# [2021q1](http://wiki.csie.ncku.edu.tw/linux/schedule) 第 15 週測驗題
###### tags: `linux2021`
:::info
目的: 檢驗學員對 Linux 記憶體管理、memfd 和 mmap 系統呼叫的認知
:::
==[作答表單](https://docs.google.com/forms/d/e/1FAIpQLSd-aG39Z3xcAcWmJaGrrOzuoZuY3g6vn2dDmMr3W0cuBp4Xew/viewform)==
### 測驗 `1`
在 [你所不知道的 C 語言:連結器和執行檔資訊](https://hackmd.io/@sysprog/c-linker-loader) 提過 ELF 執行檔格式,更多資訊可見 [Executable and Linkable Format](https://en.wikipedia.org/wiki/Executable_and_Linkable_Format),以 64 位元 ELF 來說,開頭的幾個位元組的意義:
| offset | size | Purpose |
|:------:|:----:|:-------:|
| 0x00 | 4 | 0x7F followed by ELF(`45` `4c` `46`) in ASCII; these four bytes constitute the magic number. |
| 0x04 | 1 | This byte is set to either 1 or 2 to signify 32- or 64-bit format, respectively. |
| 0x05 | 1 | This byte is set to either 1 or 2 to signify little or big endianness, respectively. This affects interpretation of multi-byte fields starting with offset `0x10`. |
| x06 | 1 | Set to 1 for the original and current version of ELF. |
| ... | ... | ...待續... |
以下程式碼嘗試在既有的 ELF 檔案內嵌另一個 ELF 檔案 (可預先加密),目的是隱匿特定的程式,避免被掃毒程式或防火牆偵測出來,或將高價值的程式嵌入到文件、圖片,甚至是影音檔案中,透過特定的載入器自檔案提取出執行檔並執行,這手法在 [Digital rights management (DRM)](https://en.wikipedia.org/wiki/Digital_rights_management) 和 [Digital watermarking](https://en.wikipedia.org/wiki/Digital_watermarking) 領域不算少見。
假設即將被嵌入的程式碼名為 `payload.c`:
```cpp
#include <stdio.h>
int main() { puts("Hello world!"); return 0; }
```
編譯並移去除錯用的符號:
```shell
$ gcc -Os payload.c -o payload
$ strip -s payload
```
接著我們要開發得以載入 ELF 的程式,在這之前,先探討以下函式及系統呼叫:
* [memfd_create](http://man7.org/linux/man-pages/man2/memfd_create.2.html): 詳見 [解析 Linux 共享記憶體機制](https://hackmd.io/@sysprog/linux-shared-memory) 一文
* [memmem](http://man7.org/linux/man-pages/man3/memmem.3.html): GNU extension,在給定的記憶體範圍找到「非」C-style 字串 (仍為連續記憶體)
* [fexecve](http://man7.org/linux/man-pages/man3/fexecve.3.html): 類似 execve 系統呼叫,但由給定的 file descriptor 載入程式並執行
假定程式載入器檔名為 `loader.c`,內容如下:
```cpp
/* A program that executes a second (embedded) ELF */
#define _GNU_SOURCE
#include <errno.h>
#include <fcntl.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/stat.h>
#include <unistd.h>
/* No glibc wrappers exist for memfd_create(2), so provide our own. */
#include <sys/syscall.h>
static inline int memfd_create(const char *name, unsigned int flags)
{
return syscall(__NR_memfd_create, name, flags);
}
/* ELF format
* https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
*/
static bool valid_elf(char *ptr)
{
return (ptr[4] == 1 || ptr[4] == 2) /* offset 0x4: 32/64-bit format */ &&
(ptr[5] == 1 || ptr[5] == 2) /* offset 0x5: endianness */ &&
(ptr[6] == 1); /* offset 0x6: current version */
}
int main(int argc, char *argv[], char **envp)
{
int pid = getpid();
int ret = 0;
char proc_path[32];
sprintf(proc_path, "/proc/%d/exe", pid);
int filedesc = open(proc_path, O_RDONLY);
if (filedesc < 0) {
printf("Invalid file descriptor for /proc: %d\n", filedesc);
return -1;
}
/* Find the size of this executable */
struct stat st;
stat(proc_path, &st);
size_t size = st.st_size;
char *entirefile = malloc(size);
if (!entirefile) {
printf("Insufficient memory.\n");
return -2;
}
read(filedesc, entirefile, size);
close(filedesc);
/* find the second ELF header, which 52 or 64 bytes long for 32-bit and
* 64-bit binaries respectively.
