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.
Related
Quoting the following paragraph from Operating Systems: Three Easy Pieces,
Note that there are two types of register saves/restores that happen
during this protocol. The first is when the timer interrupt occurs; in
this case, the user registers of the running process are implicitly
saved by the hardware, using the kernel stack of that process. The
second is when the OS decides to switch from A to B; in this case, the
kernel registers are explicitly saved by the software (i.e., the OS),
but this time into memory in the process structure of the process.
Reading other literature on context switch I understand that timer interrupt throws the cpu into kernel mode which then saves the process context into kernel stack.
Why is the author talking about a multiple context save emphasising on hardware/software?
The author emphasizes on hardware/software part of it because basically its context saving that is being done , sometimes by hardware and sometimes by software.
When a timer interrupt occurs, the user registers are saved by hardware(meaning saved by the CPU itself) on the kernel stack of that process. When the interrupt handler code is finished, the user registers will be restored using the kernel stack of that process,thereby restoring user stack and process successfully returns to user mode from kernel mode.
In case of a context switch from process A to process B, the very kernel stacks of the two processes A and B are switched,inside the kernel, which indirectly means saving and restoring of kernel registers. The term Software is used because the scheduler process after selecting which process to run next calls a function(thats why software), that does the switching of kernel stacks. The context switch code need not worry about user register values - those are already safely saved away in the kernel stack by that point.
The first is when the timer interrupt occurs; in this case, the user registers of the running process are implicitly saved by the hardware, using the kernel stack of that process.
Often, only SOME of the registers are saved and this is usually to an interrupt stack.
The second is when the OS decides to switch from A to B; in this case, the kernel registers are explicitly saved by the software (i.e., the OS), but this time into memory in the process structure of the process.
Usually this switch occurs in HARDWARE through a special instruction. Maybe they are referring to the fact that the switch is triggered through software as opposed to an interrupt that is triggered by hardware.
Also thanks for that reference. I have just started to go through it. It MUCH better than most of the OS books that only serve to confuse.
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)
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.
This questions came to my head when I was studying processes scheduling.
How does OS execute and control the execution of binary and compiled files? I thought maybe OS copies a part of the binary to some memory location, jumps there, comes back after executing that block and executes the next one. But then it wouldn't have any control on it (e.g. the program can do a jump anywhere and don't come back).
In JVM case it makes perfect sense, the VM is interpreting each instruction. But in the binary files case the instructions are real CPU executable instructions so I don't think that an OS acts like VM.
It does exactly that. The operating system, in some order,
creates an entry in the process table
creates a virtual memory space for the process
loads the program code in the process memory
points the process instruction pointer to the process entry point
creates an entry in the scheduler and sets the process thread ready for execution.
Concurrency is not handled by the program being split into blocks. Switching between tasks is done via interrupts: before a process is given CPU, a timer is set up. When the timer finishes, the CPU registers an interrupt, pushes the instruction pointer to the stack and jumps to the interrupt handler defined by the operating system. This handler stores the CPU state in memory, swaps a virtual memory table and restores some other thread that is ready for execution. The same swap occurs if the thread must pause for some other reason (waiting for user / disk / network...) or yields.
http://en.wikipedia.org/wiki/Multitasking#Preemptive_multitasking.2Ftime-sharing
Note that relying on the process yielding the CPU is possible but unreliable (the process might not yield, preventing other processes from running)
http://en.wikipedia.org/wiki/Multitasking#Cooperative_multitasking.2Ftime-sharing
Security is handled by switching the CPU into protected mode where the application code cannot run some instructions (so jumping around randomly is mostly harmless). See the link provided by #SkPhilipp
Note that modern JVM does not interpret each instruction (that would be slow). Instead it compiles into native code and runs the code or (in case of just-in-time compilation) interprets at first, but compiles the "hot spots" (the code that gets run often enough).
system call -- It is an instruction that generates an interrupt that causes OS to gain
control of processor.
so if a running process issue a system call (e.g. create/terminate/read/write etc), a interrupt is generated which cause the KERNEL TO TAKE CONTROL of the processor which then executes the required interrupt handler routine. correct?
then can anyone tell me how the processor known that this instruction is supposed to block the process, go to privileged mode, and bring kernel code.
