Why does killall (sometimes?) needs to be applied twice?

Question

Why does killall (sometimes?) need to be applied twice? Here's an example where a "double" killall is needed to end the process.

When and why does a process needed to be killed "twice"? What happens under the bonnet?

That example should not need 2 killalls. Chrome probably will start a clean up process (clearing cache etc.) that will take a bit of time. The 2nd one probably "works" because the 1st one actually ended. — Rinzwind, Sep 21 '15 at 21:44
@Rinzwind does the 2nd one possibly "override" the grace period (which I found loosely being referred to in a number of other posts)? Is there a way/flag to verbosely monitor the grace period "count-down"? Is there some hard evidence / source code to confirm the "override" hypothesis? — nutty about natty, Sep 22 '15 at 05:08
@Rinzwind the "-9" override is not the same as the double "-15" override (even if the eventual outcome is equivalent) -- plz prove me wrong. — nutty about natty, Sep 22 '15 at 07:42

score 2 · Answer 1 · answered Sep 21 '15 at 21:41

I am going to state a application-specific possibility.

When you use killall program, a SIGTERM (signal 15) is sent to the program. The usual response to SIGTERM is that the program would exit gracefully.

Now as the SIGTERM is catchable, a program can have a signal handler for SIGTERM that would do some task upon receiving the first SIGTERM (first killall) and return to a state where the second SIGTERM would just terminate it (default action). This is highly dependent on the developer of the program of course and not a general case.

Why does killall (sometimes?) needs to be applied twice?

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