Re: Weird behavior with the time command and background jobs

[Bart Schaefer <schaefer@xxxxxxxxxxxxxxxx>]

On Wed, Aug 12, 2020 at 8:34 AM Rudi C <rudiwillalwaysloveyou@xxxxxxxxx> wrote:
>
> ❯ f2() { eval "$@" & }

There's no subshell inside f2.

> ❯ time (f2 sleep 3 )
> ( f2 sleep 3; )  0.00s user 0.01s system 0% cpu 3.014 total

"time" forces its command to run synchronously, so here the subshell
you created at the command line waits for and reports the times of its
children, which includes the sleep even though the sleep itself is in
the background.

> ❯ f3() { (eval "$@" &) }

You have another subshell inside f3, so the eval'd command is a child
of that subshell, and a "grandchild" of the shell that runs f3.

> ❯ time (f3 sleep 3)
> ( f3 sleep 3; )  0.00s user 0.00s system 41% cpu 0.007 total

So here, the subshell inside f3 is not timed and not forced to be
synchronous, so it puts sleep in the background and immediately exits.
The subshell created at the command line, which is calculating the
times, only reports the time of the subshell inside f3, not of sleep.
A process can't time its grandchildren, only its children.

> Why is `f2` getting 3 seconds? It's running the job in the background, just like `f3`. In fact, their interactive behavior (without the time command) is the same.

As noted above, the time command changes the behavior of the outermost
subshell.  It has to do this in order to read the time statistics from
the operating system.