可以阿克曼函数的本次执行被称为尾递归？

问题描述

我已经写在下面C. code，我们可以把它叫做一个尾递归实现？

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;
}

如果它不是尾递归，请建议如何让它如此。到阿克曼的尾递归版本的任何引用将是C / C ++ / Java或无功能的编程语言很好。

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.

推荐答案

以与阿克曼函数的是为你直接返回递归情况下的结果的尾递归函数。由于没有进一步的逻辑取决于你的递归调用的结果，编译器可安全应用尾递归优化。

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.

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