruct));
  root_ptr -> data = 0;
  root_ptr -> next = NULL;
  cur_ptr = root_ptr;
  for (i = 1; i < NUM_STRUCTS; i ++ ) {
    tempStruct* new_ptr = malloc(sizeof(tempStruct));
    new_ptr -> data = i;
    new_ptr -> next = NULL;
    cur_ptr -> next = new_ptr;
    cur_ptr = cur_ptr -> next;
  }

  cur_ptr = root_ptr;
  while (cur_ptr != NULL) {
    printf("cur_ptr -> data = %d\n", cur_ptr -> data);
    cur_ptr = cur_ptr -> next;
  }
  return 0;
}

If you really want to create something that acts more like an array, you'll need to allocate all your memory at the same time, using something like:

the_data = malloc(NUM_STRUCTS * sizeof(tempStruct);

Then you'll have to access the data with the dot operator (i.e. '.' (no quotes in your code).

回答5:

struct foo {int bar;};

size_t i = 0, n = 8;
struct foo *arr = malloc(n * sizeof *arr);

int bar;

while ((bar = get_next_bar()) != -1) {
    if (++i == n) { // no room for new element; expand array
        arr = realloc(arr, n *= 2);
        if (arr == NULL) abort; // see note below.
    }
    arr[i] = (struct foo){bar};
}

The number of assigned elements in the array is i+1. Don’t forget to free() the array when you’re done with it.

Note: In real programs you generally do not do p = realloc(p, s) directly. Instead you assign the result of realloc() to a new pointer, then do error detection & handling before clobbering your original pointer.