Use PS0 and PS1 to display execution time of each bash command

Question

It seems that by executing code in PS0 and PS1 variables (which are eval\'ed before and after a prompt command is run, as I understand) it should be possible to record time

Answer 1

Arithmetic expansion runs in the current process and can assign to variables. It also produces output, which you can consume with something like \e[$((...,0))m (to output \e[0m) or ${t:0:$((...,0))} (to output nothing, which is presumably better). 64-bit integer support in Bash supports will count POSIX nanoseconds until the year 2262.

$ PS0='${t:0:$((t=$(date +%s%N),0))}'
$ PS1='$((( t )) && printf %d.%09ds $((t=$(date +%s%N)-t,t/1000000000)) $((t%1000000000)))${t:0:$((t=0))}\n$ '
0.053282161s
$ sleep 1
1.064178281s
$ 

$ 

PS0 is not evaluated for empty commands, which leaves a blank line (I'm not sure if you can conditionally print the \n without breaking things). You can work around that by switching to PROMPT_COMMAND instead (which also saves a fork):

$ PS0='${t:0:$((t=$(date +%s%N),0))}'
$ PROMPT_COMMAND='(( t )) && printf %d.%09ds\\n $((t=$(date +%s%N)-t,t/1000000000)) $((t%1000000000)); t=0'
0.041584565s
$ sleep 1
1.077152833s
$ 
$ 

That said, if you do not require sub-second precision, I would suggest using $SECONDS instead (which is also more likely to return a sensible answer if something sets the time).