为什么这种猜测用户输入数字平方根的递归方法不起作用

Question

出于某种原因，我一直在试图编译这段代码时出错，但似乎一切都正常，对吗？

导入java。util.*；公共类平方根{public static void input（）{

     Scanner input = new Scanner (System.in);  
     double number = 16;
     double root;
     root = square(number);
     System.out.println("Enter Guess");
     double oldguess = input.nextDouble();
     squareRoot(0, oldguess);
}
public static double square(double number){
double t;

double squareroot = number / 2;

do {
    t = squareroot;
    squareroot = (t + (number / t)) / 2;
} while ((t - squareroot) != 0);

return squareroot;}


public static double squareRoot(double newguess, double oldguess){
if (newguess == square(19)){
    return newguess;

}
else{
    newguess = (oldguess + (19/oldguess))/2;
    System.out.println(newguess);
    return squareRoot(newguess, oldguess);
}
}
}

Answer 1

因为oldguess是一个常数。因此，在第一次运行之后，此代码每次都会生成相同的结果：

newguess=（oldguess（19/oldguess））/2

你需要缓存new猜测在ther的开头，做你的计算，然后用new猜测中缓存的值替换old猜测。

像这样：

public static double squareRoot(double newguess, double oldguess) {
    double cache;
    if (newguess == square(19)) {
        return newguess;
    } else {
        cache = newguess;
        newguess = (oldguess + (19 / oldguess)) / 2;
        oldguess = cache;
        System.out.println(newguess);
        return squareRoot(newguess, oldguess);
    }
}