I want to get time of execution of a program in my terminal. I know that I should use this command:
time chmod +x ~/example
but the output is this:
real 0m0.088s
user 0m0.057s
sys 0m0.030s
But I want to access each one separately. for example just real. how can I get that?
You can use -f to format the time command:
$ /usr/bin/time -f "\t%E Elapsed Real Time" touch a
0:00.00 Elapsed Real Time
The geek stuff has a very broad documentation on this topic:
12 UNIX / Linux Time Command Output Format Option Examples.
It is also funny that calling it with time alone did not work to me, I have to use the full path.
In bash, you can influence the ouput of time with the TIMEFORMAT variable, by setting
TIMEFORMAT=%R # real
TIMEFORMAT=%U # user
TIMEFORMAT=%S # sys
before calling it. However, your problems probably don't end there -- capturing the output of time is not trivial with bash because it's not a subprocess but a shell builtin. There's an entry in the bash FAQ on the topic. Going from there, I think you ultimately want
TIMEFORMAT=%R myvar=$( { time chmod +x ~/example > /dev/null 2>&1; } 2>&1 )
Then $myvar will be the real running time of the command.
You can do:
(time chmod +x ~/example) |& awk '$1=="real"{print $2}'
0m0.003s
Related
I am having list of cron job functions.Each function run for certain period of time.Is there any way to find how long each cron function execute,Like time of start executing function and end time?
Expected output:
Cron tab start time: timestamp
cront tab end time: timestamp
Read crontab(5). A cron job is just a shell command or script. That shell could, for example, use logger(1) mixed with date(1) and time(1). See also time(7). See also GNU bash documentation.
A cron job could also be a GUILE script, or a Python one. See execve(2), fork(2) and other syscalls(2) which is called by crond
And cron is open source, so you are allowed to study its source code.... (e.g. of GNU mcron). Or use strace(1) on it to understand its behavior.
Consider reading Advanced Linux Programming and see Linux From Scratch.
For example, I am running thru cron a backup script starting with
#!/bin/bash
cd $HOME
renice -n +2 -p $$
ionice -c Best-effort -n +5 -p $$
tm=$(date +"backup-home+%Y-%b-%d-%Hh%M")
logger --id=$$ -t backup-home start $tm
sync
When I run several background jobs, I need to get a runtime for the set of jobs.
I use the following example successfully:
$ sleep 10 &
$ time wait
real 0m9.74s
user 0m0.00s
sys 0m0.00s
However, if I try to format the output to seconds, I get the following error:
$ time -f %e wait
time: cannot run wait: No such file or directory
0.00
This only seems to happen with shell builtins. Is there a workaround for this?
The problem here is two-fold.
we don't know if you're using the bash (assuming you're using the centos default shell) built-in or the binary time command.
in either case wait is a bash built-in
Ways to deal with this:
/usr/bin/time -f %e bash -c wait
TIMEFORMAT="%pR" time bash -c wait
How can I use a linux command to get the wall time in seconds spent for executing a program. In the example below,I expected to get "0.005".
$ time ls >/dev/null
real 0m0.005s
user 0m0.001s
sys 0m0.003s
Depending on your path:
/usr/bin/time -f "%e"
The normal time is given by bash's (if you happen to use bash) intern time command
type time
while you need the one,
which time
will find.
So in context of your command:
/usr/bin/time -f "%e" ls > /dev/null
But to store it in a variable, you can't use
a=$(/usr/bin/time -f "%e" ls > /dev/null)
because the output of time is written to the error stream, to not inflict with the programs output (in this example ls). See the manpage of time for further details.
I tried to measure execution time and format it with this command:
time -f "%e" ./1 1000 1
-f: command not found
real 0m0.066s
user 0m0.044s
sys 0m0.023s
But such command works:
/usr/bin/time -f "%e" ./1 1000 1
31245 212 443
0.00
I tried to determine where another time is located, but all showes to /usr/bin/time
which time
/usr/bin/time
or
whereis time
time: /usr/bin/time /usr/bin/X11/time /usr/include/time.h /usr/share/man/man7/time.7.gz /usr/share/man/man2/time.2.gz /usr/share/man/man1/time.1.gz
or
type -a time
time is a shell keyword
time is /usr/bin/time
How to define where another time is located?
Users of the bash shell need to use an explicit path in order to run
the external time command and not the shell builtin variant. On system
where time is installed in /usr/bin, the first example would become
/usr/bin/time wc /etc/hosts
OR
Note: some shells (e.g., bash(1)) have a built-in time command that
provides less functionality than the command described here. To
access the real command, you may need to specify its pathname
(something like /usr/bin/time).
http://man7.org/linux/man-pages/man1/time.1.html
The problem is when I use time in shell I get output like that:
1.350u 0.038s 0:01.45 95.1% 0+0k 0+72io 1pf+0w
And when Im using it in script I get:
real 0m1.253s
user 0m1.143s
sys 0m0.047s
I mean why? And in shell script at the beginning I write:
#!/bin/bash
Bash has a built-in command time, and your system should also have a separate binary at /usr/bin/time:
$ help time
time: time [-p] pipeline
Report time consumed by pipeline's execution.
Execute PIPELINE and print a summary of the real time, user CPU time,
...
$ which time
/usr/bin/time
The two commands produce different outputs:
$ time echo hi
hi
real 0m0.000s
user 0m0.000s
sys 0m0.000s
$ /usr/bin/time echo hi
hi
0.00user 0.00system 0:00.00elapsed 100%CPU (0avgtext+0avgdata 0maxresident)k
0inputs+0outputs (0major+199minor)pagefaults 0swaps