You discussed that kernel do not keep track of SS for process specific kernel stack as it always starts from fixed offset of task_struct, but does that mean, linux kernel do not make use of SS element in the TSS of process. I think the kernel can not by-pass the rules defined by processor. Processor expects the SS and SP elements to be right at the time of stack switching, so we need to initialize them to the right values while forking a process. I think kernel definitely keep tracks of SS and SP for all CPU levels including kernel mode also (ring 0), so that when we are switching form user space to kernel space the processor can switch the stacks automatically. At stack switching time CPU expects the SS and SP elements of TSS to be right, it's just going to copy those values in SS and SP registers of CPU, so that now we can push and pop the things on the kernel stack. Yes it definitely can happen (and I think this is the way to do it) that SS in process's TSS is initialized to the memory address just next to task_struct (as the kernel stack can grow downwards till this limit) while the forking of a process and also sets the SP to the fixed offset from task_struct from where the kernel stack starts growing downwards (as you mentioned kernel stack starts from there). Now whenever this process will be scheduled by the scheduler or a process enters the kernel mode, CPU's SS and SP registers are re-set to the SS and SP elements in TSS of that process. At the time of scheduling, scheduler also change the TSS descriptor for that CPU on which the process is going to be scheduled. It changes this TSS descriptor in GDT to point to the TSS of the selected process. Correct me if I am wrong. Cheers !! Gaurav -----Original Message----- From: Jan Hudec [mailto:bulb@xxxxxxxxxxxxxxxxxx] On Behalf Of Jan Hudec Sent: Tuesday, October 12, 2004 6:42 PM To: aq Cc: suthambhara nagaraj; Dhiman, Gaurav; main kernel; kernel Subject: Re: Kernel stack On Tue, Oct 12, 2004 at 21:30:54 +0900, aq wrote: > > > >From what you all discuss, I can say: kernel memory is devided into 2 > > > part, and the upper part are shared between processes. The below part > > > (the kernel stack, or 8K traditionally) is specifict for each process. > > > > > > Is that right? > > > > No, it's not. There is just one kernel memory. In it each process has > > it's own task_struct + kernel stack (by default 8K). There is no special > > address mapping for these, nor are they allocated from a special area. > > > > When a context of some process is entered, esp is pointed to the top of > > it's stack. That's exactly all it takes to exchange stacks. > > OK, lets say there are 20 processes running in the system. Then the > kernel must allocate 20 * 8K = 160K just for the stacks of these > processes. All of these 160K always occupy the kernel (kernel memory > is never swapped out). When a process actives, ESP would switch to > point to the corresponding stack (of that process). This is correct. > The remainding memory of kernel therefore is equally accessible to all > the processes. This is not. There is nothing like "remaining memory". **ALL* kernel memory is equally accessible to all the processes. There is noting special about the stacks and task-structs. They are normal 8K structures somewhere in kernel memory. > Is that correct ? ------------------------------------------------------------------------ ------- Jan 'Bulb' Hudec <bulb@xxxxxx> -- Kernelnewbies: Help each other learn about the Linux kernel. Archive: http://mail.nl.linux.org/kernelnewbies/ FAQ: http://kernelnewbies.org/faq/
