I was reading the book Linux Kernel Development on Chapter Process Scheduling. On Page 61, section Waking Up, the first paragraph reads:

Waking is handled via wake_up(), which wakes up all the tasks waiting on the given wait queue. It (Q1:what does this it refer to?) calls try_to_wake_up(), which sets the task’s(Q2:which task? all awoken tasks?) state to TASK_RUNNING, calls enqueue_task() to add the task to the red-black tree, and sets need_resched if the awakened task’s priority is higher than the priority of the current task.The code that causes the event to occur typically calls wake_up() itself. For example, when data arrives from the hard disk, the VFS calls wake_up() on the wait queue that holds the processes waiting for the data.

I am quite confused about the above. Let me just use the example in the above paragraph, i.e., the disk interrupted after reading data, but with a more complete picture. Please correct me if any of the following is wrong or incomplete:

  1. Some user process issued a blocking read operation, triggering a sys call and the process is in the realm of kernel.

  2. Kernel sets up the disk controller requesting the needed data and puts this process on sleep (this process is put into a wait queue). Kernel schedules another process to run.

  3. Disk interrupt occurs. CPU suspends the current executing process and jumps to the disk interrupt handling.

  4. Disk controller will kick in sometime during the interrupt handling to transfer the data read from disk to main memory (either under the direction of CPU or by DMA)

  5. (Not sure, please correct) As the paragraph says, VFS calls wake_up() on the wait queue that holds the processes waiting for the data.

My specific questions are the following:

Q1 (refer to the quoted paragraph): I assume the It in the second sentence refers to the function wake_up(). Why does the function wake_up wakes up all tasks instead of just the one waiting for this disk data?

Q2 (refer to the quoted paragraph): Does try_to_wake_up() somehow knows the specific task whose state needs to be set to TASK_RUNNING? Or try_to_wake_up() sets all awoken tasks' state to TASK_RUNNING?

Q3: How many wait queues are there for the kernel to manage? If there are more than 2 such wait queues, how does the kernel know which queue to select, such that the process waiting for the disk data is on that wait queue?

Q4: Now say we know the queue where the waiting process is on. How does the kernel know which process is waiting for the data from the disk. I can only image that some info specific to the process requesting the disk data is passed to the disk controller, like the process's PID, memory address or something. Then upon completing the interrupt handling, the disk controller(or kernel?) uses this info to pinpoint the process on the wait queue.

Please help me complete this picture of process wake_up! Thanks!

3 Answers 3


Q1: “It” is wake_up. It wakes up all tasks that are waiting for the disk data. If they weren't waiting for that data, they wouldn't be waiting on that queue.

Q2: I'm not sure I understand the question. Each wake queue entry contains a pointer to the task. try_to_wake_up receives a pointer to the task that it's supposed to wake up. It is called once per function.

Q3: There are lots of wait queues. There's one for every event that can happen. The disk driver sets up a wait queue for each request to the disk. For example, when the filesystem driver wants the content of a certain disk block, it asks the disk driver for that block, and then the request starts with the task that made the filesystem request. Other entries may be added to the wait queue if another request for the same block comes in while this one is still outstanding.

When an interrupt happens, the disk driver determines which disk has data available from the information passed by the hardware and looks up the data structure that contains the kernel data for this disk to find which request was to be filled. In this data structure, among others, are the location where the data is to be written and the corresponding wake queue indicating what to do next.

Q4: The process makes a system call, let's say to read a file. This triggers some code in the filesystem driver which determines that the data needs to be loaded from the disk. That code makes a request to the disk driver and adds the calling process to the request's wait queue. (There are actually more layers than that, but you get the idea.) When the disk read completes, the wait queue event triggers, and the process is thus removed from the disk's wait queue. The code triggered by the wait queue event is a function supplied by the filesystem driver, which copies the data to the process's memory and causes the read system call to return.

