读完后这关于通过更改PT_INTERP
为自定义解释器来执行任意程序的文章,我尝试在本地进行实验:
$ cat flag.c
#include <stdio.h>
int main(int argc, char **argv) {
printf("Hello World!\n");
return 0;
}
$ gcc -static flag.c -o flag
$ cat solution.c
const char interp_section[] __attribute__((section(".interp"))) = "./flag";
$ gcc -s -fno-ident -Wl,--build-id=none -Wl,-e,0 -static -nostdlib solution.c -o solution
$ ./solution
Segmentation fault
$ ./flag
Hello World!
这个
PT_INTERP
程序头包含一个路径,这是我们的ELF将运行的解释器(可执行文件)的路径
既然solution
requests./flag
作为其解释器,为什么不./flag
运行并打印消息“Hello World”,何时solution
执行?相反,会发生分段错误,这与本文中的行为不同。
如何成功注册并执行自定义解释器PT_INTERP
?
$ readelf -l solution
Elf file type is EXEC (Executable file)
Entry point 0x0
There are 4 program headers, starting at offset 64
Program Headers:
Type Offset VirtAddr PhysAddr
FileSiz MemSiz Flags Align
PHDR 0x0000000000000040 0x0000000000400040 0x0000000000400040
0x00000000000000e0 0x00000000000000e0 R 0x8
INTERP 0x0000000000000120 0x0000000000400120 0x0000000000400120
0x0000000000000007 0x0000000000000007 R 0x1
[Requesting program interpreter: ./flag]
LOAD 0x0000000000000000 0x0000000000400000 0x0000000000400000
0x0000000000000127 0x0000000000000127 R 0x1000
GNU_STACK 0x0000000000000000 0x0000000000000000 0x0000000000000000
0x0000000000000000 0x0000000000000000 RW 0x10
Section to Segment mapping:
Segment Sections...
00
01 .interp
02 .interp
03
答案1
如果你运行dmesg
,你应该看到类似的东西
(solution): Uhuuh, elf segment at 0000000000400000 requested but the memory is mapped already
原因是flag
和solution
都有一个位于 0x400000 的段。
避免这种情况的快速方法是使用二进制tinyelf
helloworld
作为flag
;它没有冲突的部分,一切正常。
改编 TinyELF 示例,将其写入flag.c
:
#include <asm/unistd.h>
long syscall(long n,
long a1, long a2, long a3, long a4, long a5, long a6);
void _start() {
syscall(__NR_write, 1, (long) "Hello World!\n", 13, 0, 0, 0);
syscall(__NR_exit, 0, 0, 0, 0, 0, 0);
}
long syscall(long n, long a1, long a2, long a3, long a4, long a5, long a6) {
asm volatile ("movq %4, %%r10;"
"movq %5, %%r8;"
"movq %6, %%r9;"
"syscall;"
: "=a"(n)
: "a"(n), "D"(a1), "S"(a2), "d"(a3),
"r"(a4), "r"(a5), "r"(a6)
: "%r10", "%r8", "%r9");
return n;
}
构建如下:
gcc -nostartfiles -nostdlib flag.c -o flag
然后solution
你已有的二进制文件就可以工作了。