c - Structured linked list to an array -

typedef struct node {     int value;     struct node* next; }node;  int* to_array (node* ll, int size) {     int = 0;     int* arr = malloc(size*sizeof(int));     while (ll) {           arr[i] = ll->value;           ll = ll->next;           i++;     }  return arr; } 

can please explain why

int* arr = malloc(size); 

will give array? thought when have pointers cannot change individually arr[i] = 5 or something.

your question. granted, 1 that's been asked , answered many times on so. question nonetheless.

from c/c++ faq:

http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=aryptr

arrays not pointers, though closely related (see question 6.3) , can used (see questions 4.1, 6.8, 6.10, , 6.14).

1. when declare array (e.g. int a[5]), you've allocated storage 5 "int" elements. can access each element a[i].

2. when declare pointer (e.g. int *b) haven't allocated any storage.

3. you can declare , initialize pointer @ same time:

   int *b = null;  /* initialize 0 */ ... or ... int *b = malloc (5 * sizeof (int)); /* allocate storage 5 "int" elements */ 

4. when declared array a, allocated space stack. allocation cannot changed.

   when declared b, allocated same amount of space, allocated heap. further, can change b point else, @ time. can realloc() memory change size of storage.

5. just can use index syntax a[i], use same syntax b[i].

this link might explain: http://www.geeksforgeeks.org/g-fact-5/

ps: when "have pointers", can "change individually arr[i] = 5 or something".