While working on a educational simplistic RISC processor I was wondering about how system calls work when implementing my software interrupt function. For example, hypothetically lets say our program calls sys_end which ends the current process. Now I know this would go to a vector table and then to the code to end the current process.
My question is the code that ends the process ran in supervisor mode or user mode? No where I seem to look specifies this. I'm assuming if its in normal user mode that could pose a very significant problem as a user mode process could do say do something evil like:
for (i=0; i++; i<10000){
int sys_fork //creates child process
}
which could be very bad I thought the OS would have some say on how many times a process could repeat itself and not to mention what other harmful things a process could do by changing the code in the system call itself.
system calls run in supervisor mode for the duration of the system call. The supervisor mode is necessary for accessing hardware (the screen, the keyboard), and for keeping user processes isolated from each other.
There are (or can be configured) limits on the amount of cpu, number of processes, etc. a user process may use or request, which can offer some protection against the kind of runaway program you describe.
But the default linux configuration allows 10k processes to be created in a tight loop; I've done it myself (both intentionally and accidentally)
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I'm learning computer organization and structure (I'm using Linux OS with x86-64 architecture). we've studied that when an interrupt occurs in user mode, the OS is notified and it switches between the user stack and the kernel stack by loading the kernels rsp from the TSS, afterwards it saves the necessary registers (such as rip) and in case of software interrupt it also saves the error-code. in the end, just before jumping to the adequate handler routine it zeroes the TF and in case of hardware interrupt it zeroes the IF also. I wanted to ask about few things:
the error code is save in the rip, so why loading both?
if I consider a case where few interrupts happen together which causes the IF and TF to turn on, if I zero the TF and IF, but I treat only one interrupt at a time, aren't I leave all the other interrupts untreated? in general, how does the OS treat few interrupts that occur at the same time when using the method of IDT with specific vector for each interrupt?
does this happen because each program has it's own virtual memory and thus the interruption handling processes of all the programs are unrelated? where can i read more about it?
how does an operating system keep other necessary progresses running while handling the interrupt?
thank you very much for your time and attention!
the error code is save in the rip, so why loading both?
You're misunderstanding some things about the error code. Specifically:
it's not generated by software interrupts (e.g. instructions like int 0x80)
it is generated by some exceptions (page fault, general protection fault, double fault, etc).
the error code (if used) is not saved in the RIP, it's pushed on the stack so that the exception handler can use it to get more information about the cause of the exception
2a. if I consider a case where few interrupts happen together which causes the IF and TF to turn on, if I zero the TF and IF, but I treat only one interrupt at a time, aren't I leave all the other interrupts untreated?
When the IF flag is clear, mask-able IRQs (which doesn't include other types of interrupts - software interrupts, exceptions) are postponed (not disabled) until the IF flag is set again. They're "temporarily untreated" until they're treated later.
The TF flag only matters for debugging (e.g. single-step debugging, where you want the CPU to generate a trap after every instruction executed). It's only cleared in case the process (in user-space) was being debugged, so that you don't accidentally continue debugging the kernel itself; but most processes aren't being debugged like this so most of the time the TF flag is already clear (and clearing it when it's already clear doesn't really do anything).
2b. in general, how does the OS treat few interrupts that occur at the same time when using the method of IDT with specific vector for each interrupt? does this happen because each program has it's own virtual memory and thus the interruption handling processes of all the programs are unrelated? where can i read more about it?
There's complex rules that determine when an interrupt can interrupt (including when it can interrupt another interrupt). These rules mostly only apply to IRQs (not software interrupts that the kernel won't ever use itself, and not exceptions which are taken as soon as they occur). Understanding the rules means understanding the IF flag and the interrupt controller (e.g. how interrupt vectors and the "task priority register" in the local APIC influence the "processor priority register" in the local APIC, which determines which groups of IRQs will be postponed when the IF flag is set). Information about this can be obtained from Intel's manuals, but how Linux uses it can only be obtained from Linux source code and/or Linux specific documentation.
On top of that there's "whatever mechanisms and practices the OS felt like adding on top" (e.g. deferred procedure calls, tasklets, softIRQs, additional stack management) that add more complications (which can also only be obtained from Linux source code and/or Linux specific documentation).
Note: I'm not a Linux kernel developer so can't/won't provide links to places to look for Linux specific documentation.
how does an operating system keep other necessary progresses running while handling the interrupt?
A single CPU can't run 2 different pieces of code (e.g. an interrupt handler and user-space code) at the same time. Instead it runs them one at a time (e.g. runs user-space code, then switches to an IRQ handler for very short amount of time, then returns to the user-space code). Because the IRQ handler only runs for a very short amount of time it creates the illusion that everything is happening at the same time (even though it's not).
Of course when you have multiple CPUs, different CPUs can/do run different pieces of code at the same time.
