why is the value of LD_PRELOAD on the stack

问题

I'm studying buffer overflow and solving some wargames. There was a problem that all of the stack memory above the buffer is set to 0 except return address of main, which will be:

buffer
[0000000...][RET][000000...]

and I can overwrite that RET. So I found some hints for solving this problem. It was to use LD_PRELOAD. Some people said that LD_PRELOAD's value is in somewhere of stack not only in environment variable area of stack. So I set LD_PRELOAD and search it and found it using gdb.

$ export LD_PRELOAD=/home/coffee/test.so
$ gdb -q abcde
(gdb) b main
Breakpoint 1 at 0x8048476
(gdb) r
Starting program: /home/coffee/abcde

Breakpoint 1, 0x8048476 in main ()
(gdb) x/s 0xbffff6df
0xbffff6df:      "@èC\001@/home/coffee/test.so"
(gdb) x/s 0xbffffc59
0xbffffc59:      "LD_PRELOAD=/home/coffee/test.so"
(gdb) q
The program is running.  Exit anyway? (y or n) y
$

So there is! Now I know that LD_PRELOAD's value is on stack below the buffer and now I can exploit!

But I wonder why LD_PRELOAD is loaded on that memory address. The value is also on environment variable area of the stack!

What is the purpose of this? Thanks.

回答1:

Code to explore the stack layout:

#include <inttypes.h>
#include <stdio.h>

// POSIX 2008 declares environ in <unistd.h> (Mac OS X doesn't)
extern char **environ;

static void dump_list(const char *tag, char **list)
{
    char **ptr = list;
    while (*ptr)
    {
        printf("%s[%d] 0x%.16" PRIXPTR ": %s\n",
               tag, (ptr - list), (uintptr_t)*ptr, *ptr);
        ptr++;
    }
    printf("%s[%d] 0x%.16" PRIXPTR "\n",
           tag, (ptr - list), (uintptr_t)*ptr);
}

int main(int argc, char **argv, char **envp)
{
    printf("%d\n", argc);
    printf("argv         0x%.16" PRIXPTR "\n", (uintptr_t)argv);
    printf("argv[argc+1] 0x%.16" PRIXPTR "\n", (uintptr_t)(argv+argc+1));
    printf("envp         0x%.16" PRIXPTR "\n", (uintptr_t)envp);
    printf("environ      0x%.16" PRIXPTR "\n", (uintptr_t)environ);

    dump_list("argv", argv);
    dump_list("envp", envp);
    return(0);
}

With the program compiled as x, I ran it with a sanitized environment:

$ env -i HOME=$HOME PATH=$HOME/bin:/bin:/usr/bin LANG=$LANG TERM=$TERM ./x a bb ccc
4
argv         0x00007FFF62074EC0
argv[argc+1] 0x00007FFF62074EE8
envp         0x00007FFF62074EE8
environ      0x00007FFF62074EE8
argv[0] 0x00007FFF62074F38: ./x
argv[1] 0x00007FFF62074F3C: a
argv[2] 0x00007FFF62074F3E: bb
argv[3] 0x00007FFF62074F41: ccc
argv[4] 0x0000000000000000
envp[0] 0x00007FFF62074F45: HOME=/Users/jleffler
envp[1] 0x00007FFF62074F5A: PATH=/Users/jleffler/bin:/bin:/usr/bin
envp[2] 0x00007FFF62074F81: LANG=en_US.UTF-8
envp[3] 0x00007FFF62074F92: TERM=xterm-color
envp[4] 0x0000000000000000
$

If you study that carefully, you'll see that the argv argument to main() is the start of a series of pointers to strings further up the stack; the envp (optional third argument to main() on POSIX machines) is the same as the global variable environ and argv[argc+1], and is also the start of a series of pointers to strings further up the stack; and the strings pointed at by the argv and envp pointers follow the two arrays.

This is the layout on Mac OS X (10.7.5 if it matters, which it probably doesn't), but I'm tolerably sure you'd find the same layout on other Unix-like systems.

来源：https://stackoverflow.com/questions/16121364/why-is-the-value-of-ld-preload-on-the-stack

标签

Linux

memory

stack

overflow

ld-preload