如何使用 float/double 的模数?

发布于 2024-09-04 01:56:40 字数 304 浏览 2 评论 0原文

我正在为学校项目创建 RPN 计算器,但在使用模数运算符时遇到了问题。由于我们使用的是双精度数据类型,因此模数不适用于浮点数。例如,0.5 % 0.3 应该返回 0.2,但我遇到了被零除的异常。

该指令要求使用fmod()。我到处寻找fmod(),包括javadoc,但我找不到它。我开始认为这是我必须创建的一种方法?

编辑:嗯,奇怪。我刚刚再次插入这些数字,它似乎工作正常......但以防万一。使用浮点类型时,我需要注意 Java 中 mod 运算符的使用吗?我知道这样的事情不能在 C++ 中完成(我认为)。

I'm creating an RPN calculator for a school project and having trouble with the modulus operator. Since we're using the double data type, modulus won't work on floating-point numbers. For example, 0.5 % 0.3 should return 0.2, but I'm getting a division by zero exception.

The instruction says to use fmod(). I've looked everywhere for fmod(), including javadoc, but I can't find it. I'm starting to think it's a method I'm going to have to create?

Edit: Hmmm, strange. I just plugged in those numbers again and it seems to be working fine… but just in case. Do I need to watch out for using the mod operator in Java when using floating types? I know something like this can't be done in C++ (I think).

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(4

南汐寒笙箫 2024-09-11 01:56:40

当你第一次运行它时,你可能犯了一个错字。

计算 0.5 % 0.3 按预期返回“0.2”(双精度)。

Mindprod 很好地概述了 Java 中模数的工作原理

You probably had a typo when you first ran it.

evaluating 0.5 % 0.3 returns '0.2' (A double) as expected.

Mindprod has a good overview of how modulus works in Java.

萌︼了一个春 2024-09-11 01:56:40

与 C 不同,Java 允许对整数和浮点数使用 %,并且(与 C89 和 C++ 不同)它对所有输入(包括负数)都有明确定义:

来自 JLS §15.17.3

浮点数的结果
余数运算由下式确定
IEEE算术规则:

  • 如果任一操作数为 NaN,则结果为 NaN。
  • 如果结果不是 NaN,则结果的符号等于
    股息。
  • 如果被除数为无穷大,或除数为零,或两者皆为,则
    结果为 NaN。
  • 如果被除数是有限的并且除数是无穷大,则结果
    等于股息。
  • 如果被除数为零且除数有限,则结果
    等于股息。
  • 在其余情况下,既不是无穷大,也不是零,也不是
    涉及到NaN,浮点数
    a 除法的余数 r
    定义 n 除以除数 d
    由数学关系 r=n-(d·q)
    其中 q 是负整数
    仅当 n/d 为负且为正时
    仅当 n/d 为正且其
    幅度尽可能大
    不超过的幅度
    n 和 d 的真实数学商。

因此,对于您的示例,0.5/0.3 = 1.6...。 q 与 0.5(被除数)同号(正),大小为 1(最大大小不超过大小 1.6...的整数),r = 0.5 - (0.3 * 1) = 0.2

Unlike C, Java allows using the % for both integer and floating point and (unlike C89 and C++) it is well-defined for all inputs (including negatives):

From JLS §15.17.3:

The result of a floating-point
remainder operation is determined by
the rules of IEEE arithmetic:

  • If either operand is NaN, the result is NaN.
  • If the result is not NaN, the sign of the result equals the sign of
    the dividend.
  • If the dividend is an infinity, or the divisor is a zero, or both, the
    result is NaN.
  • If the dividend is finite and the divisor is an infinity, the result
    equals the dividend.
  • If the dividend is a zero and the divisor is finite, the result
    equals the dividend.
  • In the remaining cases, where neither an infinity, nor a zero, nor
    NaN is involved, the floating-point
    remainder r from the division of a
    dividend n by a divisor d is defined
    by the mathematical relation r=n-(d·q)
    where q is an integer that is negative
    only if n/d is negative and positive
    only if n/d is positive, and whose
    magnitude is as large as possible
    without exceeding the magnitude of the
    true mathematical quotient of n and d.

So for your example, 0.5/0.3 = 1.6... . q has the same sign (positive) as 0.5 (the dividend), and the magnitude is 1 (integer with largest magnitude not exceeding magnitude of 1.6...), and r = 0.5 - (0.3 * 1) = 0.2

妄司 2024-09-11 01:56:40

我认为常规模数运算符在 Java 中可以实现此目的,但编码并不难。只需将分子除以分母,然后取结果的整数部分。将其乘以分母,然后用分子减去结果。

x = n/d
xint = Integer portion of x
result = n - d*xint

I thought the regular modulus operator would work for this in Java, but it can't be hard to code. Just divide the numerator by the denominator, and take the integer portion of the result. Multiply that by the denominator, and subtract the result from the numerator.

x = n/d
xint = Integer portion of x
result = n - d*xint
合约呢 2024-09-11 01:56:40

fmod 是用于处理浮点模数的标准 C 函数;我想你的消息来源是说 Java 处理浮点模数的方式与 C 的 fmod 函数相同。在 Java 中,您可以像整数一样对双精度数使用 % 运算符:

int x = 5 % 3; // x = 2
double y = .5 % .3; // y = .2

fmod is the standard C function for handling floating-point modulus; I imagine your source was saying that Java handles floating-point modulus the same as C's fmod function. In Java you can use the % operator on doubles the same as on integers:

int x = 5 % 3; // x = 2
double y = .5 % .3; // y = .2
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文