If you want to find out how long a program takes to execute, you can use the Linux time
command. It outputs the real, user and system resources used by the command you specify. For example, to see how long the ls command takes to execute on the current directory, you run time ls
and get something like the following:
real 0m0.005s
user 0m0.004s
sys 0m0.000s
However, by default, the time
command outputs this information to the standard error device and, as a result, it’s not always obvious how to get this information into a file.
If you’re using the GNU time
command, it’s pretty easy with the -o
or --output
option. For example, the following command writes the time information from executing ls
into the file time.log
:
time -o time.log ls
On some systems, such as a Linux server I was using recently, the only available time
command may be the one built into the shell. This time
command does not have an output option. As a result, if you want to redirect the output of both the time
command and the program it’s executing, you might think you want to do the following:
time ls > output.log 2>&1
Then, you discover that it doesn’t work as expected. All you get in the log file is the output of the ls
command while the time
command still prints to the console, even though you redirected standard error to standard out.
Why does this happen? It’s because those redirects apply only to the ls
command, not to the time
command. Fortunately, there’s an easy solution. Group the time
command and the program you want it to time using parentheses. Try changing the previous command to the following:
( time ls ) > output.log 2>&1
When you view the output.log file now, you’ll find that the output of both the ls
and time
commands went into the log file, as desired.
Hello thank’s you for you post , i’m not find solutions for (time ls) > out.txt 2>&1 now you resolved me , anew thank you so much.
Pd: Sorry by my bad english :S