Java：简单的递归函数仅返回1

发布于 2024-09-25 06:58:15 字数 600 浏览 0 评论 0原文

奇怪的是，这是我的第一个 Java 应用程序，我想实现一个任意精度的阶乘函数，我很好地完成了递归函数，但我的迭代函数仅输出“1”，没有其他内容。对我来说太晚了，我无法找出原因，我不确定我哪里出错了，这里有明显的东西吗？

public static BigInteger ifact(BigInteger n) {
    BigInteger ret = new BigInteger("1");
    BigInteger i = new BigInteger("1");
    for(i = new BigInteger("0"); i.compareTo(n) == 0; i.add(new BigInteger("1"))){
        ret.multiply(i);
    }
    return ret;
}

如果您没有注意到它使用 BigInteger 包，那么就会出现奇怪的文字。

另外，像 C 一样，您可以做一些类似于 typedef 的事情，这样我就不需要每次都输入“BigInteger”吗？

编辑：我想我的意思是将ret设置为n，可能是这样，也可能不是。

原文

This is oddly my first Java application, I was wanting to implement an arbitrary precision factorial function, I did the recursive one fine but my iterative one simply outputs "1" and nothing else.. It is so late for me I can not spot why, I am not sure where I went wrong, is there something obvious here?

public static BigInteger ifact(BigInteger n) {
    BigInteger ret = new BigInteger("1");
    BigInteger i = new BigInteger("1");
    for(i = new BigInteger("0"); i.compareTo(n) == 0; i.add(new BigInteger("1"))){
        ret.multiply(i);
    }
    return ret;
}

In case you did not notice it uses the BigInteger package, hense the odd writings..

Also, like C, can you do something similar to typedef so I don't need to type "BigInteger" each time?

EDIT: I think I meant to set ret as n, that may be it, or..maybe not.

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路还长，别太狂 2024-10-02 06:58:15

回答你的“typedef”问题，不，你不能在Java中做到这一点。

至于你的代码，它有很多错误。你的循环只会执行一次，并且仅当 n 等于 i 时。为什么在循环开始时将 i 初始化为 1，然后初始化为 0？为什么忽略i.add(...)的结果？为什么忽略 ret.multiply(i) 的结果？请记住，这些方法不会改变 BI 本身！

这包含了这些变化：

    BigInteger n = BigInteger.valueOf(10);
    BigInteger ret = BigInteger.ONE;
    for (BigInteger i = BigInteger.ONE; i.compareTo(n) <= 0; i = i.add(BigInteger.ONE)) {
        ret = ret.multiply(i);
    }
    System.out.println(ret);

In answer to your "typedef" question, no, you can't do that in Java.

As for your code, it has numerous mistakes. Your loop will only ever execute once, and only if n is equal to i. Why do you initialize i to 1 and then 0 at the beginning of the loop? Why do you ignore the result of i.add(...)? Why do you ignore the result of ret.multiply(i)? Remember, these methods don't change the BI itself!

This incorporates those changes:

    BigInteger n = BigInteger.valueOf(10);
    BigInteger ret = BigInteger.ONE;
    for (BigInteger i = BigInteger.ONE; i.compareTo(n) <= 0; i = i.add(BigInteger.ONE)) {
        ret = ret.multiply(i);
    }
    System.out.println(ret);

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極樂鬼 2024-10-02 06:58:15

尝试更改

ret.multiply(i);

为

ret = ret.multiply(i);

Try changing

ret.multiply(i);

ret = ret.multiply(i);

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陪你到最终 2024-10-02 06:58:15

BigInteger 对象是不可变的（意味着您无法更改它）。这样，它就像一个 String 对象。因此，为了使其发挥作用，您需要将线路更改

ret.multiply(i);

为

ret = ret.multiply(i);

The BigInteger object is immutable (meaning you can't change it). In this way, it's like a String object. So, for this to work, you would want to change your line

ret.multiply(i);

ret = ret.multiply(i);

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翻了热茶 2024-10-02 06:58:15

public static BigInteger ifact(BigInteger n) {
  BigInteger ret = BigInteger.ONE;
  for (BigInteger i = BigInteger.ONE; i.compareTo(n) <= 0; i = i.add(BigInteger.ONE)) {
    ret = ret.multiply(i);
  }
  return ret;
}

ret = ret.multiply
i.compareTo(n) <= 0
BigInteger.ONE;

public static BigInteger ifact(BigInteger n) {
  BigInteger ret = BigInteger.ONE;
  for (BigInteger i = BigInteger.ONE; i.compareTo(n) <= 0; i = i.add(BigInteger.ONE)) {
    ret = ret.multiply(i);
  }
  return ret;
}