Print custom text into strace. Strace comments - c

We use strace a lot. We would like to output some text into strace to mark places the code has reached. The way i see people have done it so far is to stat an non-existent file. The filename is just the text they want to see in the strace. Its pretty fast, but I'm sure there is a better way. I worry that there might be a lot of code and maybe kernel locks being hit even though the mount point is bogus. Any ideas?

write() to an out-of-range file descriptor shows up well in strace output, and should be much faster - the range check happens early, and it doesn't need to look at the data at all. (You need to pass the length of the data to be written, rather than just a 0-terminated string, but gcc will normally optimise strlen() of a constant string to a constant.)
$ cat hw.c
#include <stdio.h>
#include <string.h>
#include <unistd.h>
#define STRACE_TRACE(str) write(-1, str, strlen(str))
int main(void)
{
STRACE_TRACE("before");
printf("Hello world\n");
STRACE_TRACE("after");
return 0;
}
$ gcc -Wall -o hw hw.c
$ strace ./hw
...
write(-1, "before"..., 6) = -1 EBADF (Bad file descriptor)
fstat64(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 150), ...}) = 0
mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb77da000
write(1, "Hello world\n"..., 12Hello world
) = 12
write(-1, "after"..., 5) = -1 EBADF (Bad file descriptor)
exit_group(0) = ?
$

Related

How do the err/warn functions get the program name?

From the err/warn manpage:
The err() and warn() family of functions display a formatted error
message on the standard error output. In all cases, the last component
of the program name, a colon character, and a space are output. If the
fmt argument is not NULL, the printf(3)-like formatted error message is
output.
If I make this call: warn("message"); it will output something like this:
a.out: message: (output of strerror here)
How do the warn/err functions find the name of the program (in this case, a.out) without seemingly having any access to argv at all? Does it have anything to do with the fact that they are BSD extensions?
The err/warn functions prepend the basename of the program name. According to the answers to this SO post, there are a few ways to get the program name without access to argv.
One way is to call readlink on /proc/self/exe, then call basename on that. A simple program that demonstrates this:
#include <libgen.h>
#include <linux/limits.h>
#include <stdio.h>
#include <unistd.h>
char *
progname(void)
{
char path[PATH_MAX];
if (readlink("/proc/self/exe", path, PATH_MAX) == -1)
return NULL;
/* not sure if a basename-returned string should be
* modified, maybe don't use this in production */
return basename(path);
}
int
main(void)
{
printf("%s: this is my fancy warn message!\n", progname());
return 0;
}
You can also use the nonstandard variables __progname, which may not work depending on your compiler, and program_invocation_short_name, which is a GNU extension defined in errno.h.
In pure standard C, there's no way to get the program name passed as argv[0] without getting it, directly or indirectly, from main. You can pass it as an argument to functions, or save it in a global variable.
But system functions also have the option of using system-specific methods. On open-source operating system, you can download the source code and see how it's done. For Unix-like systems, that's libc.
For example, on FreeBSD:
The warn and err functions call the internal system function _getprogname().
_getprogname() reads the global variable __progname.
__progname is set in handle_argv which is called from _start(). This code is not in libc, but in CSU, which is a separate library containing program startup code.
_start() is the program's entry point. It's defined in the architecture-specific crt1*.c. It's also the function that calls main, and it passes the same argv to both handle_argv() and main().
_start is the first C function called in the program. It's called from assembly code that reads the argv pointer from the stack.
The program arguments are copied into the program's address space by the kernel as part of the implementation of the execve system call.
Note that there are several concepts of “program name” and they aren't always equivalent. See Finding current executable's path without /proc/self/exe for a discussion of the topic.
How do the warn/err functions find the name of the program (in this case, a.out) without seemingly having any access to argv at all? Does it have anything to do with the fact that they are BSD extensions?
Such things can easily be figured out using the strace utility, which records all system calls.
I wrote the highly complex program test.c:
#include <err.h>
int main() { warn("foo"); }
and gcc -o test -static test.c; strace ./test yields (the -static to avoid the noise from trying to load a lot of libraries):
execve("./test", ["./test"], 0x7fffcbb7fd60 /* 101 vars */) = 0
arch_prctl(0x3001 /* ARCH_??? */, 0x7ffd388d6540) = -1 EINVAL (Invalid argument)
brk(NULL) = 0x201e000
brk(0x201edc0) = 0x201edc0
arch_prctl(ARCH_SET_FS, 0x201e3c0) = 0
set_tid_address(0x201e690) = 55889
set_robust_list(0x201e6a0, 24) = 0
uname({sysname="Linux", nodename="workhorse", ...}) = 0
prlimit64(0, RLIMIT_STACK, NULL, {rlim_cur=8192*1024, rlim_max=RLIM64_INFINITY}) = 0
readlink("/proc/self/exe", "/tmp/test", 4096) = 9
getrandom("\x43\xff\x90\x4b\xa8\x82\x38\xdd", 8, GRND_NONBLOCK) = 8
brk(0x203fdc0) = 0x203fdc0
brk(0x2040000) = 0x2040000
mprotect(0x4b6000, 16384, PROT_READ) = 0
write(2, "test: ", 6) = 6
write(2, "foo", 3) = 3
write(2, ": Success\n", 10) = 10
exit_group(0) = ?
+++ exited with 0 +++
And there you have it: you can just readlink /proc/self/exe to know what you're called.

