Rounding Errors?

In my course, I am told: Continuous values are represented approximately in memory, and therefore computing with floats involves rounding errors. These are tiny discrepancies in bit patterns; thus the test e==f is unsafe if e and f are floats. Referring to Java. Is this true? I've used comparison statements with double s and float s and have never had rounding issues. Never have I r

舍入错误?

在我的课程中,我被告知: 连续值大概在内存中表示,因此使用浮点计算涉及舍入误差。 这些是位模式中的微小差异; 因此如果e和f是浮动的,测试e==f是不安全的。 引用Java。 这是真的? 我用double s和float s来比较语句,并且从来没有过四舍五入的问题。 我从来没有读过类似的教科书。 虚拟机肯定会解决这个问题? 这是真的。 这是浮点值在内存中以有限的位数表示的固有限制。 例如,这个程序打印出“false”: pu

Why does changing the sum order returns a different result?

Why does changing the sum order returns a different result? 23.53 + 5.88 + 17.64 = 47.05 23.53 + 17.64 + 5.88 = 47.050000000000004 Both Java and JavaScript return the same results. I understand that, due to the way floating point numbers are represented in binary, some rational numbers (like 1/3 - 0.333333...) cannot be represented precisely. Why does simply changing the order of the el

为什么更改总和顺序会返回不同的结果?

为什么更改总和顺序会返回不同的结果? 23.53 + 5.88 + 17.64 = 47.05 23.53 + 17.64 + 5.88 = 47.050000000000004 Java和JavaScript都返回相同的结果。 我明白,由于浮点数的方式是用二进制表示的,所以有些有理数(比如1/3 - 0.333333 ...)不能精确表示。 为什么简单地改变元素的顺序会影响结果? 也许这个问题很愚蠢,但为什么简单地改变元素的顺序会影响结果呢? 它会根据数值的大小改变数值四舍五入的点。

Why does Math.round(0.49999999999999994) return 1?

In the following program you can see that each value slightly less than .5 is rounded down, except for 0.5 . for (int i = 10; i >= 0; i--) { long l = Double.doubleToLongBits(i + 0.5); double x; do { x = Double.longBitsToDouble(l); System.out.println(x + " rounded is " + Math.round(x)); l--; } while (Math.round(x) > i); } prints 10.5 rounded is 11 1

为什么Math.round(0.49999999999999994)返回1?

在下面的程序中,您可以看到每个略小于.5值向下舍入,除了0.5 。 for (int i = 10; i >= 0; i--) { long l = Double.doubleToLongBits(i + 0.5); double x; do { x = Double.longBitsToDouble(l); System.out.println(x + " rounded is " + Math.round(x)); l--; } while (Math.round(x) > i); } 版画 10.5 rounded is 11 10.499999999999998 rounded is 10 9.5 rounded is 10

Difference between C++ and Java compilation process

Possible Duplicate: Why does C++ compilation take so long? Hi, I searched in google for the differences between C++ and Java compilation process, but C++ and Java language features and their differences are returned. I am proficient in Java, but not in C++. But I fixed few bugs in C++. From my experience, I noticed that C++ always took more time to build compared to Java for minor chang

C ++和Java编译过程之间的区别

可能重复: 为什么C ++编译需要这么长时间? 嗨, 我在谷歌搜索了C ++和Java编译过程之间的差异,但是C ++和Java语言特性及其差异被返回。 我精通Java,但不是使用C ++。 但我修正了C ++中的一些错误。 根据我的经验,我注意到与C ++相比,C ++总是花费更多时间进行构建,以便进行小的更改。 问候巴拉 我脑海中出现了一些高级别的差异。 其中一些是概括性的,应该以“经常...”或“某些编译器......”为前缀,但为了

How safe is Internal Storage?

What I need: For Android: I need to save data permanently, but also be able to edit (and obviously read) it. This data should NOT be accessible by the user - it can contain things like a highscore, which must not be edited by the user. My Problem I would have (and have already) used Internal Storage , but I´m not sure how safe it actually is. Here it says: You can save files directly on

内部存储的安全性如何?

