How to run bash function in "watch" command?

Question

Say I have a function f defined in my bashrc

function f() {
  date
}

I want to run following command to monitor the output watch f. The command failed with "sh: f: command not found". watch bash -c f gives the same.

How to get this watch command working as intended?

Answer 1

watch is an external program that starts a new program every time interval.

In your case, neither the started default shell sh nor the explicitly started bash can know about your function f – it's not being passed to them.

You include the function definition in what you run. For example, you could write

#!/bin/bash
function f() {
  date
}
f

into a text file "myscript.sh", make it executable and run it with watch ./myscript.sh.

Alternatively, you could also

function f () {
…
}
typeset -fx f
#   ^---------- modify the type of a name
#        ^----- work only on functions
#         ^---- export to environment
watch -x bash -c "f"
#     ^-------- use `exec` rather than `system` to start bash;
#               makes no sense to start a shell from a shell you
#               only start to start a shell (but omitting -x has
#               no downside other than launching an unnecessary
#               middle layer of `sh`)

to export the function declaration to an environment variable and thus make it known to bash as launched from watch, as subprocesses inherit environment variables.

I'd recommend you go with the script option - less confusing, and it doesn't mess with the environment, which might have surprising side effects. It has no performance benefits: in both cases, the shell parses the source code of your function declaration in the new subshell.

Lastly, you could do without watch:

while true ; do
  tput clear # to clear the screen
  f
  sleep 1
done

tput clear will clear the scroll buffer; if you had anything on that previously, you might want to save it before.