Java类型转换:long转double的内部处理机制

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资源摘要信息:"测试长整型转换为双精度浮点型时的参数类型" 在编程领域中,尤其是在处理数据类型转换时,我们经常会遇到各种不同的数据类型。在这个特定的例子中,我们遇到了标题为"Test_long_to_double.rar_ARGUMENT!"的资源文件,这暗示了一个测试用例,用以探讨在某种虚拟机环境下,将长整型(long)数据转换为双精度浮点型(double)的过程。 描述中提到的“Type of argument - double. Dalvik doesn't distinguish 64-bits types internally, so this conversion of double to double makes no sense but shall not crash the VM.”,意味着在Dalvik虚拟机环境下,尝试将一个64位的数据类型(在这里是double类型)转换为另一种64位的数据类型,这一转换行为在内部并没有实际的意义,因为它们本质上是相同的类型。然而,即使这种转换在逻辑上看似多余,Dalvik虚拟机(一种常用于Android平台的虚拟机)保证不会因此导致虚拟机崩溃。 在这里,我们需要详细探讨以下知识点: 1. 数据类型长整型(long)与双精度浮点型(double)的区别: - 长整型(long)是Java编程语言中的一种基本数据类型,用于存储64位的有符号整数。其取值范围为-2^63到2^63-1。 - 双精度浮点型(double)同样是Java中的一种基本数据类型,用于存储64位的双精度浮点数。它遵循IEEE 754标准,提供了比单精度(float)更高的精度和更大的数值范围。 2. 数据类型转换(Type Casting): - 数据类型转换分为两种:隐式(自动)转换和显式(强制)转换。 - 隐式转换发生在程序的变量或表达式需要不同数据类型时,如果转换不会导致数据精度损失,转换会自动进行。 - 显式转换则是程序员通过代码强制指定的转换,当数据类型的转换可能造成精度损失或者超出目标类型的范围时,需要进行显式转换。 3. Dalvik虚拟机: - Dalvik是Android平台的虚拟机,负责执行Android应用程序中的字节码。由于历史原因,Dalvik与Java虚拟机(JVM)相似但不完全相同。 - Dalvik虚拟机并不区分64位数据类型内部的差异,即使进行64位数据类型之间的转换,它也不会因为类型不匹配而崩溃。 4. 虚拟机(VM)的稳定性和异常处理: - 虚拟机的设计要考虑到稳定性和容错能力。一个设计良好的虚拟机在面对不合理的操作时,比如不必要的数据类型转换,应该有能力进行处理而不崩溃。 - 在Java和Android开发中,虚拟机的异常处理机制保障了程序在遇到错误时能够优雅地处理,例如通过抛出异常,而不是直接导致程序终止。 在压缩包子文件的文件名称列表中,我们看到了"Test_long_to_double.java"这个文件名。这是一个Java源代码文件,很可能包含了执行上述测试用例的代码。文件名中的"Test"表明这是一个测试类,而"long_to_double"则清楚地表示了测试的目标是关于长整型到双精度浮点型的转换。 总结来说,给定文件的信息涉及了编程中的数据类型转换,特别是长整型到双精度浮点型的转换在Dalvik虚拟机中的行为,以及虚拟机设计上的稳定性和异常处理机制。这些内容对于理解虚拟机运行时的类型系统和异常处理机制非常关键,对于设计和编写稳定高效的代码具有重要的意义。

帮我写出以下java代码:Add a class Bubble that extends Shape. The Bubble class has an instance variable called radius of type double that represents the radius of the bubble. The constructor of the Bubble class takes an x and a y as arguments, which represent the position of the new bubble. The radius of a new bubble is always 10 and never changes after that. The isVisible method indicates whether the bubble is currently visible inside a window of width w and height h (position (0, 0) is in the upper-left corner of the window). The bubble is considered visible if at least one pixel of the bubble is visible. Therefore a bubble might be visible even when its center is outside the window, as long as the edge of the bubble is still visible inside the window. The code of the isVisible method is a little bit complex, mostly because of the case where the center of the circle is just outside one of the corners of the window. So here is the code of the isVisible method, which you can directly copy-paste into your assignment: // Find the point (wx, wy) inside the window which is closest to the // center (x, y) of the circle. In other words, find the wx in the // interval [0, w - 1] which is closest to x, and find the wy in the // interval [0, h - 1] which is closest to y. // If the distance between (wx, wy) and (x, y) is less than the radius // of the circle (using Pythagoras's theorem) then at least part of // the circle is visible in the window. // Note: if the center of the circle is inside the window, then (wx, wy) // is the same as (x, y), and the distance is 0. public boolean isVisible(int w, int h) { double x = getX(); double y = getY(); double wx = (x < 0 ? 0 : (x > w - 1 ? w - 1 : x)); double wy = (y < 0 ? 0 : (y > h - 1 ? h - 1 : y)); double dx = wx - x; double dy = wy - y; return dx * dx + dy * dy <= radius * radius; } The isIn method indicates whether the point at coordinates (x, y) (which are the arguments of the method) is currently inside the bubble or not. The edge of the bubble counts as being inside of the bubble. HINT: use Pythagoras's theorem to compute the distance from the center of the bubble to the point (x, y). The draw method uses the graphics object g to draw the bubble. HINT: remember that the color of the graphics object g is changed in the draw method of the superclass of Bubble. Also add a testBubble method to test all your methods (including inherited methods, but excluding the isVisible method, which I provide, and excluding the draw method since it requires as argument a graphics object g that you

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