C ++更改作为函数参数传递的指针值

Question

I working with a library where I need to send a pointer to an object to a function.. the problem is that I need to change what this pointer points to in the function itself, and I don't really know how I can overcome this problem since pointers are passed-as-value..

struct.cpp

struct MyStruct {
    Node* previous;
    ...
};

main.cpp

int main(...) {
    MyStruct* m = new MyStruct;
    m->previous = NULL;
    ...
    while (traversing) {
        library_function((AFUNPTR) myFunction, (void*) m->previous);
    }

function.cpp

void myFunction(void* node_from_lib, void* point_to_previous) {
    Node* current = (Node*) node_from_lib;
    Node* previous = (Node*) point_to_previous;
    if (previous != NULL) {
        ...
    }
    previous = current;
}

My problem is that I need to traverse through all the nodes, but this way previous will be sent into myFunction pointing to NULL all the time.. I've tried to use Node** previous and assign the "new" previous like *previous = current; but I'm still getting a seg-fault.

The library I'm using is Intel-PIN and I'm trying to instrument an instruction and chain them into a graph, although I've stripped away everything PIN from the example since I think this is a general c++ problem?

---

Here is my attempt with a pointer to a pointer...

struct.cpp

struct MyStruct {
    Node** previous;
    ...
};

main.cpp

int main(...) {
    MyStruct* m = new MyStruct;
    *(m->previous) = NULL;
    ...
    while (traversing) {
        library_function((AFUNPTR) myFunction, (void*) m->previous);
    }

function.cpp

void myFunction(void* node_from_lib, void** point_to_previous) {
    Node* current = (Node*) node_from_lib;
    Node** previous = (Node**) point_to_previous;
    if ((*previous) != NULL) {
        (*previous)->memberFunc();
        ...
    }
    *previous = current;
}

Answer 1

C:

int main(...) {
    MyStruct* m = new MyStruct;
    m->previous = NULL;
    ...
    while (traversing) {
        library_function((AFUNPTR) myFunction, &m->previous);
    }

here, pass the m->previous via pointer (the pointer is then of type Node** , that can be reinterpreted to void* and back without losing anything. The function expects void* .)

function.cpp

void myFunction(void* node_from_lib, void* point_to_previous) {
    Node* current = (Node*) node_from_lib;
    Node* previous = * ((Node**)point_to_previous); // cast for reading - assume point_to_previous is pointer to pointer, and retrieve the pointer
    if (previous != NULL) {
        ...
    }
    previous = current;
    *((Node**)point_to_previous) = previous; // assume, point_to_previous is pointer to pointer, and change the pointer
}

Here, you can cast back the void* to Node** .

Here, you pass the pointer to node pointer ( Node** ) as a second parameter. It is reinterpreted as void* , but it is just semantic.