Is it possible to have stack overflow by using recursive function?

Question

This function has some problems ?

 unsigned long factorial(unsigned long m)
 {
     return (m == 0) ? 1 : m * factorial(m - 1);
 }

If I add another function:

  int myCombina(int q, int p)
  {
      return factorial(q) / ( factorial(q) * factorial(q-p) );
  }

This myCombina() is not efficient because a greatest common divisor should be cancelled in it to find combinatorial.

max(factorial(q), factorial(qp)) can be cancelled out. We only need to compute qx (q-1) ... x (q -k +1).

Are there other problems ?

Any comments are welcome.

Thanks

Answer 1

if m is very large, it may have stack overflow.

A stack overflow is not the main problem with your code... if m is very large you will get an integer overflow before you get a stack overflow.

You need to use some sort of Bignum type if you want this to work for m larger than about 12 (depending on the size of unsigned long on your platform).

Answer 2

它不是以尾部递归形式编写的，因此即使编译器支持适当的尾部调用优化，您也不会从中受益。

Answer 3

The function can actually cause an stack overflow (each level of recursion will consume a bit of the stack until it get's all consumed).

As others have mentioned, you can convert the recursive function into a loop, which in this case would be simple, or you can modify the recursion to allow for tail-call optimization (have the compiler transform the recursion into a loop).

Just for the sake of it, to transform into a tail recursive call, the last statement of the function must be a return with the result obtained from the recursive call. Your code cannot be optimized away because your return statement contains m*factorial(n-1) , that is, you don't return the value of the recursion, but actually operate on it before returning.

The transformation to make it tail recursive require pushing the multiplication down to the recursive call, and that is usually performed as an extra argument that keeps the temporary result:

unsigned long factorial_tail_recursive( 
                        unsigned long n,           // number to calculate factorial
                        unsigned long temp_result  // accumulated result
                      ) 
{
   return n == 0? tmp_result 
                : factorial_tail_recursive( n-1, n*temp_result );
}

// syntactic sugar to offer the same interface
unsigned long factorial( unsigned long n ) {
   return factorial_tail_recursive( n, 1 );    // accumulated result initialized to 1
}

But then again, for that particular problem an efficient iterative solution is probably easier to get right. Just maintain the accumulated result in a temporary variable and loop until the argument is reduced to 1.