c++ - How to understand function recursion precisely?

Question

I am currently programming some divide-conquer algorithms, where function recursions are used everywhere, but I have very vague idea or no idea how exactly it works, and that's why I post it here and hope you don't mind it's too basic.

For example, if we have the following code:

#include<iostream>
using namespace std;
void Recursion(int n)
{
  cout << n << endl;
  if(n > 0)
  {
    Recursion(n-1);
  }
  cout<<n<<endl;
}

int main()
{
  Recursion(3);
  return 0;
}

I tested Recursion(3) and the print out in the terminal is:

3
2
1
0
0
1
2
3

I can understand the concept of recursive call of the function but I don't understand the mechenism how it works. For example, what will they do after they can't call the function again? For example, here, I can understand it prints from 3 to 0 but why it also prints from 0 to 3 again? I heard it's because function recursion is stored in a stack for one recursion and when it reaches the "bottom" it also has to delete.

But anyway, I don't know about it. So, can anyone help me out and clearly tell me what happened here and the exact flow of function call?

Thanks for your help!

Answer 1

You are right, I also find recursive functions difficult to understand. Here is what i do, if i see a recursive function: run all the code step by step in your mind. This advice may seem trivial, but most of the time it works for me. Let's look at your code: You call Recursion() function with parameter 3. It prints n and n>0 that's why it calls Recursion(2) (note that we didn't return from the Recursion(3) call we are still in it and now we are also in Recursion(2). The same is for recursion(1) and 0. Now n>0 conditional is false. It prints 0. and we return from recursion(0) We print 1 and return from recursion(1) and it goes on the recursion(3)

Recursion(3)
    Recursion(2)
        Recursion(1)
            Recursion(0)  
            return from Recursion(0)
        return from Recursion(1)
    return from Recursion(2)
return from Recursion(3)