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execve("/bin/sh", 0, 0); in a pipe

I have the following example program:

#include <stdio.h>

int
main(int argc, char ** argv){
    char buf[100];

    printf("Please enter your name: ");
    fflush(stdout);
    gets(buf);
    printf("Hello \"%s\"\n", buf);

    execve("/bin/sh", 0, 0);
}

I and when I run without any pipe it works as it should and returns a sh promt:

bash$ ./a.out
Please enter your name: warning: this program uses gets() which is unsafe.
testName
Hello "testName"
$ exit
bash$

But this does not work in a pipe, i think I know why that is, but I cannot figure out a solution. Example run bellow.

bash$ echo -e "testName\npwd" | ./a.out
Please enter your name: warning: this program uses gets() which is unsafe.
Hello "testName"
bash$

I figure this has something to do with the fact that gets empties stdin in such a way that /bin/sh receives a EOF and promtly quits without an error message.

But how do I get around this (without modifying the program, if possible, and not removing gets, if not) so that I get a promt even though I supply input through a pipe?

P.S. I am running this on a FreeBSD (4.8) machine D.S.

like image 414
A. Nilsson Avatar asked Dec 14 '11 17:12

A. Nilsson


1 Answers

You can run your program without any modifications like this:

(echo -e 'testName\n'; cat ) | ./a.out

This way you ensure that your program's standard input doesn't end after what echo outputs. Instead, cat continues to supply input to your program. The source of that subsequent input is your terminal since this is where cat reads from.

Here's an example session:

bash-3.2$ cc stdin_shell.c 
bash-3.2$ (echo -e 'testName\n'; cat ) | ./a.out 
Please enter your name: warning: this program uses gets(), which is unsafe.
Hello "testName"
pwd
/home/user/stackoverflow/stdin_shell_question
ls -l
total 32
-rwxr-xr-x  1 user  group  9024 Dec 14 18:53 a.out
-rw-r--r--  1 user  group   216 Dec 14 18:52 stdin_shell.c
ps -p $$
  PID TTY           TIME CMD
93759 ttys000    0:00.01 (sh)
exit

bash-3.2$

Note that because shell's standard input is not connected to a terminal, sh thinks it is not executed interactively and hence does not display the prompt. You can type your commands normally, though.

like image 154
Adam Zalcman Avatar answered Oct 05 '22 04:10

Adam Zalcman