Bash reopen tty on simple program

#include <stdio.h>
#include <stdlib.h>
int main()
{
char buf[512];
fgets(buf, 512, stdin);
system("/bin/sh");
}
Compile with cc main.c
I would like a one-line command that makes this program run ls without it waiting for user input.
# This does not work - it prints nothing
(echo ; echo ls) | ./a.out
# This does work if you type ls manually
(echo ; cat) | ./a.out
I'm wondering:
Why doesn't the first example work?
What command would make the program run ls, without changing the source?
My question is shell and OS-agnostic but I would like it to work at least on bash 4.
Edit:
While testing out the answers, I found out that this works.
(python -c "print ''" ; echo ls) | ./a.out
Using strace:
$ (python -c "print ''" ; echo ls) | strace ./a.out
...
read(0, "\n", 4096)
...
This also works:
(echo ; sleep 0.1; echo ls) | ./a.out
It seems like the buffering is ignored. Is this due to the race condition?
strace shows what's going on:
$ ( echo; echo ls; ) | strace ./foo
[...]
read(0, "\nls\n", 4096) = 4
[...]
clone(child_stack=NULL, flags=CLONE_PARENT_SETTID|SIGCHLD, parent_tidptr=0x7ffdefc88b9c) = 9680
In other words, your program reads a whole 4096 byte buffer that includes both lines before it runs your shell. It's fine interactively, because by the time the read happens, there's only one line in the pipe buffer, so over-reading is not possible.
You instead need to stop reading after the first \n, and the only way to do that is to read byte by byte until you hit it. I don't know libc well enough to know if this kind of functionality is supported, but here it is with read directly:
#include <unistd.h>
#include <stdlib.h>
int main()
{
char buf[1];
while((read(0, buf, 1)) == 1 && buf[0] != '\n');
system("/bin/sh");
}

can /proc/self/exe be mmap'ed?

Can a process read /proc/self/exe using mmap? This program fails to mmap the file:
$ cat e.c
#include <fcntl.h>
#include <unistd.h>
#include <sys/mman.h>
int main()
{
int f=open("/proc/self/exe",O_RDONLY);
char*p=mmap(NULL,0,PROT_READ,0,f,0);
return 0;
}
$ cc e.c -o e
$ strace ./e
[snip]
open("/proc/self/exe", O_RDONLY) = 3
mmap(NULL, 0, PROT_READ, MAP_FILE, 3, 0) = -1 EINVAL (Invalid argument)
exit_group(0) = ?
+++ exited with 0 +++
You are making 2 mistakes here:
Mapped size must be > 0. Zero-size mappings are invalid.
You have to specify, if you want to create a shared (MAP_SHARED) or a private (MAP_PRIVATE) mapping.
The following should work for example:
char *p = mmap(NULL, 4096, PROT_READ, MAP_SHARED, f, 0);
If you wish to map the full executable, you will have to do a stat() on it first, to retrieve the correct file size and then use that as the second parameter to mmap().

