How to force vdso memory to be in fixed address? - linux

It appears that there is no information in the executable about where vdso should be. Assuming that I can control how the program is compiled, linked and written, how can I force it to be at the address I want it to be?

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why non-pic code can't be totally ASLR using run-time fixups?

I understand that PIC code makes ASLR randomization more efficient and easier since the code can be placed anywhere in memory with no change in code. But if i understand right according to Wikipedia relocation dynamic linker can make "fixups" at runtime so a symbol can be located although code being not position-independent. But according to many answers i saw here non-pic code can't ASLR sections except the stack(so cant randomize program entry point). If that is correct then what are runtime fixups are used for and why can't we just fixup all locations in code at runtime before program start to make program entry point randomized.
TL:DR: Not all uses of absolute address will have relocation info in a non-PIE executable (ELF type EXEC, not DYN). Therefore the kernel's program-loader can't find them all to apply fixups.
Thus there's no way to retroactively enable ASLR for executables built as non-PIE. There's no way for a traditional executable to flag itself as having relocation metadata for every use of an absolute address, and no point in adding such a feature either since if you want text ASLR you'd just build a PIE.
Because ELF-type EXEC Linux executables are guaranteed to be loaded / mapped at the fixed base address chosen by the linker at link time, it would be a waste of space in the executable to make symbol-table entries for internal symbols. So toolchains didn't do that, and there's no reason to start. That's simply how traditional ELF executables were designed; Linux switched from a.out to ELF back in the mid 90s before stack ASLR was a thing, so it wasn't on people's radar.
e.g. the absolute address of static char buf[100] is probably embedded somewhere in the machine code that uses it (if we're talking about 32-bit code, or 64-bit code that puts the address in a register), but there's no way to know where or how many times.
Also, for x86-64 specifically, the default code model for non-PIE executables guarantees that static addresses (text / data / bss) will all be in the low 2GiB of virtual address space, so 32-bit absolute signed or unsigned addresses can work, and rel32 displacements can reach anything from anything. That's why non-PIE compiler output uses mov $symbol, %edi (5 bytes) to put an address in a register, instead of lea symbol(%rip), %rdi (7 bytes). https://godbolt.org/z/89PeK1
So even if you did know where every absolute address was, you could only ASLR it in the low 2GiB, limiting the number of bits of entropy you could introduce. (I think Windows has a mode for this: LargeAddressAware = no. But Linux doesn't. 32-bit absolute addresses no longer allowed in x86-64 Linux? Again, PIE is a better way to allow text ASLR, so people (distros) should just compile for that if they want its benefits.)
Unlike Windows, Linux doesn't spend huge effort on things that can be handled better and more efficiently by recompiling binaries from source.
That being said, GNU/Linux does support fixup relocations for 64-bit absolute addresses even in PIC / PIE ELF shared objects. That's why beginner code like NASM mov rdi, BUFFER can work even in a shared library: use objdump -drwC -Mintel to see the relocation info on that use of the symbol in a mov reg, imm64 instruction. An lea rdi, [rel BUFFER] wouldn't need any relocation entry if BUFFER wasn't a global symbol. (Equivalent of C static.)
You might be wondering why metadata is essential:
There's no reliable way to search text/data for possible absolute addresses; false positives would be possible. e.g. /usr/bin/ld probably contains 0x401000 as the default start address for an x86-64 executable. You don't want ASLR of ld's code+data to also change its defaults. Or that integer value could have come up in any number of ways in many programs, e.g. as a bitmap. And of course x86-64 machine code is variable length so there's no reliable way to even distinguish opcodes from immediate operands in the most general case.
And also potentially false negatives. Not super likely that an x86 program would construct an absolute address in a register with multiple instructions, but it's certainly possible. However in non-x86 code, that would be common.
RISC machines with fixed-length instructions can't put a 32-bit address into a 32-bit instruction; there'd be no room left for anything else. So to load from static storage, the absolute addresses would have to be split across multiple instructions, like MIPS lui $t0, %hi(0x612300) / lw $t1, %lo(0x612300)($t0) to load from a static variable at absolute address 0x612300. (There would normally be a symbol name in the asm source, but it wouldn't appear in the final linked binary unless it was .globl, so I used numbers as a reminder.) Instructions like that don't have to come in pairs; the same high-half of the address could be reused by other accesses into the same array or struct in later instructions.
Let's first have a look at Windows before having a look at Linux:
Windows' .EXE files (programs) typically have a so-called "base relocation table" and they have an "image base".
The "image base" is the "desired" start address of the program; if Windows loads the program to that address, no relocation needs to be done.
The "base relocation table" contains a list of all values in a program which represent addresses. If the program is loaded to a different address than the "image base", Windows must add the difference to all values listed in that table.
If the .EXE file does not contain a "base relocation table" (as far as I know some 32-bit GCC versions generate such files), it is not possible to load the file to another address.
This is because the following C code statements will result in exactly the same machine code (binary code) if the variable someVariable is located at the address 12340000, and it is not possible to distinguish between them:
long myVariable = 12340000;
And:
int * myVariable = &someVariable;
In the first case, the value 12340000 must not be changed in any situation; in the second case, the address (which is 12340000) must be changed to the real address if the program is loaded to another address.
If the "base relocation table" is missing, there is no information if the value 12340000 is an integer value (which must not be changed) or an address (which must be changed).
So the program must be loaded to some fixed address.
I'm not sure about the latest 32-bit Linux releases, but at least in older 32-bit Linux versions there was nothing like a "base relocation table" and programs did not use PIC. This means that these programs had to be loaded to their "favorite" address.
I don't know about 64-bit Linux programs, but if a program is compiled the same way as the (older) 32-bit programs, they also must be loaded to a certain address and ASLR is not possible.

