这种Ackermann函数的实现可以称为尾递归吗？

Question

I have written following code in C. Can we call it a tail recursive implementation?

#include <stdio.h>

int ackermann(unsigned int *m, unsigned int *n, unsigned int* a, int* len)
{
  if(!*m && *len == -1) {
    return ++*n;
  }
  else if(!*m && *len >= 0) {
    ++*n;
    *m = a[(*len)--];
  }
  else if(*n == 0) {
    --*m;
    *n = 1;
  } else {
    ++*len;
    a[*len] = *m - 1;
    --*n;
  }
  return ackermann(m, n, a, len);
}

int main()
{
  unsigned int m=4, n=1;
  unsigned int a[66000];
  int len = -1;

  for (m = 0; m <= 4; m++)
    for (n = 0; n < 6 - m; n++) {
      unsigned int i = m;
      unsigned int j = n;
      printf("A(%d, %d) = %d\n", m, n, ackermann(&i, &j, a, &len));
    }

  return 0;
}

If it is not tail-recursive please suggest ways to make it so. Any reference to a tail recursive version of Ackermann would be nice in C/C++/Java or non-functional programming language.

Answer 1

Your function uses a data structure to do it's backtracking so while it is a tail recursive function it definitely isn't a simple recursive or iterative process. The array a takes the role as a recursion stack. You could write out the recursive call alltogether:

int ackermann(unsigned int *m, unsigned int *n, unsigned int* a, int* len)
{
  while (*m || *len != -1) {
      if(!*m && *len >= 0) {
        *n++;
        *m = a[(*len)--];
      } else if(*n == 0) {
        *m--;
        *n = 1;
      } else {
        ++*len;
        a[*len] = *m - 1;
        *n--;
      }
  }
  return ++*n;
}

Still without any recursive call I consider this a recursive process.

Answer 2

By definition your ackermann function is a tail-recursive function as you're directly returning the result of the recursive case. Since no further logic depends on the result of your recursive call, the compiler can safely apply tail-recursion optimization.