An interactive shell is one started without non-option arguments
and without the -c option whose standard input and error are both
connected to terminals (as determined by isatty(3)),or one started
with the -i option. PS1 is set and $- includes i if bash is
interactive,allowing a shell script or a startup file to test this
state.

然而,使用ssh执行命令也会导致运行.bashrc,这与我期望的相反,因为该命令不是以交互方式运行的.所以,这种行为看起来像一个bug,但它似乎在我尝试过的Red Hat和bash的所有版本上都很普遍.有人能解释为什么这种行为是正确的吗？

还有一点：即使运行.bashrc,也会设置$ – 和$PS,就好像shell是非交互式的(正如我所料).

$grep USER /etc/passwd
USER:x:UID:GID:UNAME:/home/USER:/bin/bash

$cat ~/.bashrc
echo bashrc:$-,$PS1

$bash -c 'echo $-'
hBc

$ssh localhost 'echo $-' </dev/null 2>/dev/null
USER@localhost's password:
bashrc:hBc,hBc

$ssh localhost 'ps -ef | grep $$' </dev/null 2>/dev/null
USER@localhost's password:
bashrc:hBc,USER 28296 28295 0 10:04 ? 00:00:00 bash -c ps -ef | grep $$
USER 28297 28296 0 10:04 ? 00:00:00 ps -ef
USER 28298 28296 0 10:04 ? 00:00:00 grep 28296

我目前正在通过在.bashrc中测试[[$ – = * i *]]来解决这个问题,但似乎我不应该这样做.

一个示例服务器,在我的主目录中不包含.bashrc(和.ssh)中的其他文件具有以下配置：

$cat /etc/redhat-release
Red Hat Enterprise Linux Server release 5.7 (Tikanga)

$bash --version
GNU bash,version 3.2.25(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2005 Free Software Foundation,Inc.

我尝试过的bash版本：3.00.15,3.1.17,3.2.25,4.1.2(后者在Red Hat 6.3上).