I mean as a programmer i would just type stream1=system.io.readfile(ABC) or something, which translates to open and read file ABC.
Now what is monitoring the execution of this process, is there a magical power in the cpu to detect this?
As from what i have read a PROCESSOR can only execute only process at a time, so WHERE IS THE MONITOR PROGRAM RUNNING?
How can the KERNEL monitor if a system call is made or not when IT IS NOT IN RUNNING STATE!!
or does the computer have a SYSTEM CALL INSTRUCTION TABLE which it compares with before executing any instruction?
please help
thanku
The kernel doesn't monitor the process to detect a system call. Instead, the process generates an interrupt which transfers control to the kernel, because that's what software-generated interrupts do according to the instruction set reference manual.
For example, on Unix the process stuffs the syscall number in eax and runs an an int 0x80 instruction, which generates interrupt 0x80. The CPU reacts to this by looking in the Interrupt Descriptor Table to find the kernel's handler for that interrupt. This handler is the entry point for system calls.
So, to call _exit(0) (the raw system call, not the glibc exit() function which flushes buffers) in 32-bit x86 Linux:
movl $1, %eax # The system-call number. __NR_exit is 1 for 32-bit
xor %ebx,%ebx # put the arg (exit status) in ebx
int $0x80
Let's analyse each questions you have posed.
Yes, your understanding is correct.
See, if any process/thread wants to get inside kernel there are only two mechanisms, one is by executing TRAP machine instruction and other is through interrupts. Usually interrupts are generated by the hardware, so any other process/threads wants to get into kernel it goes through TRAP. So as usual when TRAP is executed by the process it issues interrupt (mostly software interrupt) to your kernel. Along with trap you will also mentions the system call number, this acts as input to your interrupt handler inside kernel. Based on system call number your kernel finds the system call function inside system call table and it starts to execute that function. Kernel will set the mode bit inside cs register as soon as it starts to handle interrupts to intimate the processor as current instruction is a privileged instruction. By this your processor will comes to know whether the current instruction is privileged or not. Once your system call function finished it's execution your kernel will execute IRET instruction. Which will clear mode bit inside CS register to inform whatever instruction from now inwards are from user mode.
There is no magical power inside processor, switching between user and kernel context makes us to think that processor is a magical thing. It is just a piece of hardware which has the capability to execute tons of instructions at a very high rate.
4..5..6. Answers for all these questions are answered in above cases.
I hope I've answered your questions up to some extent.
The interrupt controller signals the CPU that an interrupt has occurred, passes the interrupt number (since interrupts are assigned priorities to handle simultaneous interrupts) thus the interrupt number to determine wich handler to start. The CPu jumps to the interrupt handler and when the interrupt is done, the program state reloaded and resumes.
[Reference: Silberchatz, Operating System Concepts 8th Edition]
What you're looking for is mode bit. Basically there is a register called cs register. Normally its value is set to 3 (user mode). For privileged instructions, kernel sets its value to 0. Looking at this value, processor knows which kind of instruction it is. If you're interested digging more please refer this excellent article.
Other Ref.
Where is mode bit
Modern hardware supports multiple user sessions. If your hw supports multi user mode, i provides a mechanism called interrupt. An interrupt basically stops the execution of the current code to execute other code (e.g kernel code).
Which code is executed is decided by parameters, that get passed to the interrupt, by the code that issues the interrupt. The hw will increase the run level, load the kernel code into the memory and forces the cpu to execute this code. When the kernel code returns, it again directly informs the hw and the run level gets decreased.
The HW will then restore the cpu state before the interrupt and set the cpu the the next line in the code that started the interrupt. Done.
Since the code is actively calling the hw, which again actively calls the kernel, no monitoring needs to be done by the kernel itself.
Side note:
Try to keep your question short. Make clear what you want. The first answer was correct for the question you posted, you just didnt phrase it well. Make clear that you are new to the topic and need a detailed explanation of basic concepts instead of explaining what you understood so far and don't use caps lock.
Please accept the answer cnicutar provided. thank you.