On Sun, 21 Jan 2007 11:50:40 +0100
Manfred Rebentisch wrote:

> Hello,
> I try to call ssh from my Linux-Daemon (or for testing with a simple
> C-program) with fork() and execvp() from the forked client. This works
> great with all programs I tested, but not with ssh.

Try setting -o 'BatchMode yes', either on the command line, or in your
options. Also try doing an ssh -vvv from your program and view the
output. It sounds like it is waiting for input.

Doug

--
For UNIX, Linux and security articles
visit http://SecurityBulletins.com/

Hello Doug,

Doug Spencer wrote:

> On Sun, 21 Jan 2007 11:50:40 +0100
> Manfred Rebentisch wrote:
>
>> Hello,
>> I try to call ssh from my Linux-Daemon (or for testing with a simple
>> C-program) with fork() and execvp() from the forked client. This works
>> great with all programs I tested, but not with ssh.
>
> Try setting -o 'BatchMode yes', either on the command line, or in your
> options. Also try doing an ssh -vvv from your program and view the
> output. It sounds like it is waiting for input.
>
I forgot to tell: testet with BachMode yes too. Same result.

Thank you for thinking about my problem.

Manfred

On Sun, 21 Jan 2007 17:33:37 +0100
Manfred Rebentisch wrote:

> Hello Doug,
>
> Doug Spencer wrote:

> > Try setting -o 'BatchMode yes', either on the command line, or in your
> > options. Also try doing an ssh -vvv from your program and view the
> > output. It sounds like it is waiting for input.
> >
> I forgot to tell: testet with BachMode yes too. Same result.

Do you get the same result if you try the command in a shell script
rather than a C program?

You should also try doing an strace and check the call and the result.
Do an strace -o /tmp/output.strace -f ./a.out and check the result
in /tmp/output.strace.

Doug

--
For UNIX, Linux and security articles
visit http://SecurityBulletins.com/

This program works for me:

---------------------------------------------------------------
#include
#include
#include

main ()
{
int status;
char *argv[] = {"ssh","remote_host","echo","foo",NULL};

if (fork()) {
wait(&status);
} else {
execvp("/usr/bin/ssh", argv);
}
}
---------------------------------------------------------------

Since the log shows successful authentication, something is happening
during the login. Try -n in case something waiting for input, as well as
-v as Doug suggested. Also, you might try putting aside any shell init
files on the remote side.

--
Richard Silverman
res@qoxp.net

Hello Richard,

Richard E. Silverman wrote:

>
> This program works for me:
>
your simple source does work for me too, including public key.
So the error is on my side, an executing object with a side effect...

Thank you!

Manfred

On 2007-01-21 19:28:38 +0100, "Richard E. Silverman" said:

> ---------------------------------------------------------------
> #include
> #include
> #include
>
> main () {
> int status;
> char *argv[] = {"ssh","remote_host","echo","foo",NULL};
>
> if (fork()) {
> wait(&status);
> } else {
> execvp("/usr/bin/ssh", argv);
> }
> }
> ---------------------------------------------------------------

Just a slight modification. The main function must return an int
according to the standard, use:

int main(int argc, char* argv[])

or

int main(void)


And don't forget the return statement. Just my 2 cents to avoid
undefined behaviors

--
Sensei

Research (n.): a discovery already published by a chinese guy one month
before you, copying a russian who did it in the 60s.

Hallo,
I found the reason for my problem. After I put the lines:

if( open("/dev/null", O_RDONLY) != 0) {
syslog(LOG_ERR, "failed to open /dev/null");
exit(EXIT_FAILURE);
}

before the call of execvp() and after closing STDIN anything runs!

Thank you to all!

Manfred