void pointer as argument [duplicate] - c

Question

This question already has an answer here:
Dynamic memory access only works inside function
(1 answer)
Closed 3 years ago.
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

As this is C, you cannot pass the pointer by reference without passing in a pointer to the pointer (e.g., void ** rather than void * to point to the pointer). You need to return the new pointer. What is happening:
f(a1);
Pushes the value of the pointer (NULL) as the stack parameter value for a. a picks up this value, and then reassigns itself a new value (the malloced address). As it was passed by value, nothing changes for a1.
If this were C++, you could achieve what you want by passing the pointer by reference:
void f(void *&a);

Answer 2

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));
// ...

Answer 3

To change a variable via a function call, the function needs to have reference semantics with respect to the argument. C doesn't have native reference variables, but can implement reference semantics by means of taking addresses and passing pointers.
Generally:
void mutate_thing(Thing * x) // callee accepts pointer
{
*x = stuff; // callee derefences ("*")
}
int main()
{
Thing y;
mutate_thing(&y); // caller takes address-of ("&")
}
In your case, the Thing is void *:
void f(void ** pv)
{
*pv = malloc(12); // or whatever
}
int main()
{
void * a1;
f(&a1);
}

Answer 4

Based on Kerrek SB's example I came with this to demonstrate void pointer-to-pointer as an argument and how it may be used.
#include <stdio.h>
void test(void ** jeez)
{
*jeez = (void *) (int) 3;
}
int main (int argc, char* argv[])
{
void *a;
test(&a);
int b = *(int *)&a;
printf("value returned = %d\n", b);
return 0;
}