android - 必须始终到达每个 WakeLock.release() 调用吗？安卓 Lint 警告

Question

Android lint 对以下 [Wakelock] 发出警告:

public static void acquire(Context ctx, long timeout) {
    if (wakeLock != null) {
        wakeLock.release();
    }

    PowerManager powerManager
        = (PowerManager) ctx.getSystemService(Context.POWER_SERVICE);
    wakeLock = powerManager.newWakeLock(PowerManager.FULL_WAKE_LOCK |
                    PowerManager.ACQUIRE_CAUSES_WAKEUP |
                    PowerManager.ON_AFTER_RELEASE, 
                    Common.TAG);
    wakeLock.setReferenceCounted(false);

    if ( timeout <= 0 ) {
        wakeLock.acquire();
    } else {
        wakeLock.acquire(timeout);
    }
}

public static synchronized void release() {
    if ( wakeLock != null ) {
        if ( wakeLock.isHeld() ) {
            wakeLock.release(); 
        }
        wakeLock = null;
    }
}

第一次出现时给出警告

[lint] [...]/WakeLocker.java: Warning: The release() call is not always reached [Wakelock]

然而，它并不是真的需要每次都释放，因为有超时。

将其包装在 try-catch block 中的默认解决方案，例如 android-wakelock-not-released-after-getactivenetworkinfo ，或在 @rainash's answer below , 不解决导致使用这种方法的问题，即 that the device can go back to sleep .

这是如何解决的？还是应该忽略它？

Answer 1

使用 try-catch block 包装你在设备唤醒时执行的任务

PowerManager pm = (PowerManager) context.getSystemService(Context.POWER_SERVICE);
    final WakeLock wakeLock = pm.newWakeLock(PowerManager.PARTIAL_WAKE_LOCK, "tag");
    wakeLock.acquire();
            try {
                do something
            } finally {
                wakeLock.release();
            }
    });