Doing memcpy without allocating memory with malloc before

Question

I am asking myself why this piece of code works well when I haven't allocated memory for fptr. I would expect that it has rather an Undefined Behavior because of doing memcpy without allocating memory for fptr or?

struct conf *pconf = NULL; 
void (*fptr)(char *, struct conf **);
void *temp = dlsym(dlptr, "config_run_all"); 
memcpy(&fptr, &temp, sizeof fptr); 
fptr("test.conf", &pconf);

Answer 1

You have allocated memory for fptr :

void (*fptr)(char *, struct conf **);`

This declares fptr as a pointer to function.

The memcpy() assigns the value from temp into fptr , making it so that fptr points to the function that temp points to.

What would be problematic would be omitting the & from the memcpy() ; then you'd be trying to copy to memory when fptr has not been set to point to anything.

Answer 2

Your code is basically equivalent to:

struct conf *pconf = NULL; 
void (*fptr)(char *, struct conf **) = (void (*)(char*, struct conf**)) dlsym(dlptr, "config_run_all"); 
fptr("test.conf", &pconf);

witout temp .

Answer 3

The clue is in the memcpy invocation.

memcpy(&fptr, &temp, sizeof fptr); 

You are copying to the address of the pointer, not dereferencing it. This is overwriting the memory of the pointer variable itself, not what it points to (which is nothing).