C++ Recursion Segmentation Fault - c++

Question

#include <iostream>
using namespace std;
int findSumofOdds(int n);
int main()
{
int n = 88;
int x;
x = findSumofOdds(n);
cout << x << endl;
return 0;
}
int findSumofOdds(int n)
{
if (n != 1)
{
if( n % 2 == 0)
n = (n - 1);
return(findSumofOdds(n-1) + 1);
}
else
return 1;
}
Why isn't this recursion working? It tries to run and then crashes. Please let me know. My teacher said that it would work but doesn't.

Answer 1

When n is even, you are decrementing n by two. If it skips over n == 1, it will recurse until it causes a stack overflow. Since n starts out at 88, that's what's happening.
int findSumofOdds(int n)
{
if (n != 1)
{
if( n % 2 == 0)
n = (n - 1); // <== first decrement
return(findSumofOdds(n-1) + 1); // <== second decrement
}
else
return 1;
}
Also, you seem to be counting the number of odd numbers, not adding them. My guess is that you actually want something like:
int findSumofOdds(int n)
{
if (n != 1)
{
if( n % 2 == 0)
return(findSumofOdds(n - 1));
return(findSumofOdds(n-1) + n); // or + 1 to just count
}
else
return 1;
}
If you want to practice recursion, that's fine. But there's a much simpler way to write a function to sum the odd numbers up to and including n:
int fundSumofOdds(int n) {
n = (n + 1) / 2;
return n * n;
}
This is because there's a general formula:
1 + 3 + 5 + ... + 2n-1 = n2

Answer 2

You have to make it
if (n > 1)
Consider n = 2 here
if (n != 1)
{
if( n % 2 == 0) // Yes
n = (n - 1); // n = 1
return(findSumofOdds(n-1) + 1); // n = 0 <-- will not stop.
and change this too. right now it is just counting the number of odd numbers. You need to sum them.
return(n + findSumofOdds(n - 1));
}
else
return 0;

Answer 3

If you print n in findSumofOdds you can see what is happening -- n becomes negative and you get infinite recursion. If your program doesn't crash earlier, you can get integer overflow (when n goes below the minimum value for int) which yields undefined behavior.
To correct this you can do this:
int findSumofOdds(int n)
{
if(n < 1)
{
return 0;
}
if(n % 2 == 0)
{
return findSumofOdds(n - 1) + 1;
}
return findSumofOdds(n - 2) + 1;
}
You can subtract 2 from n in the last statement, because you only need odd numbers and you know that n can't be even at that point (because of if(n % 2 == 0)).
Also, do you need to find the sum of all odd numbers smaller than n (e.g. 4 (== 1 + 3) for n=5) or do you just need to count them (which is what you are doing now)?
If you want to sum the numbers, you have do add n instead of 1 when returning.