Directing Shellcode Output to a File - C

I'm using this code from here: Read and Execute Shellcode from a .txt File
#include <stdio.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <stdlib.h>
int main(void)
{
FILE *file = fopen("text.txt", "r");
unsigned char *buf;
int length = 0;
struct stat st;
int v;
// get file size and allocate. We're going to convert to bytes
// from text, so this allocation will be safely large enough
fstat(fileno(file), &st);
buf = valloc(st.st_size);
while (fscanf(file, "\\x%02x", &v) == 1)
{
buf[length++] = v;
}
fclose(file);
mprotect(buf, length, PROT_EXEC);
int (*ret)() = (int (*)())buf;
ret();
return 0;
}
And I'm compiling with: gcc -fno-stack-protector -z execstack testshell.c -o testshell
It runs just fine but the shellcode it executes writes to the terminal, but I would like to somehow redirect that to a file.
I tried:
./testshell > output.txt
but couldn't seem to get that to capture the results of the shellcode either.
How can I capture the output of any shellcode it runs, and if possible redirect that to a file?
Update: The shellcode I am using, and it outputs with a sys_write syscall output to a file descriptor (It does a calculation and prints to the screen) -
\xeb\x4d\x5e\x66\x83\xec\x0c\x48\x89\xe0\x48\x31\xc9\x68\x33\x09\x00\x7c\x48\x89\xcf\x80\xc1\x0c\x40\x8a\x3e\x40\xf6\xd7\x40\x88\x38\x48\xff\xc6\x68\x16\x96\xd0\xd9\x48\xff\xc0\xe2\xea\x2c\x0c\x48\x89\xc6\x68\xf2\xf5\x44\x48\x48\x31\xc0\x48\x89\xc7\x04\x01\x48\x89\xc2\x80\xc2\x0b\x0f\x05\x48\x31\xc0\x04\x3c\x0f\x05\xe8\xae\xff\xff\xff\x85\xa7\xaa\xc7\x9e\x87\xa5\xa5\x8e\xb7\x87\xba\x31\x80\xe0\x55\xef\xa1\x93\x0c\x4e\x1c\xdc\x34\x53\xb3\x8b\x43\x48\x68\x30\x1d\x4b\x65\x5b\x52\x41\x4e\x44\x53\x54\x52\x32\x5d
Transferring comments into an answer, giving credit where credit's due.
Deanie said:
This should work if the shellcode is writing to stdout and not stderr. Try:
./testshell > output.txt 1>&2
To which user2059300, the OP, responded:
No dice on the 1>&2, the output still occurs in the terminal and not in output.txt
And David C. Rankin said:
I think he meant ./testshell > output.txt 2>&1 to redirect both stdout & stderr to output.txt.
But user2059300 stated:
Still a no-go on that, … I provided the shellcode I'm testing.
Then I asked:
How does the shell code do the writing? What is it doing? I'm not going to dissect it for you; shell code tends to be very platform specific, and you've not identified which system you're using. Granted, it is reasonably easy to guess that you're using Linux on an Intel machine, but it might be 32-bit or 64-bit Linux, and what the shell code does still needs to be understood to know how to redirect its output. If it opens the terminal and writes to it, you'll be hard put to avoid the output appearing on the terminal, for example.
And user2059300 stated:
It's a sys_write syscall output to a file descriptor.
prompting me to ask:
Which file descriptor? 0, 1, 2, or another?
David C. Rankin noted:
That's still up in the air. Dumping to assembly shows a call to fprintf, but redirecting both stderr and stdout does nothing. It's almost like a kernel printf is being used.
And I countered with:
That's why I listed 0 as a candidate file descriptor. File descriptor 0 is usually a dup of file descriptors 1 and 2, so you can both write to 0 (standard input) and read from 1 and 2 (standard output and standard error). Of course, doing so is totally unsupported and non-portable (and you can't do it via the file streams stdin, stdout, stderr), but then shell code in general is non-portable.
And that seems to have been the key. David C. Rankin confirmed:
Hah! You nailed it. ./testshell > output.txt 0>&1 worked just fine. Learning has occurred… Thanks. That is the first, and hopefully the last time, I'll run across that. I'm too old to be learning those tricks.
and so did user2059300:
Thank you guys so much. Learning has indeed occurred. ./testshell > output.txt 0>&1
Obviously, I didn't know that was going to be the solution, but when the I/O redirection of standard output and standard error failed, it became a possibility. That behaviour has a very long history (7th Edition Unix, probably before that, too), though seldom if ever explicitly documented.

When non-blocking I/O is turned on for stdout, is it correct for the OS to turn it on for stdin too?

I have noticed some unexpected behaviour on both OS X and Linux. Turning on non-blocking I/O (using O_NONBLOCK) for standard output turns it on for standard input too!
Are these OSes behaving correctly? If so, is this behaviour defined by POSIX? Please point me to the relevant documentation if this is the case.
Here's a example program I used to test this:
#include <fcntl.h>
#include <stdio.h>
#include <string.h>
#include <unistd.h>
int main (int argc, char *argv[]) {
int flags = fcntl(STDOUT_FILENO, F_GETFL);
if (argc > 1 && strcmp(argv[1], "1") == 0) {
fcntl(STDOUT_FILENO, F_SETFL, flags | O_NONBLOCK);
}
printf("stdout O_NONBLOCK is: %d\n", fcntl(STDOUT_FILENO, F_GETFL) & O_NONBLOCK);
printf("stdin O_NONBLOCK is: %d\n", fcntl(STDIN_FILENO, F_GETFL) & O_NONBLOCK);
return 0;
}
On OS X:
$ clang -o fd-test fd-test.c
$ ./fd-test
stdout O_NONBLOCK is: 0
stdin O_NONBLOCK is: 0
$ ./fd-test 1
stdout O_NONBLOCK is: 4
stdin O_NONBLOCK is: 4
On Linux:
$ gcc -o fd-test fd-test.c
$ ./fd-test
stdout O_NONBLOCK is: 0
stdin O_NONBLOCK is: 0
$ ./fd-test 1
stdout O_NONBLOCK is: 2048
stdin O_NONBLOCK is: 2048
When a process is started from the shell, stdin, stdout and stderr point to the same file description. This file description is marked as O_NONBLOCK by your fcntl(1) call and therefore expectedly both file descriptors are marked as O_NONBLOCK.
If you want to indeed write to the same file from two file descriptors but want one to be marked as O_NONBLOCK, you need to create a new file description for the same file. If you know the file name of the file in question, you can achieve this by simply calling open() with the desired file name and flags. On some operating systems, you can find the file name using a platform-specific API, such as the /proc/fd virtual file system (many Unices including Linux) or the fd2path() function from Plan 9. Refer to this question for more details.
Note that simply calling open("/dev/fd/0", ...) may not work as intended as it returns a pointer to the same file description on some platforms if I recall correctly.

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