6

As far as my understanding goes, User space programs run in the unprivileged mode, and thus do not have direct access to memory or I/O.

Then how exactly can we directly access memory or I/O locations when we mmap /dev/mem in user space programs?

For example:

int fd = 0;
u8 leds = 0;
fd = open("/dev/mem", O_RDWR|O_SYNC);
leds = (u8 *)mmap(0, getpagesize(), PROT_READ|PROT_WRITE, MAP_SHARED, fd, 0x80840000);

This is a hack very commonly used in embedded devices.

Now the variable leds can be used on the fly to access any device that could be present at 0x80840000.

We won't be using any system call to access that address anymore.

Even something like

leds[0x20] = val;

would work.

But privileged operations, such as reading/writing directly to/from an I/O address should be possible only by putting the processor to privileged mode through a system call.

Source.

Improve this question
4
  • Show an example where that is happening as an unprivileged user?
    – wurtel
    Nov 14, 2014 at 12:09
  • 2
    I think you're missing a * in the declaration of leds, but that's just code, no evidence of this actually working as an unprivileged user; in my (limited) experience, everything runs as root on embedded devices.
    – wurtel
    Nov 14, 2014 at 14:54
  • @wurtel - my bad, might've copy pasted poorly... Yes, everything does work as root, but my question goes down to the very heart of the OS. So does this mean that if you are root, you get just as much privileges as the kernel itself?
    – Stark07
    Nov 16, 2014 at 10:41
  • The root user has many privileges, which include being allowed to override traditional filesystem permissions. Some "files" are really access to devices (like disk or memory), which are off-limits for regular users. Not root. But root runs in userspace, so it doesn't enjoy the full privileges of system mode (execute privileged instructions, mostly).
    – vonbrand
    Feb 28, 2016 at 23:26

3 Answers 3

Reset to default
9

Allowing access to /dev/mem by unprivileged processes would indeed be a security problem and should not be permitted.

On my system, ls -l /dev/mem looks like this:

crw-r----- 1 root kmem 1, 1 Sep  8 10:12 /dev/mem

So root can read and write it, members of the kmem group (of which there happen to be none) can read it but not write it, and everyone else cannot open it at all. So this should be secure.

If your /dev/mem is anything like mine, your unprivileged process should not even have been able to open the file at all, let alone mmap it.

Check the permissions of /dev/mem on your system to make sure they are secure!

Improve this answer
11
  • But assuming that I am doing this as root... Does it mean that a process run by root gets kernel level access privileges?
    – Stark07
    Nov 16, 2014 at 10:41
  • 1
    Well, if you're doing this as root then you have access to everything. I'm not sure what you mean by "kernel level access privileges", but root can certainly do anything it wants to, one way or another (in particular, for example, by crafting and dynamically loading a new kernel module).
    – Celada
    Nov 17, 2014 at 0:37
  • 1
    Yes, that's correct: when you map a file, you afterwards get to read & write the file without the use of any ststem calls. That's true whether you've mapped a regular disk file or a special device like /dev/mem . If you prefer, you could open /dev/mem and issue read() and write() systems calls. Then you would not be sending I/O without the use of any system calls. The end result is the same, but if you have lots of small I/O to do, mmap() and direct access will probably perform better specifically because you don't need system calls! This is true for regular files too!
    – Celada
    Nov 17, 2014 at 7:20
  • 1
    @Stark07 I think you've missed the point. The mmap (and /dev/mem) themselves does NOT provide direct access to the system memory, and it is impossible to do so in Ring 4. What it does instead is just mapping a file (or resource) to a specific virtual address in the calling process, just like what happens when a program is loaded/execed (in this case, the program image is mmaped to the virtual memory).
    – minmaxavg
    Apr 8, 2016 at 2:09
  • 2
    TL;DR An unprivileged process does not have direct access to system resource, but can do so when those resource is mapped to their virtual address space. There's no difference between accessing its stack/heap/rodata/whatsoever and accessing kernel memory - They eventually both access the actual physical memory, while the latter case just happens to be under the program's control.
    – minmaxavg
    Apr 8, 2016 at 2:13
6

The addresses visible to a user process (whether running as root or an unpriviledged user) are virtual addresses, which get mapped to physical addresses by the MMU through the page tables. Setting up the page tables is a priviledged operation, and can only be performed by kernel code; however, once the page tables are set, accessing the memory is allowed in user mode.

Concretely, your code uses mmap to request that the kernel set up the page tables to allow access to a given range of physical memory. The kernel checks the process's priviledges (it has read/write access to /dev/mem) and sets up the page tables to allow it to access physical memory.

Improve this answer
2

The value of leds is a virtual address. As long as it is in the user-space of the current process, the process can access it directly via instructions like leds[0] = val without having to be in privileged mode, no matter where in the RAM this virtual address is mapped to

Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.