Storing time and program output into separate variables

Question

With bash, how do I store the time it takes for my program to execute in one variable, and the output of my program in another variable? I know that time sends its output to stdout; the closest I've got is this:

exec 3>&1 4>&2
time_output=$( { time echo hello world 1>&3 2>&4; } 2>&1 )

but "hello world" is printed to the terminal. How to I capture "hello world" to another variable?

I've also tried:

prog_output=$(time_output=$( { time echo hello world 1>&3 2>&4; } 2>&1 ) )

but this doesn't work. prog_output contains nothing and "hello world" is printed to the terminal.

possible duplicate of Capturing the output of bash time in script variable — tripleee
– tripleee, Commented Feb 14, 2015 at 16:42

Maria Zverina · Accepted Answer · 2012-05-16 18:45:35Z

1

Would this work for you?

prog_output=`( time ls ) 2> time_output`
time_output=`cat time_output`

answered May 16, 2012 at 18:45

Maria Zverina

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