  • I assume there can be multiple processes on the disk's wait queue. When the disk read completes, which process is removed from the disk's wait queue? The front of the queue?
    – Rich
    Mar 6, 2016 at 20:35
  • @Rich All of them. Well, all the ones that have their read completed. I don't know how disk drivers are organized exactly: whether there's a single wait queue for the disk (with all processes that are waiting for the disk) or one per request (each with the processes that are waiting for that particular sector). Mar 6, 2016 at 20:37
  • OH? But what if the data read from disk was specific for only one of the process on the wait queue?
    – Rich
    Mar 6, 2016 at 20:38
  • So it is the disk driver's job to figure out which processes' request are fulfilled? Could it happen that only some of the processes waiting on the wait queue can be woken up, due to other processes' requests not completely fulfilled? I guess my general confusion is that wake_up wakes up all processes but in the end it is possible that only a subset of processes are truly runnable.
    – Rich
    Mar 6, 2016 at 20:50
  • @Rich There are many conditions that can make a process not runnable, for example if it's suspended. I don't know the Linux kernel well enough to know whether they'd be removed from the disk's wait queue in that case; I think they generally are, because otherwise the queue and all similar queues would tend to fill up with processes that can't be woken up and that would hurt performance. Generally speaking, as an OS designer, you don't want processes to be woken up unnecessarily; we try to have that happen only on race conditions (when the reason for the wake up has just now disappeared). Mar 6, 2016 at 21:02

To build on Gilles’s answer,

Q2: I’m not sure, but I would interpret the passage from the book to mean that the interrupt handler calls wake_up() once (passing the queue identifier as an argument), and wake_up() calls try_to_wake_up() for each process that is on that queue.  (This reiterates Gilles’s answer to Q1: it doesn’t wake all tasks, just the ones that are on the wait queue associated with the event that called wake_up().)

Q3: Each wait queue is “owned” by some piece of kernel code — mostly device drivers, some others.  The routine that owns a queue assigns it a unique identifier, based on some unique characteristic of the event that the queue is for.  When it (the driver/other module) puts a process to sleep, it specifies (by ID) what queue to put it on.  The routine that calls wake_up() (typically an interrupt handler) must be part of the same module that put the process to sleep, so it knows the identifier for the queue that corresponds to the event that happened.

Last time I looked at Unix kernel source code (which was many years ago), the disk drivers had a different event ID for each I/O request. As Gilles says, multiple processes can be waiting for the same event if they are reading the same file at the same time.  (This, of course, also relates to Q1.)

Q4: When I hear the phrase “disk controller”, I think of hardware.  But, aside from that, you’re right; the disk driver (a software module in the kernel) has (at least potentially) access to all information about any process that invokes it (i.e., by doing disk I/O).  So, when the disk driver puts a process to sleep because it has initiated a physical I/O that takes time to complete, it (the driver) puts “the process’s PID, memory address or something” into the wait queue.  Whatever this is, it is enough for try_to_wake_up() to wake the process.

The last sentence of the passage that you quoted says, “… VFS calls wake_up() on the wait queue …”.  I question whether this is literally accurate.  Filesystem code is a layer above the disk driver.  I would expect the interrupt (a signal from the disk hardware to the CPU) to be handled by the disk interrupt handler (part of the disk driver) which would wake up the process(es) that are waiting (by calling wake_up()).  The driver would subsequently wake the filesystem code.  (This terminology might be imprecise, too.  It might be better to say that the driver does something to allow the filesystem code to resume processing.)  The filesystem code might then return to the user process, or it might invoke the disk driver again, resulting in the process being put to sleep again.

I quibble with your step #4.  If a device is using DMA, it won’t interrupt until after the data transfer has completed.


The problem you cite, waking up all processes waiting is what is called the "thundering herd" problem (many processes are woken up when a resource becomes available, they fight over which one gets exclusive access to the resource, the others go back to sleep). This becomes a problem when there are many processors, and thus there are many processors fighting here. Newer versions of Linux solve this by just waking up one of the processes waiting.

Most of the time there will be one (or at just a few) processes waiting for a particular event (there isn't a fixed, much less small, number of events on which a process can be waiting).

Gilles' answer goes through your points one by one, not much to add to that here.

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