*/
const char elf_magic[] = {0x7F, 'E', 'L', 'F'};
char *newelf = memmem(entirefile + 52, size - 52, elf_magic, 4);
if (newelf && !valid_elf(newelf)) /* forcely find again for real ELF */
newelf = memmem(newelf + 6, size - (intptr_t) newelf - 6, elf_magic, 4);
if (!newelf || !valid_elf(newelf)) {
printf("No second ELF header found.\n");
ret = -3;
goto cleanup;
}
int newsize = AAA;
int memfd = memfd_create("hidden", 0);
if (memfd < 0) {
printf("Invalid memfd.\n");
ret = -4;
goto cleanup;
}
/* Write ELF to temporary memory file */
write(memfd, newelf, newsize);
// Deploy the payload as a different process
fork();
if (BBB) {
ret = fexecve(memfd, argv, envp); /* Execute the in-memory ELF */
/* The above will only return if there is an error. */
printf("Fail to execute payload. ret=%d (%s)\n", ret, strerror(errno));
}
cleanup:
free(entirefile);
return ret;
}
```
編譯、嵌入上述 `payload` 執行檔,然後再執行: (你沒看錯,真的用 `cat` 命令)
```shell
$ gcc -Wall loader.c -o loader
$ cat payload >> loader
$ ./loader
```
在 x86_64 GNU/Linux (核心版本: 4.15+) 預期輸出為:
```
Hello world!
```
> 注意:只有一行 "Hello world!" 字串
請補完程式碼,只要考慮 x86_64 硬體架構即可。
==作答區== (注意: 複選題,儘量選取有效的答案)
AAA = ?
* `(a)` `newelf - entirefile`
* `(b)` `size - newelf`
* `(c)` `entirefile - newelf`
* `(d)` `size - newelf - entirefile`
* `(e)` `size - newelf + entirefile`
* `(f)` `newelf - entirefile + size`
* `(g)` `entirefile - newelf - size`
`size - (newelf - entirefile)`
BBB = ?
* `(a)` `getpid()`
* `(b)` `getpid() != pid`
* `(c)` `getpid() == pid`
* `(d)` `0`
* `(e)` `1`
:::success
延伸問題:
1. 解釋上述程式碼運作原理,指出其中不足處並改進;
2. 參照 [Embedding binary data in executables](https://csl.name/post/embedding-binary-data/) 和 [incbin](https://github.com/graphitemaster/incbin),將 payload 加密並嵌入到給定的 C 程式中,允許在執行時期解密再載入 payload 並執行
3. 學習 [Digital rights management (DRM)](https://en.wikipedia.org/wiki/Digital_rights_management) 手法,實作一個電子書程式,將特定的文字檔案加密再嵌入於執行檔中,只有在特定的機器 (例如偵測 [MAC address](https://simple.wikipedia.org/wiki/MAC_address)) 才能開啟閱讀,過程中不會在檔案系統出現明文的文字檔案暫存檔。
:::
---
### 測驗 `2`
考慮以下透過 mmap 實作快速檔案複製的程式碼: `mmap-filecopy.c`
```cpp
/* copy modified blocks of source file to destination file efficiently
* using mmap.
