What does this recursion function output? - recursion

Question

I'm just starting to learn recursion and I don't understand the output, which is 39493. Could you explain?
struct Node {
int value;
Node* next;
}
head --> 3 --> 9 --> 4 --> NULL
void f4(Node* n) {
if(n == NULL)
return;
if(n->next == NULL)
cout << n->value << “ ”;
else {
cout << n->value << “ ”;
f4(n->next);
cout << n->value << “ ”;
}
}

Answer 1

Your f4 function prints out a linked list of integers, forwards (from the head/root to the tail) and then backwards. It does this because the inner call of f4 within the else clause is surrounded by two cout statements, which will both print the current node's value, adding a stack frame (new execution context with the next node) in between. The last node (the tail) is captured by if(n->next == NULL), and there's only one cout statement in this clause, so the tail only prints once.
I added a main method to illustrate this:
#include <iostream>
using namespace std;
struct Node {
int value;
Node* next;
};
void f4(Node* n) {
// This will never be reached
// except if f4 is called with NULL directly
if(n == NULL)
return;
if(n->next == NULL)
cout << n->value << endl;
else {
cout << n->value << endl;
f4(n->next);
cout << n->value << endl;
}
}
int main() {
cout << "Starting!" << endl;
Node n1 = {1, 0};
Node n2 = {2, &n1};
Node n3 = {3, &n2};
// get pointer to n3 -> n2 -> n1
f4(&n3);
return 0;
}
Output:
Starting!
3
2
1
2
3
To understand the stack frame (recursive call) handling mechanism:
https://www.bottomupcs.com/elements_of_a_process.xhtml.
You might understand it better by adding additional markers to differentiate the cout << n->value << endl on different lines.