bash中的奇怪行为(可能还有其他shell？)

Question

当我做:

/bin/bash -c 'cat /proc/$$/cmdline'

我得到的输出是:

cat/proc/25050/cmdline

而我预期的输出是:

/bin/bash -c 'cat /proc/$$/cmdline'

---

另一方面,当我这样做时:

/bin/bash -c 'echo $$; cat /proc/$$/cmdline'

我得到了预期的输出,即:

28259
/bin/bash-cecho $$; cat /proc/$$/cmdline

---

似乎$$是猫的pid而不是bash/sh的pid.
为什么是这样？
shell是否进行某种解析和execve()样式替换？如果是这样,它甚至在取代之前如何知道猫的PID？

Answer 1

为了理解这种行为，必须弄清楚如何bash执行在命令行上传递给它的命令。关键点是，如果命令足够简单，就没有fork（或clone类似的东西）。

$ strace -f -e clone,execve /bin/bash -c 'cat /proc/$$/cmdline'
execve("/bin/bash", ["/bin/bash", "-c", "cat /proc/$$/cmdline"], [/* 80 vars */]) = 0
execve("/bin/cat", ["cat", "/proc/2942/cmdline"], [/* 80 vars */]) = 0
cat/proc/2942/cmdline+++ exited with 0 +++
$

OTOH 如果命令更复杂，则bash分叉：

$ strace -f -e clone,execve /bin/bash -c 'echo $$; cat /proc/$$/cmdline'
execve("/bin/bash", ["/bin/bash", "-c", "echo $$; cat /proc/$$/cmdline"], [/* 80 vars */]) = 0
2933
clone(child_stack=0, flags=CLONE_CHILD_CLEARTID|CLONE_CHILD_SETTID|SIGCHLD, child_tidptr=0x7ff64e6779d0) = 2934
Process 2934 attached
[pid  2934] execve("/bin/cat", ["cat", "/proc/2933/cmdline"], [/* 80 vars */]) = 0
/bin/bash-cecho $$; cat /proc/$$/cmdline[pid  2934] +++ exited with 0 +++
--- SIGCHLD {si_signo=SIGCHLD, si_code=CLD_EXITED, si_pid=2934, si_uid=1000, si_status=0, si_utime=0, si_stime=0} ---
+++ exited with 0 +++
$

看起来$$是cat的pid而不是bash/sh的pid。

实际上两者都是。bash execvescat直接，所以一个变成另一个。

要了解无分叉行为到底需要什么，我们需要查看源代码。有这样一条评论：

      /*
       * IF
       *   we were invoked as `bash -c' (startup_state == 2) AND
       *   parse_and_execute has not been called recursively AND
       *   we're not running a trap AND
       *   we have parsed the full command (string == '\0') AND
       *   we're not going to run the exit trap AND
       *   we have a simple command without redirections AND
       *   the command is not being timed AND
       *   the command's return status is not being inverted
       * THEN
       *   tell the execution code that we don't need to fork
       */

来源