Open a shell in the second process of a pipe - shell

I'm having problems understanding what's going on in the following situation. I'm not familiar with UNIX pipes and UNIX at all but have read documentation and still can't understand this behaviour.
./shellcode is an executable that successfully opens a shell:
seclab$ ./shellcode
$ exit
seclab$
Now imagine that I need to pass data to ./shellcode via stdin, because this reads some string from the console and then prints "hello " plus that string. I do it in the following way (using a pipe) and the read and write works:
seclab$ printf "world" | ./shellcode
seclab$ hello world
seclab$
However, a new shell is not opened (or at least I can't see it and iteract with it), and if I run exit I'm out of the system, so I'm not in a new shell.
Can someone give some advice on how to solve this? I need to use printf because I need to input binary data to the second process and I can do it like this: printf "\x01\x02..."

When you use a pipe, you are telling Unix that the output of the command before the pipe should be used as the input to the command after the pipe. This replaces the default output (screen) and default input (keyboard). Your shellcode command doesn't really know or care where its input is coming from. It just reads the input until it reaches the EOF (end of file).
Try running shellcode and pressing Control-D. That will also exit the shell, because Control-D sends an EOF (your shell might be configured to say "type exit to quit", but it's still responding to the EOF).
There are two solutions you can use:
Solution 1:
Have shellcode accept command-line arguments:
#!/bin/sh
echo "Arguments: $*"
exec sh
Running:
outer$ ./shellcode foo
Arguments: foo
$ echo "inner shell"
inner shell
$ exit
outer$
To feed the argument in from another program, instead of using a pipe, you could:
$ ./shellcode `echo "something"`
This is probably the best approach, unless you need to pass in multi-line data. In that case, you may want to pass in a filename on the command line and read it that way.
Solution 2:
Have shellcode explicitly redirect its input from the terminal after it's processed your piped input:
#!/bin/sh
while read input; do
echo "Input: $input"
done
exec sh </dev/tty
Running:
outer$ echo "something" | ./shellcode
Input: something
$ echo "inner shell"
inner shell
$ exit
outer$
If you see an error like this after exiting the inner shell:
sh: 1: Cannot set tty process group (No such process)
Then try changing the last line to:
exec bash -i </dev/tty

Related

How to read user's input from bash (when catting a script) [duplicate]

I have a simple Bash script:
#!/usr/bin/env bash
read X
echo "X=$X"
When I execute it with ./myscript.sh it works. But when I execute it with cat myscript.sh | bash it actually puts echo "X=$X" into $X.
So this script prints Hello World executed with cat myscript.sh | bash:
#!/usr/bin/env bash
read X
hello world
echo "$X"
What's the benefit of executing a script with cat myscript.sh | bash? Why doesn't do it the same things as if I execute it with ./myscript.sh?
How can I avoid Bash to execute line by line but execute all lines after the STDIN reached the end?
Instead of just running
read X
...instead replace it with...
read X </dev/tty || {
X="some default because we can't read from the TTY here"
}
...if you want to read from the console. Of course, this only works if you have a /dev/tty, but if you wanted to do something robust, you wouldn't be piping from curl into a shell. :)
Another alternative, of course, is to pass in your value of X on the command line.
curl https://some.place/with-untrusted-code-only-idiots-will-run-without-reading \
| bash -s "value of X here"
...and refer to "$1" in your script when you want X.
(By the way, I sure hope you're at least using SSL for this, rather than advising people to run code they download over plain HTTP with no out-of-band validation step. Lots of people do it, sure, but that's making sites they download from -- like rvm.io -- big targets. Big, easy-to-man-in-the-middle-or-DNS-hijack targets).
When you cat a script to bash the code to execute is coming from standard input.
Where does read read from? That's right also standard input. This is why you can cat input to programs that take standard input (like sed, awk, etc.).
So you are not running "a script" per-se when you do this. You are running a series of input lines.
Where would you like read to read data from in this setup?
You can manually do that (if you can define such a place). Alternatively you can stop running your script like this.