I am wondering if pie does anything if the aslr is turned off on the system? Or is pie dependant on aslr?

I am also wondering what does pie and what does aslr effect in memory as I understand, aslr randomizes addresses of libc base,stack and heap. And pie randomizes elf base and with that .text,.data,.bss,.rodata...
Is that correct or am I getting it wrong?
PIE requires position-independent code, costing a small amount of performance. (Or a large amount like 15% on ISAs like 32-bit x86 that don't easily support PC-relative addressing). See 32-bit absolute addresses no longer allowed in x86-64 Linux?.
With ASLR disabled, e.g. when running under GDB or with it disabled system-wide, Linux chooses a base address of 0x0000555555555000 to map the executable, so objdump -d addresses relative to the file start like 0x4000 end up at that high virtual address.
A PIE executable is an ELF shared object (like a .so), as opposed to an ELF "executable". An ELF executable has a base address in the ELF headers, set by the linker, so absolute addresses can be hard-coded into the machine code and data without needing relocations for fixups. (That's why there's no way to ASLR a proper ELF executable, only a PIE).
The mechanism that supports PIE was originally just a fun hack of putting an entry point in a library. Later people realized it would be useful for ASLR for static code/data in executables to be possible, so this existing support became official. (Or something like that; I haven't read up on the history.)
But anyway, ASLR is enabled by PIE, but PIE is a thing even with ASLR disabled, if you want the most general non-technical description.

How can we tell an instruction is from application code or library code on Linux x86_64

