java - 在java中模拟具有两个最大幂的无符号数

Question

我有一些来自 (C++) 应用程序的输出，该应用程序以一种在 2³³ 处回绕为零的类型存储一个 tickcount 值。 (8,589,934,592) (没有密码)

我需要以同样的方式编写自己的输出。我通过 JNA 从 C 库中检索了滴答计数，但是如果我将它存储在一个 int 中，它会在约 25 天的滴答计数后包装到 -2³¹ (-2,147,483,648) 并且如果我将它存储在一个 long它一直超过 2³³。

如何在 Java 中存储(或写入)值，使其在 2³³ 处回绕(归零)？

(最好是 JRE7- & JRE8- on Win32- & Win64- & Linux-compatible solution)

要获取滴答计数，我使用以下命令:

import com.sun.jna.*;
public interface Kernel32 extends Library {
    Kernel32 INSTANCE = (Kernel32) Native.loadLibrary((Platform.isWindows() ? "kernel32" : "c"), Kernel32.class);

    /**
     * Retrieves the number of milliseconds that have elapsed since the system was started.
     *
     * @return number of milliseconds that have elapsed since the system was started.
     * @see http://msdn2.microsoft.com/en-us/library/ms724408.aspx
     */
    Long GetTickCount();
}

public interface Clib extends Library {
    Clib INSTANCE = (Clib) Native.loadLibrary((Platform.isWindows() ? "kernel32" : "c"), Clib.class);

    /**
     * Retrieves the number of milliseconds that have elapsed since the system was started.
     *
     * @return number of milliseconds that have elapsed since the system was started.
     */
    Long clock();
}

// And later
Number n = (Platform.isWindows() ? Kernel32.INSTANCE.GetTickCount() : Clib.INSTANCE.clock()).toString());

Answer 1

您可以将值存储在 long 中，然后通过执行以下操作将值截断为 33 位(将 2³³ 环绕为 0):

n &= (1L << 33) - 1;

这与以下内容完全相同:

n &= 0x1_ffff_ffffL;

也等同于:

n &= 8_589_934_591L;