How does basecase influence which lines of recursive functions are used? - c++

Question

I have the following code in C++ for a recursive function. I don't understand why everything above the basecase is used before basecase is met (stacking?), and then only the lines below the basecase are used as its unstacking.
#include <iostream>
using namespace std;
void printnum(int begin)
{
cout << "upanddown";
cout << begin << endl; //Why is everything this line and above cout'd as it builds the stack and ignored on the unstacking.
if (begin <= 9) { printnum(begin + 1); } // Once the program starts unstacking the functions, why doesn't it console out "upanddown"?
e
cout << begin << endl; //Why is everything below the base case only shown as it unstacks?
}
int main()
{
printnum(1); //First function call, so it starts at one
system("PAUSE");
}
Which results in:
upanddown1
upanddown2
upanddown3
upanddown4
upanddown5
upanddown6
upanddown7
upanddown8
upanddown9
upanddown10
10
9
8
7
6
5
4
3
2
1
Press any key to continue . . .
Why doesn't upanddown execute below the base case ase well?

Answer 1

int main()
{
printnum(1); //First function call, so it starts at one
system("PAUSE");
}
Here is your main function. It has two statements. system("PAUSE"), which is the tail call¸ executed after printnum(1) has returned. To know where to continue it uses the stack for both indicating where to execute next and it's used to give functions arguments. Every call uses the stack so a recursive function is nothing special. The only special thing about it is that it happens to call itself, but the call to itself is independent and the current function is suspended until it has ended just like main. Consider your recursive function:
void printnum(int begin)
{
cout << "upanddown"; // 1
cout << begin << endl; // 2
if (begin <= 9) // 3
{
printnum(begin + 1); // 4
}
cout << begin << endl; // 5
}
So what happens if begin is 10? It will do 1, 2, 3, not 4, then 5.
Output is "upanddowdn10\n10\n".
For begin = 9 it will do 1, 2 which is "upanddown9\n", then the result of printnum(10) is printed since it hits your default case, then "9\n" is printed last. Thus the printed result is "upanddown9\nupanddowdn10\n10\n9\n"
For begin = 8 it will do 1, 2 which is "upanddown8\n", then the result of printnum(9) is printed since it hits your default case, then "8\n" is printed last. Thus the printed result is "upanddown8\nupanddown9\nupanddowdn10\n10\n9\n8\n"
...
For begin = 1 it will do 1, 2 which is "upanddown1\n", then the result of printnum(2) is printed since it hits your default case, then "1\n" is printed last. Thus the printed result is "upanddown1\nupanddown2\nupanddown3\nupanddown4\nupanddown5\nupanddown6\nupanddown7\nupanddown8\nupanddown9\nupanddowdn10\n10\n9\n8\n7\n6\n5\n4\n3\n2\n1\n"
So to answer your question. The "Upanddownx\n" output is done first because it is the first thing to do in your function. If begin is below 10 it will output the whole result of printnum(begin+1). Last it will print "x\n" in the end.

Answer 2

When the flow execution reaches a recursive function call, a function call is made and it starts from the top (That's why you don't see the numbers printed here). When a recursive call reaches the last statement of the function, the execution flow is transferred to the next statement below caller statement (in this case that statement is cout << begin << endl;) That's why you only see the number when backtracking.
Some references about recursion:
Forum
Tutorial

Answer 3

Just remove this line from your code.It is getting those outputs while unstacking.
cout << begin << endl; //Why is everything below the base case only shown as it unstacks?