bash 'time' function

Soren Jacobsen devsoren at attbi.com
Fri Dec 21 00:42:47 PST 2001


On Fri, Dec 21, 2001 at 01:26:26AM -0500, Chris F.A. Johnson wrote:
> On Thu, 20 Dec 2001, Soren Jacobsen wrote:
>... 
> No, I am redirecting stderr, and using process substitution.

Neat, I didn't know about that.

> > soren:~$ { time ls -R /usr/local &>/dev/null; } 2>&1 | grep sys
> > sys     0m0.600s
> 
> If the command sends anything to stdout, this will lose it, or give
> unwanted output if it happens to contain the string "sys".

It is supposed to lose the output, that's why &>/dev/null is there, so
grep doesn't get anything other than what it's supposed to. 

Ex:
test.c is:

#include <stdio.h>
main()
{
fprintf(stdout, "standard output. blef.\n");
fprintf(stderr, "standard error. sys.\n");
}

soren:~$ ./test
standard output. blef.
standard error. sys.

# to show time outputs to stderr
soren:~$ { time ./test &>/dev/null; } 1>/dev/null

real    0m0.005s
user    0m0.010s
sys     0m0.000s

# send stderr to stdout and grep it.
soren:~$ { time ./test &>/dev/null; } 2>&1 | grep sys
sys     0m0.000s

tada.

Sure, $TIMEFORMAT is easier, and you could just as easily use 2> >( grep
sys ) where I redirected stderr to stdout and piped, that's more of a
matter of choice I guess. Gotta stand by my method, though. :p

-- 
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|(__) Soren Jacobsen|
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