更优雅的内部方法声明(没有所有的混乱)?

More elegant inner methods declarations (without all the clutter)?

我有一个递归的方法,用某种算法计算x^n,但这在这里并不重要。重要的是我的辅助函数,它跟踪该算法的递归调用。

public class FastPot {

    public static double fastPotRek(double x, int n) {
        class Aux {
            private double aux(double x, int n, int c) {
                if (n == 0) {System.out.print("It took "+c+" recursive calls to compute "); return 1;}
                else return (n % 2 == 0) ? aux(x*x, n/2, c+1)
                        : x * aux(x, n-1, c+1);
            }
        }
        Aux a = new Aux();
        return a.aux(x, n, 0);
    }
}

为了稍微组织一下,我想在 fastPotRek 内声明 aux,为此您必须使用内部 class,我不能将其方法声明为静态的。因此,我实例化 Aux a = new Aux(); 以便能够调用 aux.

请告诉我有一种方法可以使它更优雅,并告诉我我忽略了什么......就像蜜蜂能够以某种方式使 aux 静态或不需要实例化 Aux

不需要内部 class 也不需要使其成为静态的:

public class FastPot {
    //Static, use only from within FastPot
      private static double aux(double x, int n, int c) {
                if (n == 0) {
                  System.out.print("It took "+c+" recursive calls to compute "); 
                  return 1;
                } else {
                  return (n % 2 == 0) ? aux(x*x, n/2, c+1)
                        : x * aux(x, n-1, c+1);
                }
            }
    }

    //Your outward interface
    public static double fastPotRek(double x, int n) {
        return aux(x, n, 0);
    }
}

或者如果你坚持使用内部 class:

public class FastPot {
    //Static, use only from within FastPot
    private static class Aux {
      private static double aux(double x, int n, int c) {
                if (n == 0) {
                  System.out.print("It took "+c+" recursive calls to compute "); 
                  return 1;
                } else {
                  return (n % 2 == 0) ? aux(x*x, n/2, c+1)
                        : x * aux(x, n-1, c+1);
                }
            }
    }


    //Your outward interface
    public static double fastPotRek(double x, int n) {
        return Aux.aux(x, n, 0);
    }
}

你有这个:

Aux a = new Aux();
return a.aux(x, n, 0);

我会这样写(一样):

return new Aux().aux(x, n, 0);

编辑: 我用完了 Whosebug。你们不需要帮助吗?我不会给它。以后我只问我的个问题,不回答

我会 post 这个答案,尽管很多人(包括我)对此并不满意。请不要使用这种代码:

public static double fastPotRek(double x, int n) {
    return new Cloneable() {
        private double aux(double x, int n, int c) {
            if (n == 0) {System.out.print("It took "+c+" recursive calls to compute "); return 1;}
            else return (n % 2 == 0) ? aux(x*x, n/2, c+1) : x * aux(x, n-1, c+1);
        }
    }.aux(x, n, 0);
}

同样,我强烈建议使用私有静态方法,例如:

public static double fastPotRek(double x, int n) {
    return aux(x,n,0);
}
private static double aux(double x, int n, int c) {
    ...
}