void pointer as argument

Question

The following C snippet:

[...] 
void f1(void* a){
  printf("f(a) address = %p \n",a);
  a = (void*)(int*)malloc(sizeof(int));

  printf("a address = %p \n",a);
  *(int*)a = 3;

  printf("data = %d\n",*(int*)a);
}

void f(void){
  void* a1=NULL;
  printf("a1 address = %p \n",a1);

  f1(a1);

  printf("a1 address = %p \n",a1);
  printf("Data.a1 = %d\n",*(int*)a1);
}
[...]

results in

a1 address = (nil) 
f(a) address = (nil) 
a address = 0xb3f010 
data = 3
a1 address = (nil) 
Segmentation fault (core dumped)

Why doesn't a1 keep the address that has been assigned to it in the function?

Answer 1

Passing a pointer to a1 to your function, you can't change where a1 points. The pointer is passed by value, so in f1 you're only changing a copy of the address held by a. If you want to change the pointer, i.e. allocate new memory for the pointer passed in, then you'll need to pass a pointer to a pointer:

void f1(void **a)
{
    // ...
    *a = malloc(sizeof(int));
    // ...