Let's rearrange this expression p->setName(p, "John Doe"); (p->setName) (p, "John Doe"); Here can you see that (p->setName) is entirely independent of (

Let's rearrange this expression p->setName(p, "John Doe"); (p->setName) (p, "John Doe"); Here can you see that (p->setName) is entirely independent of (p, "John Doe"), so no, there is no way to pass p as an argument automatically. You say that you want to "get the ptr to a caller of a function". However there's nothing that is a "caller". You have a function pointer. Where you got that from (inside a struct perhaps) is nothing to do with the call that you later make. However, what you could do is char *setNameImp( struct Person *self, const char *name) { strcpy(self->name, name ); } char * (*setName)( struct Person *p, const char *name) { p->setName(p, name); } int main(void) { struct Person *p = (struct Person *)malloc(sizeof(struct Person)); p->setName = &setNameImp; setName(p, "John Doe"); } So have a static wrapper that calls the virtual method for you. Also, you should not be using strcpy into a buffer without checking that the buffer is large enough.