我需要的: 对于Android:我需要永久保存数据,但也能够编辑(并明显读取)它。 这些数据不应该由用户访问 - 它可以包含高分等内容,用户不得编辑。 我的问题 我会(并已经)使用Internal Storage ,但我不确定它实际上有多安全。 它在这里说: 您可以将文件直接保存在设备的内部存储器上。 默认情况下,保存到内部存储的文件对于您的应用程序是私人的,其他应用程序无法访问它们(用户也无法访问)。 当用户卸载您

Multiprocessing on Android

I've been executing some tests on Android in order to verify how good the performance of an algorithm (like FFT) can be improved if it is parallelized. I've implemented the algorithms by using pthread with JNI (FFTW) and Java threads (from JTransforms). Instead of getting a better performance by using threads as expected, I've got better results using serial algorithm. It is unclea

Android上的多重处理

我一直在Android上执行一些测试,以验证算法(如FFT)的性能如果能够并行化,可以提高性能。 我通过使用带有JNI(FFTW)和Java线程(来自JTransforms)的pthread来实现算法。 通过使用预期的线程来获得更好的性能,而不是使用串行算法获得更好的结果。 我不清楚为什么自从我在多核设备上执行这些测试后,我得到了这些结果。 看起来,Android系统使用的调度算法与Linux使用的调度算法有点不同,如果您想要使用多个CPU在Androi

Trilateration of 3 Calculated Distances from WiFI Strength Signals

I am using android to scan WIFI AP's every frame of time, I am getting from each AP the Strength of Signal (RSSI in dbm) and I am calculating the distance with this formula: public double calculateDistance(double levelInDb, double freqInMHz) { double exp = (32.44 - (20 * Math.log10(freqInMHz)) + Math.abs(levelInDb)) / 20.0; return Math.pow(10.0, exp); } That is working fine, So I ha

3 WiFI强度信号计算距离的三边测量

我正在使用android来扫描WIFI AP的每一帧时间,我从每个AP获得信号强度(RSSI in dbm),并使用此公式计算距离: public double calculateDistance(double levelInDb, double freqInMHz) { double exp = (32.44 - (20 * Math.log10(freqInMHz)) + Math.abs(levelInDb)) / 20.0; return Math.pow(10.0, exp); } 这是工作正常,所以我有三个或更多的距离,现在我需要在地图上绘制所有AP的固定位置,我做了一些在互联网上

Trilateration with distance measurement errors

In this link, the process of trilateration is given. And this is my java code for the localization process: public static double[] localize(final double[] p1, final double[] p2, final double[] p3, final double[] p4, final double r1, final double r2, final double r3, final double r4) { double[] ex = normalize(difference(p2,p1)); double i = dot(ex, difference(p3, p1)); double[] ey =

具有距离测量误差的三边测量

在这个环节中,给出了三边测量的过程。 这是我的本地化过程的Java代码: public static double[] localize(final double[] p1, final double[] p2, final double[] p3, final double[] p4, final double r1, final double r2, final double r3, final double r4) { double[] ex = normalize(difference(p2,p1)); double i = dot(ex, difference(p3, p1)); double[] ey = normalize(difference(difference(p3,p1),

Find location of an object using 2D trilateration

I am trying to use trilateration in a program to simulate how it would be in real life. I have 3 transmitters which send out a signal that gets weaker after each square that the signal moves. I am using a 2D array and code that generates a crude circle of numbers. This shows one of the transmitters 00000000000000000000 00000000000000000000 00000000000000000000 00000000001000000000 00000000112

使用2D三边测量查找对象的位置

我试图在程序中使用三边测试来模拟它在现实生活中的情况。 我有3个发射器发出一个信号,在信号移动的每个正方形后变弱。 我正在使用一个2D数组和代码来生成一个粗略的数字圆。 这显示了其中一个发射器 00000000000000000000 00000000000000000000 00000000000000000000 00000000001000000000 00000000112110000000 00000001223221000000 00000001234321000000 00000012344432100000 00000001234321000000 0000000122322100000

Project: Creating a Chess Game with GUI

I'm a fresh programmer with limited experience looking to expand upon it. This is my first project I will be attempting outside of school so my resources are rather fluctuated due to me simply not knowing what it is I need to know. However, I have some small foundation regarding basic GUI using Java. I want to create a chess game, and later implement an AI. At the moment however, I am ju

项目:使用GUI创建国际象棋游戏

我是一位有经验的新程序员,希望能够扩展它。 这是我在校外尝试的第一个项目,所以我的资源很波动,因为我根本不知道我需要知道什么。 但是,对于使用Java的基本GUI,我有一些小的基础。 我想创建一个国际象棋游戏,然后实现一个AI。 但是,现在我只是在构建游戏本身。 现在,我想要一个简单的GUI框架,它不必看起来漂亮或复杂。 只需显示棋盘,棋子在哪里,点击棋子,它会突出显示你可以走到哪里,然后点击广场,瞧,移