separating a looping algorithm from actions - c++

Question

Suppose we have a singly linked list of integers and we are given the following requirements:
For a given node, menu option 1 should display the data field of the node and any subsequent nodes in the console output.
For a given node, menu option 2 should display the data field of the node and any subsequent nodes in the console output only if data is multiple of 2.
For a given node, menu option 3 should display the data field of the node and any subsequent nodes in the console output only if data is multiple of 3.
A possible solution is as follows:
struct node {
int data;
node *next;
};
void insertBeginning(node **list, int data)
{
node *n = new node;
n->data = data;
n->next = *list;
*list = n;
}
void option1(node *node)
{
if (node != NULL) {
cout << node->data << endl;
option1(node->next);
}
}
void option2(node *node)
{
if (node != NULL) {
if (node->data % 2 == 0)
cout << node->data << endl;
option2(node->next);
}
}
void option3(node *node)
{
if (node != NULL) {
if (node->data % 3 == 0)
cout << node->data << endl;
option3(node->next);
}
}
void main()
{
node *root = new node;
root->data = 5;
root->next = NULL;
insertBeginning(&root, 4);
insertBeginning(&root, 3);
insertBeginning(&root, 2);
insertBeginning(&root, 1);
cout << "OPTION 1" << endl;
option1(root);
cout << endl;
cout << "OPTION 2" << endl;
option2(root);
cout << endl;
cout << "OPTION 3" << endl;
option3(root);
}
But a closer inspection reveals that in each menu option there is an algorithm that loop through nodes, which is the same in all options, and then a specific action. So it might be good to separate the looping algorithm from the action. How would you do it?
In this case the looping algorithm as well as the actions are very simple, so doing a separation would be killer, but in my situation the looping algorithm and the actions are more complex, and I want to avoid copying and pasting the looping algorithm. Also, this case uses a singly linked list, but it could be a tree. Anyway I do not want to distract you with these details.
Finally consider a fourth option:
For a given node, menu option 4 should display the sum of the data field of the node and any subsequent nodes in the console output.
A possible solution is as follows:
void loopNodes(node * n, void action(node * n))
{
if (n != NULL) {
action(n);
loopNodes(n->next, action);
}
}
void option1(node * node)
{
cout << node->data << endl;
}
void option2(node * node)
{
if (node->data % 2 == 0)
cout << node->data << endl;
}
void option3(node * node)
{
if (node->data % 3 == 0)
cout << node->data << endl;
}
int sum;
void option4(node * node)
{
sum += node->data;
}
void main()
{
node *root = new node;
root->data = 5;
root->next = NULL;
insertBeginning(&root, 4);
insertBeginning(&root, 3);
insertBeginning(&root, 2);
insertBeginning(&root, 1);
cout << "OPTION 1" << endl;
loopNodes(root, option1);
cout << endl;
cout << "OPTION 2" << endl;
loopNodes(root, option2);
cout << endl;
cout << "OPTION 3" << endl;
loopNodes(root, option3);
cout << endl;
cout << "OPTION 4" << endl;
sum = 0;
loopNodes(root, option4);
cout << "SUM = " << sum << endl;
}
Option 4 introduced a challenge; the need to maintain state. I used a variable with file scope, but an alternative could be (and this is my final solution):
void loopNodes(node * n, void action(node * n, void *state), void *state)
{
if (n != NULL) {
action(n, state);
loopNodes(n->next, action, state);
}
}
void option1(node * node, void *state)
{
cout << node->data << endl;
}
void option2(node * node, void *state)
{
if (node->data % 2 == 0)
cout << node->data << endl;
}
void option3(node * node, void *state)
{
if (node->data % 3 == 0)
cout << node->data << endl;
}
void option4(node * node, void *state)
{
*(int *) state += node->data;
}
void main()
{
node *root = new node;
root->data = 5;
root->next = NULL;
insertBeginning(&root, 4);
insertBeginning(&root, 3);
insertBeginning(&root, 2);
insertBeginning(&root, 1);
cout << "OPTION 1" << endl;
loopNodes(root, option1, NULL);
cout << endl;
cout << "OPTION 2" << endl;
loopNodes(root, option2, NULL);
cout << endl;
cout << "OPTION 3" << endl;
loopNodes(root, option3, NULL);
cout << endl;
cout << "OPTION 4" << endl;
int *sum = new int;
*sum = 0;
loopNodes(root, option4, sum);
cout << "SUM = " << *sum << endl;
}
What do you think?
Any feedback is really appreciated!
Note: I have to use core language facilities (I cannot use the standard library, boost, etc.).

Answer 1

In terms of OOP, your option mechanism depends on some specific handling bound to each option input, which are in this case positive integers. You may consider having a Handle class which takes two inputs during construction, one is option number (i.e an integer) and second is a lambda function or a function pointer which applies the logic. Your linked list then can have a AssignHandles() method to construct the menu behavior automatically.

Answer 2

You could use a templated version for your loop function. This could be used with functions or function objects, which implement the operator ():
#include <iostream>
int list[] = {1, 2, 3, 4, 5};
template<class Option>
void loop(Option & option) {
for (size_t i = 0; i < sizeof(list)/sizeof(list[0]); ++i) {
option(list[i]);
}
}
void print(int e) {
std::cout << e << std::endl;
}
struct Sum {
int sum = 0;
void operator () (int e) {
sum += e;
}
};
int main(int argc, char** argv) {
loop(print);
Sum s;
loop(s);
std::cout << s.sum << std::endl;
return 0;
}