SynchronizedRGB是表示颜色的类，每一个对象代表一种颜色，使用三个整形数表示颜色的三基色，字符串表示颜色名称。
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public class SynchronizedRGB {
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// Values must be between 0 and 255.
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private int red;
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private int green;
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private int blue;
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private String name;
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private void check(int red,
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int green,
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int blue) {
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if (red < 0 || red > 255
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|| green < 0 || green > 255
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|| blue < 0 || blue > 255) {
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throw new IllegalArgumentException();
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}
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}
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public SynchronizedRGB(int red,
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int green,
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int blue,
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String name) {
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check(red, green, blue);
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this.red = red;
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this.green = green;
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this.blue = blue;
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this.name = name;
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}
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public void set(int red,
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int green,
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int blue,
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String name) {
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check(red, green, blue);
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synchronized (this) {
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this.red = red;
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this.green = green;
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this.blue = blue;
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this.name = name;
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}
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}
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public synchronized int getRGB() {
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return ((red << 16) | (green << 8) | blue);
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}
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public synchronized String getName() {
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return name;
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}
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public synchronized void invert() {
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red = 255 - red;
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green = 255 - green;
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blue = 255 - blue;
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name = "Inverse of " + name;
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}
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} |
使用SynchronizedRGB时需要小心，避免其处于不一致的状态。例如一个线程执行了以下代码：
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SynchronizedRGB color = |
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new SynchronizedRGB(0, 0, 0, "Pitch Black");
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... |
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int myColorInt = color.getRGB(); //Statement 1
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String myColorName = color.getName(); //Statement 2
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如果有另外一个线程在Statement 1之后、Statement 2之前调用了color.set方法，那么myColorInt的值和myColorName的值就会不匹配。为了避免出现这样的结果，必须要像下面这样把这两条语句绑定到一块执行：
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synchronized (color) {
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int myColorInt = color.getRGB();
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String myColorName = color.getName();
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} |
这种不一致的问题只可能发生在可变对象上。
文章转自 并发编程网-ifeve.com
时间： 2024-09-20 03:48:12