I know that an interrupt causes the OS to change a CPU from its current task and to run a kernel routine. I this case, the system has to save the current context of the process running on the CPU.
However, I would like to know whether or not a context switch occurs when any random process makes a system call.
I would like to know whether or not a context switch occurs when any random process makes a system call.
Not precisely. Recall that a process can only make a system call if it's currently running -- there's no need to make a context switch to a process that's already running.
If a process makes a blocking system call (e.g, sleep()), there will be a context switch to the next runnable process, since the current process is now sleeping. But that's another matter.
There are generally 2 ways to cause a content switch. (1) a timer interrupt invokes the scheduler that forcibly makes a context switch or (2) the process yields. Most operating systems have a number of system services that will cause the process to yield the CPU.
well I got your point. so, first I clear a very basic idea about system call.
when a process/program makes a syscall and interrupt the kernel to invoke syscall handler. TSS loads up Kernel stack and jump to syscall function table.
See It's actually same as running a different part of that program itself, the only major change is Kernel play a role here and that piece of code will be executed in ring 0.
now your question "what will happen if a context switch happen when a random process is making a syscall?"
well, nothing will happen. Things will work in same way as they were working earlier. Just instead of having normal address in TSS you will have address pointing to Kernel stack and SysCall function table address in that random process's TSS.
I find that neither my textbooks or my googling skills give me a proper answer to this question. I know it depends on the operating system, but on a general note: what happens and why?
My textbook says that a system call causes the OS to go into kernel mode, given that it's not already there. This is needed because the kernel mode is what has control over I/O-devices and other things outside of a specific process' adress space. But if I understand it correctly, a switch to kernel mode does not necessarily mean a process context switch (where you save the current state of the process elsewhere than the CPU so that some other process can run).
Why is this? I was kinda thinking that some "admin"-process was switched in and took care of the system call from the process and sent the result to the process' address space, but I guess I'm wrong. I can't seem to grasp what ACTUALLY is happening in a switch to and from kernel mode and how this affects a process' ability to operate on I/O-devices.
Thanks alot :)
EDIT: bonus question: does a library call necessarily end up in a system call? If no, do you have any examples of library calls that do not end up in system calls? If yes, why do we have library calls?
Historically system calls have been issued with interrupts. Linux used the 0x80 vector and Windows used the 0x2F vector to access system calls and stored the function's index in the eax register. More recently, we started using the SYSENTER and SYSEXIT instructions. User applications run in Ring3 or userspace/usermode. The CPU is very tricky here and switching from kernel mode to user mode requires special care. It actually involves fooling the CPU to think it was from usermode when issuing a special instruction called iret. The only way to get back from usermode to kernelmode is via an interrupt or the already mentioned SYSENTER/EXIT instruction pairs. They both use a special structure called the TaskStateSegment or TSS for short. These allows to the CPU to find where the kernel's stack is, so yes, it essentially requires a task switch.
But what really happens?
When you issue an system call, the CPU looks for the TSS, gets its esp0 value, which is the kernel's stack pointer and places it into esp. The CPU then looks up the interrupt vector's index in another special structure the InterruptDescriptorTable or IDT for short, and finds an address. This address is where the function that handles the system call is. The CPU pushes the flags register, the code segment, the user's stack and the instruction pointer for the next instruction that is after the int instruction. After the systemcall has been serviced, the kernel issues an iret. Then the CPU returns back to usermode and your application continues as normal.
Do all library calls end in system calls?
Well most of them do, but there are some which don't. For example take a look at memcpy and the rest.
I just read this in "Operating System Concepts" from Silberschatz, p. 18:
A bit, called the mode bit, is added to the hardware of the computer
to indicate the current mode: kernel(0) or user(1). With the mode bit,
we are able to distinguish between a task that is executed on behalf
of the operating system and one that is executed on behalf of the
user.
Where is the mode bit stored?
(Is it a register in the CPU? Can you read the mode bit? As far as I understand it, the CPU has to be able to read the mode bit. How does it know which program gets mode bit 0? Do programs with a special adress get mode bit 0? Who does set the mode bit / how is it set?)
Please note that your question depends highly on the CPU itselt; though it's uncommon you might come across certain processors where this concept of user-level/kernel-level does not even exist.
The cs register has another important function: it includes a 2-bit
field that specifies the Current Privilege Level (CPL) of the CPU. The
value 0 denotes the highest privilege level, while the value 3 denotes
the lowest one. Linux uses only levels 0 and 3, which are respectively
called Kernel Mode and User Mode.
(Taken from "Understanding the Linux Kernel 3e", section 2.2.1)
Also note, this depends on the CPU as you can clearly see and it'll change from one to another but the concept, generally, holds.
