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?
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);