System.nanotime给出负数

Question

我写了一个使用System.nanotime的安卓定时器应用。问题是它给了我预测不足的结果和负数。在摄影机的每个帧上更新redVal、blueVal和greenVal。

结果

504455566
-95947265
9063721
61035
-99487305
-98937988
12664795
-75317382

代码

        for (testAmount = 0; testAmount < 80; testAmount++) {
            runOnUiThread(new Runnable() {
                public void run() {
                    lagSquare.setBackgroundColor(Color.rgb(255, 255, 255));
                    lagStartTime = System.nanoTime(); //start lagTimer start
                }
            });
            while (redVal <= 100.0 && blueVal <= 100.0 && greenVal <= 100.0) {
                x=0;
            }
            runOnUiThread(new Runnable() {
                public void run() {
                    lagEndTime = System.nanoTime(); //start lagTimer end
                    lagSquare.setBackgroundColor(Color.rgb(000, 000, 000));//set lagSquare black
                }
            });
            lagTimeResult = (lagEndTime - lagStartTime);
            timeArray[testAmount] = lagTimeResult;
            Log.i("LTR", String.valueOf(lagTimeResult));
            try {
                Thread.sleep(60);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }

Answer 1

您正在尝试输出一个时间差，该时间差依赖于在不同线程中设置的值，而不需要任何同步。这几乎总是以错误的值结束：

for (testAmount = 0; testAmount < 80; testAmount++) {

            // this will schedule the Runnable to run *sometime* in
            // the future - but not necessarily right now
            runOnUiThread(new Runnable() {
                public void run() {
                    lagStartTime = System.nanoTime(); //start lagTimer start
                }
            });

            // this will also schedule this Runnable to run *sometime* in
            // the future - but not necessarily after the first one
            runOnUiThread(new Runnable() {
                public void run() {
                    lagEndTime = System.nanoTime(); //start lagTimer end
                }
            });

            // this will probably execute first (before either of
            // the other Runnables have completed)
            lagTimeResult = (lagEndTime - lagStartTime);
        }

您不能简单地依赖于线程按照您编写的顺序执行的顺序--而且绝对不能在循环中执行。我不能从你的问题中理解你在尝试计时，但是带回家的规则是，当你有多个线程时，你可以在不使用某种形式的同步的情况下执行的顺序。