I'm performing a set of activities to make sure Redis runs well in a set of embedded systems, including the Raspberry PI. In order to fix certain code paths of Redis where unaligned memory accesses are performed (due to a change introduced in Redis 3.2) I'm trying to force the PI to either log a message on unaligned memory accesses or send a signal to the process when this happens. In this way I can both make sure that Redis will run well where unaligned accesses are a violation, and that it will run faster in platforms where instead such accesses can be performed but are slower. ARM v6, the one used in the PI v1, is apparently able to deal with unaligned memory accesses, so if I use following command to configure Linux in order to sent a signal to the process performing the unaligned access:
echo 4 > /proc/cpu/alignment
And then run the following program:
#include <stdio.h>
#include <stdint.h>
int main(int argc, char **argv) {
char *buf = "foobareklsjdfklsjdfslkjfskdljfskdfjdslkjfdslkjfsd";
uint32_t *l = (uint32_t*) (buf+1);
printf("%p\n", l);
printf("%d\n", (int)*l);
return 0;
}
I can't see any signal received by the process, or the counters at /proc/cpu/alignment incrementing.
My guess is that this is due to ARM v6 ability to deal with unaligned addresses automatically, if a given CPU configuration flag is set. My question is, is my hypothesis correct? And if so, how to force a PI version 1 to actually raise an exception in case of unaligned accesses so that the Linux kernel can trap it and send a signal, log the access, and so forth, according to /proc/cpu/alignment settings?
EDIT: It is worth to note that not all the instructions can perform unaligned accesses even in ARM v6. For instance STMDB, STMFD, LDMDB, LDMEA and similar multiple words instructions will indeed raise an exception and will be trapped by the Linux kernel.
I think I eventually found my answers:
Yes I'm correct, up to the word size ARM v6 (or greater) can silently handle the unaligned accesses so no trap is generated and is completely transparent for the Linux kernel. Nothing will be logged, nor the traps counter in /proc/cpu/alignment will be incremented.
AFAIK there is no way I can force the kernel to trap word-sized unaligned accesses, since to do that apparently the CPU should be configured in order to trap the unaligned addresses in every case, but the Linux kernel does not do that AFAIK, probably because there is alignment unsafe code inside the kernel itself. Checking the Linux kernel source code indeed one can see:
if (cpu_is_v6_unaligned()) {
set_cr(__clear_cr(CR_A));
ai_usermode = safe_usermode(ai_usermode, false);
}
What this means is that the SCTLR.A bit is always cleared, so no trap
will be generated for unaligned accesses ARM v6 can handle.
There are a great deal of instructions that will still generate traps when used with unaligned addresses, for example multi store/load instructions, loading and storing of double values.
However, there are instructions that GCC (the version shipped in the default Raspberry Linux distribution) will happily produced that are not handled by the Linux kernel correctly, that will result in a SIGBUS generated even when /proc/cpu/alignment is set to fix the access.
So point number 4 basically means that, it is not a good idea to fix programs to run in ARM v6 just letting the Linux kernel handle unaligned addresses for us, even when the performance implications of unaligned addresses are not a problem: the program can still crash since not all the instructions are handled.
How to reliably find all the unaligned accesses in a program remains an open question AFAIK, since unfortunately, the otherwise wonderful valgrind program, never implemented this feature. In the past I had to use QEMU emulating Sparc, however this is a very slow process. Valgrind would be the trivial way to do that.
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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 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.
Here's a passage from the book
When executing kernel code, the system is in kernel-space execut-
ing in kernel mode.When running a regular process, the system is in user-space executing
in user mode.
Now what really is a kernel code and user code. Can someone explain with example?
Say i have an application that does printf("HelloWorld") now , while executing this application, will it be a user code, or kernel code.
I guess that at some point of time, user-code will switch into the kernel mode and kernel code will take over, but I guess that's not always the case since I came across this
For example, the open() library function does little except call the open() system call.
Still other C library functions, such as strcpy(), should (one hopes) make no direct use
of the kernel at all.
If it does not make use of the kernel, then how does it make everything work?
Can someone please explain the whole thing in a lucid way.
There isn't much difference between kernel and user code as such, code is code. It's just that the code that executes in kernel mode (kernel code) can (and does) contain instructions only executable in kernel mode. In user mode such instructions can't be executed (not allowed there for reliability and security reasons), they typically cause exceptions and lead to process termination as a result of that.
I/O, especially with external devices other than the RAM, is usually performed by the OS somehow and system calls are the entry points to get to the code that does the I/O. So, open() and printf() use system calls to exercise that code in the I/O device drivers somewhere in the kernel. The whole point of a general-purpose OS is to hide from you, the user or the programmer, the differences in the hardware, so you don't need to know or think about accessing this kind of network card or that kind of display or disk.
Memory accesses, OTOH, most of the time can just happen without the OS' intervention. And strcpy() works as is: read a byte of memory, write a byte of memory, oh, was it a zero byte, btw? repeat if it wasn't, stop if it was.
I said "most of the time" because there's often page translation and virtual memory involved and memory accesses may result in switched into the kernel, so the kernel can load something from the disk into the memory and let the accessing instruction that's caused the switch continue.
If a Windows executable makes use of SYSENTER and is executed on a processor implementing AMD64 ISA, what happens? I am both new and newbie to this topic (OSes, hardware/software interaction) but from what I've read I have understood that SYSCALL is the AMD64 equivalent to Intel's SYSENTER. Hopefully this question makes sense.
If you try to use SYSENTER where it is not supported, you'll probably get an "invalid opcode" exception.
Note that this situation is unusual - generally, Windows executables do not directly contain instructions to enter kernel mode.
As far as i know AM64 processors using different type of modes to handle such issues.
SYSENTER works fine but is not that fast.
A very useful site to get started about the different modes:
Wikipedia
They got rid of a bunch of unused functionality when they developed AMD64 extensions. One of the main ones is the elimination of the cs, ds, es, and ss segment registers. Normally loading segment registers is an extremely expensive operation (the CPU has to do permission checks, which could involve multiple memory accesses). Entering kernel mode requires loading new segment register values.
The SYSENTER instruction accelerates this by having a set of "shadow registers" which is can copy directly to the (internal, hidden) segment descriptors without doing any permission checks. The vast majority of the benefit is lost with only a couple of segment registers, so most likely the reasoning for removing the support for the instructions is that using regular instructions for the mode switch is faster.