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我正在查看GNU coreutils的源代码,特别是圆形检测。cat他们正在比较设备和 inode 并且工作正常,但是如果输入为空,他们允许输出作为输入的额外情况。查看代码,这是必须的lseek (input_desc, 0, SEEK_CUR) < stat_buf.st_size)部分。我阅读了从 中找到的联机帮助页和讨论git blame,但我仍然不太明白为什么需要此调用lseek

这是如何cat检测的要点,如果它会无限耗尽磁盘(请注意,为简洁起见,还删除了一些错误检查,完整的源代码在上面链接):

struct stat stat_buf;
fstat(STDOUT_FILENO, &stat_buf);
out_dev = stat_buf.st_dev;
out_ino = stat_buf.st_ino;
out_isreg = S_ISREG (stat_buf.st_mode) != 0;

// ...
// for <infile> in inputs {
    input_desc = open (infile, file_open_mode); // or STDIN_FILENO
    fstat(input_desc, &stat_buf);
    /* Don't copy a nonempty regular file to itself, as that would
       merely exhaust the output device.  It's better to catch this
       error earlier rather than later.  */
    if (out_isreg 
        && stat_buf.st_dev == out_dev && stat_buf.st_ino == out_ino
        && lseek (input_desc, 0, SEEK_CUR) < stat_buf.st_size)         // <--- This is the important line
    {
      // ...
    }
// } (end of for)

我有两种可能的解释,但似乎都有些奇怪。

  1. 根据某些标准(posix),文件可能是“空的”,尽管它仍然包含一些信息(用 计数st_size)和lseek/或open通过默认偏移来尊重这些信息。我不知道为什么会这样,因为空意味着空,对吧?
  2. 这种比较确实是两个条件的“聪明”组合。首先这对我来说是有意义的,因为如果input_descSTDIN_FILENO并且不会有文件传送到stdin,lseek会失败ESPIPE(根据手册页)并返回-1。那么,整个语句将是lseek(...) == -1 || stat_buf.st_size > 0. 但这不可能是真的,因为只有在设备和 inode 相同的情况下才会进行此检查,并且只有在 a) stdin 和 stdout 指向相同的 pty 时才会发生这种情况,但随后out_isreg会是false或者 b) stdin 和 stdout 指向同一个文件,但随后lseek无法返回-1,对吗?

我还编写了一个小程序,可以打印出返回值和errno重要部分,但对我来说没有什么突出的:

#include <errno.h>
#include <fcntl.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/stat.h>
#include <unistd.h>

int main(int argc, char **argv) {
  struct stat out_stat;
  struct stat in_stat;

  if (fstat(STDOUT_FILENO, &out_stat) < 0)
    exit(1);

  printf("this is written to stdout / into the file\n");

  int fd;
  if (argc > 1)
    fd = open(argv[1], O_RDONLY);
  else
    fd = STDIN_FILENO;

  fstat(fd, &in_stat);
  int res = lseek(fd, 0, SEEK_CUR);
  fprintf(stderr,
          "errno after lseek = %d, EBADF = %d, EINVAL = %d, EOVERFLOW = %d, "
          "ESPIPE = %d\n",
          errno, EBADF, EINVAL, EOVERFLOW, ESPIPE);

  fprintf(stderr, "input:\n\tlseek(...) = %d\n\tst_size = %ld\n", res,
          in_stat.st_size);

  printf("outsize is %ld", out_stat.st_size);
}

$ touch empty
$ ./a.out < empty > empty
errno after lseek = 0, EBADF = 9, EINVAL = 22, EOVERFLOW = 75, ESPIPE = 29
input:
        lseek(...) = 0
        st_size = 0
$ echo x > empty
$ ./a.out < empty > empty
errno after lseek = 0, EBADF = 9, EINVAL = 22, EOVERFLOW = 75, ESPIPE = 29
input:
        lseek(...) = 0
        st_size = 0

因此,我的研究没有触及我的最终问题:如何从源代码lseek中确定此示例中的文件是否为空?cat

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1 回答 1

3

这是我对此进行逆向工程的尝试——我找不到任何公开讨论来解释为什么lseek()放在那里(GNU coreutils 中没有其他地方这样做)。

指导性问题是:条件何时为lseek (input_desc, 0, SEEK_CUR) < stat_buf.st_size假?

测试用例:

#!/bin/bash
# (edited based on comments)

set -x

# arrange for cat to start off past the end of a non-empty file

echo abcdefghi > /tmp/so/catseek/input
# get the shell to open the input file for reading & writing as file descriptor 7
exec 7<>/tmp/so/catseek/input
# read the whole file via that descriptor (but leave it open)
dd <&7
# ask linux what the current file position of file descriptor 7 is
# should be everything dd read, namely 10 bytes, the size of the file
grep ^pos: /proc/self/fdinfo/7
# run cat, with pre and post content so that we know how to locate the interesting part
# "-" will cause cat to reuse its file descriptor 0 rather than creating a new file descriptor
# the redirections tell the shell to redirect file descriptors 1 and 0 to/from our open file descriptor 7
# which, as you'll remember, already has a file position of 10 bytes
strace -e lseek ./src/cat /tmp/so/catseek/pre - /tmp/so/catseek/post <&7 >&7
# now let's see what's in the file
cat /tmp/so/catseek/input

和:

$ cat /tmp/so/catseek/pre
pre
$ cat /tmp/so/catseek/post
post

catlseek (input_desc, 0, SEEK_CUR) < stat_buf.st_size

+ test.sh:8:echo abcdefghi
+ test.sh:10:exec
+ test.sh:12:dd
abcdefghi
0+1 records in
0+1 records out
10 bytes copied, 2.0641e-05 s, 484 kB/s
+ test.sh:15:grep '^pos:' /proc/self/fdinfo/7
pos:    10
+ test.sh:20:strace -e lseek ./src/cat /tmp/so/catseek/pre - /tmp/so/catseek/post
lseek(0, 0, SEEK_CUR)                   = 14
+++ exited with 0 +++
+ test.sh:22:cat /tmp/so/catseek/input
abcdefghi
pre
post

cat0 < stat_buf.st_size

+ test.sh:8:echo abcdefghi
+ test.sh:10:exec
+ test.sh:12:dd
abcdefghi
0+1 records in
0+1 records out
10 bytes copied, 3.6415e-05 s, 275 kB/s
+ test.sh:15:grep '^pos:' /proc/self/fdinfo/7
pos:    10
+ test.sh:20:strace -e lseek ./src/cat /tmp/so/catseek/pre - /tmp/so/catseek/post
./src/cat: -: input file is output file
+++ exited with 1 +++
+ test.sh:22:cat /tmp/so/catseek/input
abcdefghi
pre
post

如您所见,当cat开始时,文件位置可能已经在文件结尾之后,并且仅检查文件大小会cat跳过文件,但也会触发失败,因为if语句中的代码是:

error (0, 0, _("%s: input file is output file"), infile);
ok = false;
goto contin;

使用lseek()允许cat说“哦,文件是相同的,并且不是空的,但我们的读取仍然会变为空,因为这就是读取过去 EOF 的工作方式,所以我们可以允许这种情况”。

于 2021-01-12T21:38:44.957 回答