I was surprised with this comment in other question:
Sending dd the USR1 signal too soon after it has started (i.e. in a bash script, the line after you started it) will in fact terminate it
Can anybody explain why?
|
Each signal has a "default disposition" -- what a process does by default when it receives that signal. There's a table in the
|
|||
|
|
{ dd if=/dev/zero of=/dev/null & }; kill -USR1 $!; jobs; sleep 1; jobsto reproduce the effect you're describing. – jippie May 13 '12 at 22:26