I have a process I am running via dtach. I would like to measure it's time and write it to a file. let's call the process ls -l.
I've tried several things I saw around but couldn't make it work..
Can anyone help?
time command may help you:
time dtach -c /tmp/foofoo -Ez ls -l
output:
...
...
[EOF - dtach terminating]
real 0m0.005s
user 0m0.000s
sys 0m0.000s
Related
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 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
I want to find just how long my program takes to run from start to finish in order to compare it with a past version.
How would I go about finding the time it takes for both of these versions? I'm running ubuntu 12.04LTS
Use the time command:
time yourprogram
By default it will output something similar to this:
real 0m0.020s
user 0m0.004s
sys 0m0.000s
real means the total time your program runs. user means the time your program spent in user land code and sys is the time your program spent in kernel calls.
Run time myprogram
The time command will display all the details you need to know.
Example:
rh63-build(greg)~>time ls >/dev/null
real 0m0.003s
user 0m0.001s
sys 0m0.002s
Here is more about the time command: http://linux.die.net/man/1/time
Linux comes with the 'time' program.
$time ./myapp
real 0m0.002s
user 0m0.000s
sys 0m0.000s
My original problem was to kill a process & its children when timeout. And I found GNU timeout quite a good choice.
However, in this testcase, things become weird:
Assume that we have a test1.sh like this:
#!/bin/sh
# test1.sh
output=`timeout 2 ./run.sh`
echo $output
and run.sh like this:
#!/bin/sh
# run.sh
sleep 8s&
Intuitively we should expect test1.sh exits instantly, since init will take charge of that silly sleep process and run.sh will then exits.
However:
sh-4.2$ time ./test1.sh
real 0m8.022s
user 0m0.013s
sys 0m0.003s
And if I create this test2.sh:
#!/bin/sh
# test2.sh
timeout 2 ./run.sh
sh-4.2$ time ./test2.sh
real 0m0.014s
user 0m0.003s
sys 0m0.007s
So, obviously we hit something wrong during command substitution, but why?
It may be the way you have in shell script --
`timeout 2 ./run.sh`
-- you are using command substitution, so as long as the command has not finished execution, substitution cannot be done, because the output is not there...this may explain the output you are seeing.
Try this to see a similar outcome....
echo "hello `sleep 2 &`"
Another script of interest --
$ cat y.sh
echo "hi"
sleep 2 &
echo "bye"
sleep 2 &
Run using
echo "hello `sh y.sh`"
$time sh y.sh
hi
bye
real 0m0.006s
user 0m0.000s
sys 0m0.004s
$time echo "hello `sh y.sh`"
hello hi
bye
real 0m2.008s
user 0m0.004s
sys 0m0.000s
This page explains more about the relationship between background process and file descriptor. Basicly:
Background (better: forked) processes inherit the file descriptors,
and Running a command in backticks means to collect its stdout until
its stdout is closed
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