UNIX for Advanced & Expert Users

hi

I have created a application which uses SIGUSR2. It send this signal to server and waits for signal SIGUSR2 from server after server performing some operation server sends SIGUSR2 back to the application. The application then quits.

This works fine which ran from terminal , but when I use this application in a shell script, shrll script stop execution after execution of this application with USER SIGNAL 2 Message.

Is is a normal behaviour, is tere some way to stop this message

kindly some one clarify me with this

with hope

Yes, the default action for SIGUSR1 and SIGUSR2 is to terminate the process and this is what ksh, at least, does. I guess whatever shell you are using works the same way. Did you send the signal to the wrong process? The shell can only respond to signal it receives. Depending on your shell, it may be possible to ignore the signal. But it sounds like the shell got a signal intended for another process.

thanks for reply

You are right...i too thought that shell script is recieveing the signal,and i checked for that, but my application is also recieveing SIGUSR2, as server takes
the PID of the process which sends SIGUSR2 to it(in this case,my application),and server then sends back the same signal to application. I am not sure why Shell script is recieveing this signal, Is there any possibility that the signal is propagated from my application to shell in which its running?

I don't see how. I wrote a quick script that invoked "sleep 100". I killed the sleep process with usr2 and the script continued fine. But it detected the death due to a signal and printed a message about that. The shell is simply using wait() to determine how the sleep process died. It did not get the signal itself.