In bash print to line above terminal output

EDIT: Corrected process/thread terminology
My shell script has a foreground process that reads user input and a background process that prints messages. I would like to print these messages on the line above the input prompt rather than interrupting the input. Here's a canned example:
sleep 5 && echo -e "\nINFO: Helpful Status Update!" &
echo -n "> "
read input
When I execute it and type "input" a bunch of times, I get something like this:
> input input input inp
INFO: Helpful Status Update!
ut input
But I would like to see something like this:
INFO: Helpful Status Update!
> input input input input input
The solution need not be portable (I'm using bash on linux), though I would like to avoid ncurses if possible.
EDIT: According to #Nick, previous lines are inaccessible for historical reasons. However, my situation only requires modifying the current line. Here's a proof of concept:
# Make named pipe
mkfifo pipe
# Spawn background process
while true; do
sleep 2
echo -en "\033[1K\rINFO: Helpful Status Update!\n> `cat pipe`"
done &
# Start foreground user input
echo -n "> "
pid=-1
collected=""
IFS=""
while true; do
read -n 1 c
collected="$collected$c"
# Named pipes block writes, so must do background process
echo -n "$collected" >> pipe &
# Kill last loop's (potentially) still-blocking pipe write
if kill -0 $pid &> /dev/null; then
kill $pid &> /dev/null
fi
pid=$!
done
This produces mostly the correct behavior, but lacks CLI niceties like backspace and arrow navigation. These could be hacked in, but I'm still having trouble believing that a standard approach hasn't already been developed.
The original ANSI codes still work in bash terminal on Linux (and MacOS), so you can use \033[F where \033 is the ESCape character. You can generate this in bash terminal by control-V followed by the ESCape character. You should see ^[ appear. Then type [F. If you test the following script:
echo "original line 1"
echo "^[[Fupdated line 1"
echo "line 2"
echo "line 3"
You should see output:
updated line 1
line 2
line 3
EDIT:
I forgot to add that using this in your script will cause the cursor to return to the beginning of the line, so further input will overwrite what you have typed already. You could use control-R on the keyboard to cause bash to re-type the current line and return the cursor to the end of the line.

Reading input while also piping a script via stdin

I have a simple Bash script:
#!/usr/bin/env bash
read X
echo "X=$X"
When I execute it with ./myscript.sh it works. But when I execute it with cat myscript.sh | bash it actually puts echo "X=$X" into $X.
So this script prints Hello World executed with cat myscript.sh | bash:
#!/usr/bin/env bash
read X
hello world
echo "$X"
What's the benefit of executing a script with cat myscript.sh | bash? Why doesn't do it the same things as if I execute it with ./myscript.sh?
How can I avoid Bash to execute line by line but execute all lines after the STDIN reached the end?
Instead of just running
read X
...instead replace it with...
read X </dev/tty || {
X="some default because we can't read from the TTY here"
}
...if you want to read from the console. Of course, this only works if you have a /dev/tty, but if you wanted to do something robust, you wouldn't be piping from curl into a shell. :)
Another alternative, of course, is to pass in your value of X on the command line.
curl https://some.place/with-untrusted-code-only-idiots-will-run-without-reading \
| bash -s "value of X here"
...and refer to "$1" in your script when you want X.
(By the way, I sure hope you're at least using SSL for this, rather than advising people to run code they download over plain HTTP with no out-of-band validation step. Lots of people do it, sure, but that's making sites they download from -- like rvm.io -- big targets. Big, easy-to-man-in-the-middle-or-DNS-hijack targets).
When you cat a script to bash the code to execute is coming from standard input.
Where does read read from? That's right also standard input. This is why you can cat input to programs that take standard input (like sed, awk, etc.).
So you are not running "a script" per-se when you do this. You are running a series of input lines.
Where would you like read to read data from in this setup?
You can manually do that (if you can define such a place). Alternatively you can stop running your script like this.

How can I automatically send commands and capture output with a program that is still running?

I am running a program that accepts commands and provides terminal output, but does not terminate once it has finished executing a command, and instead waits for the next command. I want to use a bash script to read lines from a file, send those lines as commands to the program, and then write the program's output to a file.
I've been trying something like this:
program
while read line
do
$line # I want this sent as a command to the program
$variable=(`another command to program`)
echo $variable >> $2
done <$1
But this fails at line 3; the command is not sent to the program.
In bash 4.x:
coproc program
while read -u "${COPROC[0]}" line; do
echo "something" >&"${COPROC[1]}"
done
Older shells can (somewhat more painfully) be made to do this with named pipes.

Bash process substitution: what does `echo >(ls)` do?

Here is an example of Bash's process substitution:
zjhui#ubuntu:~/Desktop$ echo >(ls)
/dev/fd/63
zjhui#ubuntu:~/Desktop$ abs-guide.pdf
Then I get a cursor waiting for a command.
/dev/fd/63 doesn't exist. I think what happens is:
Output the filename used in /dev/fd
Execute the ls in >(ls)
Is this right? Why is there a cursor waiting for input?
When you execute echo >(ls), bash replaces this with echo /dev/fd/63 and runs ls with /dev/fd/63 connected to its standard input. echo does not use its arguments as file names, and ls does not use standard input. Bash will write your standard prompt but the output of ls comes after it. You can type in any Bash command there, you just lost the visual cue from the prompt, which is further up the screen.
echo >(ls) is not something that is likely to ever be useful.

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