I wanted to know whether an instruction is from the application itself or from the library code.
I observed some application code/data are located at about 0x000055xxxx while libraries and mmaped regions are by default located at 0x00007fcxxxx. Can I use for example, 0x00007f00...00 as a boundary to tell instruction is from the application itself or from the library?
How can I configure this boundary in Linux kernel?
Updated.
Can I prevent (or detect) a syscall instruction being issued from application code (only allow it to go through libc). Maybe we can do a binary scan, but due to the variable length of instruction, it's hard to prevent unintended syscall instruction.
Do it the other way. You need to learn a lot.
First, read a lot more about operating systems. So read the Operating Systems: Three Easy Pieces textbook.
Then, learn more about ASLR.
Read also Drepper's How to write shared libraries and Levine's Linkers and loaders book.
You want to use pmap(1) and proc(5).
You probably want to parse the /proc/self/maps pseudo-file from inside your program. Or use dladdr(3).
To get some insight, run cat /proc/$$/maps and cat /proc/self/maps in a Linux terminal
I wanted to know whether an instruction is from userspace or from library code.
You are confused: both library code and main executable code are userspace.
On Linux x86_64, you can distinguish kernel addresses from userpsace addresses, because the kernel addresses are in the FFFF8000'00000000 through FFFFFFFF'FFFFFFFF range on current (48-bit) implementations. See canonical form address description here.
I observed some application code/data are located at about 0x000055xxxx while libraries and mmaped regions are by default located at 0x00007fcxxxx. Can I use for example, 0x00007f00...00 as a boundary to tell instruction is from the application itself or from the library?
No, in general you can't. An application can be linked to load anywhere within canonical address space (though most applications aren't).
As Basile Starynkevitch already answered, you'll need to parse /proc/$pid/maps, or know what address the executable is linked to load at (for non-PIE binary).

Is it possible to force a range of virtual addresses?

I have an Ada program that was written for a specific (embedded, multi-processor, 32-bit) architecture. I'm attempting to use this same code in a simulation on 64-bit RHEL as a shared object (since there are multiple versions and I have a requirement to choose a version at runtime).
The problem I'm having is that there are several places in the code where the people who wrote it (not me...) have used Unchecked_Conversions to convert System.Addresses to 32-bit integers. Not only that, but there are multiple routines with hard-coded memory addresses. I can make minor changes to this code, but completely porting it to x86_64 isn't really an option. There are routines that handle interrupts, CPU task scheduling, etc.
This code has run fine in the past when it was statically-linked into a previous version of the simulation (consisting of Fortran/C/C++). Now, however, the main executable starts, then loads a shared object based on some inputs. This shared object then checks some other inputs and loads the appropriate Ada shared object.
Looking through the code, it's apparent that it should work fine if I can keep the logical memory addresses between 0 and 2,147,483,647 (32-bit signed int). Is there a way to either force the shared object loader to leave space in the lower ranges for the Ada code or perhaps make the Ada code "think" that it's addresses are between 0 and 2,147,483,647?
Is there a way to either force the shared object loader to leave space in the lower ranges for the Ada code
The good news is that the loader will leave the lower ranges untouched.
The bad news is that it will not load any shared object there. There is no interface you could use to influence placement of shared objects.
That said, dlopen from memory (which we implemented in our private fork of glibc) would allow you to do that. But that's not available publicly.
Your other possible options are:
if you can fit the entire process into 32-bit address space, then your solution is trivial: just build everything with -m32.
use prelink to relocate the library to desired address. Since that address should almost always be available, the loader is very likely to load the library exactly there.
link the loader with a custom mmap implementation, which detects the library of interest through some kind of side channel, and does mmap syscall with MAP_32BIT set, or
run the program in a ptrace sandbox. Such sandbox can again intercept mmap syscall, and or-in MAP_32BIT when desirable.
or perhaps make the Ada code "think" that it's addresses are between 0 and 2,147,483,647?
I don't see how that's possible. If the library stores an address of a function or a global in a 32-bit memory location, then loads that address and dereferences it ... it's going to get a 32-bit truncated address and a SIGSEGV on dereference.

Is a linux application loaded in the loader's address space?

When ld-linux (Linux's loader) loads an application, it loads its ELF data structures to memory, builds some structures (e.g., GOT), and passes the execution to the entry point of the loaded application.
Is the loading of this application's code and data done into the loader's address space? Does the execution of the application's code occur in the loader's address space?
If not, what is the mechanism ld-linux uses to pass the execution to the loaded instructions?
Answer (EDIT): The application's code is loaded into the loader's address space. The application code and loader are ran on the same address space.
http://grahamwideman.wordpress.com/2009/02/09/the-linux-loader-and-how-it-finds-libraries/ http://www.tenouk.com/ModuleW.html basically there are assemblers and linkers too.The hiearchy of ld-linux (loader's linux is very well explained in the second url.
Thanks & Regards,
Alok

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