segmentation fault on a recursive function

Question

I simply want to test something. I am wondering what I did wrong?

   #include <iostream>
   using namespace std;
   unsigned long pwr(unsigned long n, unsigned long m)
   {
          if(m == 0)
            n = 1;
          if(m == 1)
            n = n;
          n = pwr(n, m/2) * pwr(n, m/2);
          return n;
   }

   int main ()
   {
          unsigned long n(2), m(16);
          cout << pwr(n, m);
          return 0;
   }

output is

Segmentation fault

Have you tried printf debugging to see on what line it fails and how deeply it's recursed? — Michael Price
– Michael Price, Commented Nov 20, 2011 at 15:20

Lol4t0 · Accepted Answer · 2011-11-20 15:19:27Z

7

There is no exit from recursion.

You may wanted

          if(m == 0)
             n = 1;
          else if(m == 1)
             n = n;
          else 
             n = pwr(n, m/2) * pwr(n, m/2);
          return n;

answered Nov 20, 2011 at 15:19

Lol4t0

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01d55 · Accepted Answer · 2011-11-20 15:21:32Z

3

Infinite recursion: The recursive call is executed unconditionally, and so the call stack grows until an error stops it.

This is a stack overflow.

answered Nov 20, 2011 at 15:21

01d55

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Comments

bshields · Accepted Answer · 2011-11-20 15:20:05Z

2

You're not ending the recursion when you hit your base case. Even when m == 0 or m == 1 are true, you still recursively call pwr. So you've got infinite recursion.

answered Nov 20, 2011 at 15:20

bshields

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OSH · Accepted Answer · 2011-11-20 15:18:36Z

-1

you are dividing by 0: let's say m starts from 1, in the next iteration m = 1/2 = 0, and you will get the fault. what you probably want to do it return 1 if m = 0 instead of going through the method.

answered Nov 20, 2011 at 15:18

OSH

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2 Comments

Michael Price Over a year ago

m is never used as a divisor, only 2 is.

OSH Over a year ago

that is not a base since you continue running the method. you need to return something at that point to stop the recursion.

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