UNIX for Dummies Questions & Answers

I'm learning about the trap command from my bash book. I tried out the little script they gave:

Code:

trap "echo 'You hit control-C!' " INT
while true; do
	sleep 60
done

But when I type control-c, the script just stops and the message is not displayed. I checked stty all and saw that control-c is intr, but the same thing happened whet I tried that. Can anyone tell me what I'm missing?

I'm using terminal (BSD) in OS X

kill -l (lowercase L) lists the names of your signals. stty -a shows terminal settings.

Look for something like these in login profiles:

Code:

trap "" 2 3
trap 2 3
trap 'exit' 2
trap 'kill $$' 2

SIGINT = 2 on almost all systems.

The first turns off the signal in your login profile(s). trap 2 3 enables them. trap 'exit' 2
specificaly converts SIGINT into an exit command. trap 'kill $$' converts SIGINT into SIGTERM the default for kill.

I don't follow. Why would my book use int instead of numbers?

And when I run the script and control-c all I get is

^C

on the screen and the cursor keeps blinking

The syntax of the command is

Code:

trap  "command "  [signal name or signal number]

So, that said, you want to see if a trap already exists in your shell. I would say that yes, one does. And it ignores SIGINT (or 2). Why it exists I can't say. You need to work on that one. So, that is why I suggest looking at all of the various login profiles you have on your box. /etc/profile, $HOME/.profile, .kshrc, or .bashrc ---- whatever.

I agree with the rest that the signal is somehow disabled.

Whether you have INT or 2 in the signal name, this could be a better construct for a test script because it lets you exit the script.

Code:

trap 'echo You hit control-C!;exit' INT
while true; do
sleep 60
done

Perhaps try positively setting the key as the first line of the script.

Code:

stty intr '^C'

Methyl:
I tried setting the key but it didn't work:

stty: illegal option -- INT
usage: stty [-a|-e|-g] [-f file] [options]
trap: usage: trap [-lp] [arg signal_spec ...]


I wish I had a unix manual of some sort --

I typed trap -l and got this (on my iMac at home -- still the same result when I ran the script) :

1) SIGHUP 2) SIGINT 3) SIGQUIT 4) SIGILL
5) SIGTRAP 6) SIGABRT 7) SIGEMT 8) SIGFPE
9) SIGKILL 10) SIGBUS 11) SIGSEGV 12) SIGSYS
13) SIGPIPE 14) SIGALRM 15) SIGTERM 16) SIGURG
17) SIGSTOP 18) SIGTSTP 19) SIGCONT 20) SIGCHLD
21) SIGTTIN 22) SIGTTOU 23) SIGIO 24) SIGXCPU
25) SIGXFSZ 26) SIGVTALRM 27) SIGPROF 28) SIGWINCH
29) SIGINFO 30) SIGUSR1 31) SIGUSR2

so I'm guessing that the interrupt signal is SIGINT.

Even with this, the same thing happens -- ^C is displayed on the screen and it goes right back to the script. If I type control z it says this :

[1]+ Stopped sleep 60

and goes right back to the script -- and the job number increases by one every time I type it. I can't stop the script from running. The only way I can is to quit terminal, at which point I get a message saying the following processes will be stopped -- bash, and however many sleep jobs it's running.

Methyl -- and anyone else:

It's been a while since I've been here, but I've got something else.

I fooled around with some other scripts and got it to print a message -- are you getting it to display the message?

Here's a script that I did today:


It works fine -- the numbers zip by and get larger and larger. If I hit control - c I can see the message -- it's only a flash. If I put sleep 10 in the script after the echo command,and hit control-c, all I get is ^C displayed on the screen, but no message. I'm wondering if the message is displayed briefly, but so quickly that I can't see it. Am I not typing the command correctly and not noticing it, maybe?


If i try this script

It works fine. The message is displayed:

I think it has something to do with the sleep command, but I can't figure out why. Do you have any ideas?