*/
#include <assert.h>
#include <fcntl.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <sysexits.h>
#include <unistd.h>
int main(int argc, char *argv[])
{
if (argc != 3) {
printf("Usage: %s <source> <destination>\n", argv[0]);
return EX_USAGE;
}
const char *src_name = argv[1];
const char *dst_name = argv[2];
int src_fd, dst_fd;
struct stat dst_stat = {0};
off_t src_len, dst_len;
src_fd = open(src_name, O_RDONLY);
if (src_fd == -1) {
perror(src_name);
return EX_DATAERR;
}
dst_fd = open(dst_name, O_RDWR | O_CREAT,
S_IRUSR | S_IWUSR | S_IRGRP | S_IWGRP | S_IROTH | S_IWOTH);
if (dst_fd == -1 || fstat(dst_fd, &dst_stat) != 0) {
perror(dst_name);
return EX_DATAERR;
}
src_len = lseek(src_fd, 0, SEEK_END);
if (src_len < 0) {
perror(src_name);
return EX_DATAERR;
}
dst_len = lseek(dst_fd, 0, SEEK_END);
if (dst_len < 0) {
perror(dst_name);
return EX_DATAERR;
}
if (dst_len > src_len) {
printf("Destination file is larger (%zd) than input file (%zd)\n",
dst_len, src_len);
return EX_DATAERR;
}
const size_t page_size =
dst_stat.st_blksize > 0 ? dst_stat.st_blksize : BUFSIZ;
const size_t len = src_len;
if (ftruncate(dst_fd, len) != 0) {
perror(dst_name);
return EX_DATAERR;
}
size_t read_count = 0;
size_t write_count = 0;
if (len > 0) {
const uint8_t *src;
uint8_t *dst;
src = mmap(NULL, len, PROT_READ, MAP_SHARED, src_fd, 0);
if (src == NULL ||
posix_madvise((void *) src, len, POSIX_MADV_SEQUENTIAL) != 0) {
perror(src_name);
return EX_UNAVAILABLE;
}
dst = mmap(NULL, len, PROP, MAP_SHARED, dst_fd, 0);
if (dst == NULL ||
posix_madvise(dst, len, POSIX_MADV_SEQUENTIAL) != 0) {
perror(dst_name);
return EX_UNAVAILABLE;
}
for (size_t i = 0; i < len; i += page_size) {
size_t block_size = (len - i) >= page_size ? page_size : (len - i);
if (memcmp(src + i, dst + i, block_size)) {
memcpy(dst + i, src + i, block_size);
write_count += block_size;
}
read_count += block_size;
}
if (munmap((void *) src, len) != 0) {
perror(src_name);
return EX_UNAVAILABLE;
}
if (msync(dst, len, MS_SYNC) != 0 || munmap(dst, len) != 0) {
perror(dst_name);
return EX_UNAVAILABLE;
}
}
if (close(src_fd) != 0) {
perror(src_name);
return EX_UNAVAILABLE;
}
if (close(dst_fd) != 0) {
perror(dst_name);
return EX_UNAVAILABLE;
}
printf("%zu bytes read\n", read_count);
printf("%zu bytes written\n", write_count);
return EXIT_SUCCESS;
}
```
編譯方式:
```shell
$ gcc -std=c11 -D_POSIX_C_SOURCE=200809L -o mmap-filecopy mmap-filecopy.c
```
假設原本已有檔名為 `in` 的檔案,且 `out` 不存在目前的路徑,可執行以下命令:
```shell
$ ./mmap-filecopy in out
```
這樣即可達成快速的檔案複製。
請補完程式碼,使得符合預期。
==作答區==
PROP = ?
* `(a)` `PROT_READ | PROT_WRITE`
* `(b)` `PROT_READ`
:::success
延伸問題:
1. 解釋上述程式碼運作原理,並指出其缺失
2. 探討 sendfile 和 splice 等系統系統在上述程式的應用
* 參見 [以 sendfile 和 splice 系統呼叫達到 Zero-Copy](https://hackmd.io/@sysprog/linux2020-zerocopy)
:::