Who sets it? Typically, the kernel/cpu and a user-process cannot change it but let me explain something here.
**This is an over-simplification, do not take it as it is**
Let's assume that the kernel is loaded and the first application has just started(the first shell), the kernel loads everything for this application to start, sets the bit in the cs register(if you are running x86) and then jumps to the code of the Shell process.
The shell will continue to execute all of its instructions in this context, if the process contains some privileged instruction, the cpu will fetch it and won't execute it; it'll give an exception(hardware exception) that tells the kernel someone tried to execute a privileged instruction and here the kernel code handles the job(CPU sets the cs to kernel mode and jumps to some known-location to handle this type of errors(maybe terminating the process, maybe something else).
So how can a process do something privileged? Talking to a certain device for instance?
Here comes the System Calls; the kernel will do this job for you.
What happens is the following:
You set what you want in a certain place(For instance you set that you want to access a file, the file location is x, you are accessing for reading etc) in some registers(the kernel documentation will let you know about this) and then(on x86) you will call int0x80 instruction.
This interrupts the CPU, stops your work, sets the mode to kernel mode, jumps the IP register to some known-location that has the code which serves file-IO requests and moves from there.
Once your data is ready, the kernel will set this data in a place you can access(memory location, register; it depends on the CPU/Kernel/what you requested), sets the cs flag to user-mode and jumps back to your instruction next to the it int 0x80 instruction.
Finally, this happens whenever a switch happens, the kernel gets notified something happened so the CPU terminates your current instruction, changes the CPU status and jumps to where the code that handles this thing; the process explained above, roughly speaking, applies to how a switch between kernel mode and user-mode happens.
It's a CPU register. It's only accessible if you're already in kernel mode.
The details of how it gets set depend on the CPU design. In most common hardware, it gets set automatically when executing a special opcode that's used to perform system calls. However, there are other architectures where certain memory pages may have a flag set that indicates that they are "gateways" to the kernel -- calling a function on these pages sets the kernel mode bit.
These days it's given other names such as Supervisor Mode or a protection ring.
As I can understand, every OS need to have some mechanism to periodically check if it should run some tasks and suspend others.
One way would be some kind of timer on whose expiry the OS will check if it should run/suspend some task.
Generally, say on a ARM system that would probably be some kind of ISR.
My real question, is that I've been ABLE to only visualize this and not see it somewhere. Could some one point to some free/open RTOS code where I can actually see the code that handles the preemption/scheduling?
freertos.org. The entire OS is open source, and right there for you to see. And there are dozens of different ports to compare and contrast. For the context switch code, you will want to look in the ports directory, in any one of many files called port.c, port.asm, etc. And yes, in the case of freertos all context switches are performed in interrupts (a tick timer ISR, or any other SysCall interrupt).
A context switch is very-much processor specific, as the list of registers to save and the assembly code to save them varies between processor families, and sometimes within a given family. As a result each port has a separate file for this code.
The scheduling (selection of next task to run), on the other hand, is done in a file called tasks.c, which is common to all ports and references the port-specific code.
It is not the case than an RTOS simply context switches periodically - that is how most GPOS work. In an RTOS the scheduler runs on any scheduling event. These include system-tick, but also message post, event trigger, semaphore give, or mutex unlock for example.
On ARM Cortex-M the CMSIS 3.x includes an RTOS API (intended primarily for RTOS developers rather than a complete RTOS itself), the source for this will include a context switching mechanism.
If you want a detailed description for a simple RTOS you might consider reading µC/OS-II: The Real-Time Kernel or the slightly more sophisticated µC/OS-III: The Real-Time Kernel .
FreeRTOS is increasingly popular, though perhaps a little unconventional architecturally. A more complete (in that it is not just a scheduling kernel but a more complete OS) and very powerful option is eCos.
You can take a look at xv6.
Its not an RTOS, it is just a skeleton OS(based on V6 unix) meant for academic purpose.
In the XV6 book take a look at chapter 4, there is explanation along with the code as to how scheduling is done on a small OS like xv6.XV6 puts a process to sleep when it is waiting for disk or some I/O operation, there is also timer interupt every 100msec to switch a process.
There is also explanation with code on how the context switching takes place, what information is saved( context frame of a process), how the switch from user to kernel mode happens when the scheduler has to run.
The best part is that the amount of reading you have to do to understand these concepts is very less unlike some reference book on OS :) The code is relatively small, you can infact run the XV6 on qemu set breakpoints in the sched , swtch and other functions and actually see the information saved during a context switch.(how to run xv6 in this link)
You dont have to read previous chapters to understand the chapter4. There isnt much dependency,xv6 uses struct proc to identify a process, ptable for all the current running process in the system, proc->conext -refers to the state the process is in (register value etc) , this is saved by the scheduler.
Cheers :)