Re: TIMEFORMAT bug

[Chet Ramey]

Марк Коренберг wrote:
> Configuration Information [Automatically generated, do not change]:
> Machine: i486
> OS: linux-gnu
> Compiler: gcc
> Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='i486'
> -DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i486-pc-linux-gnu'
> -DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash'
> -DSHELL -DHAVE_CONFIG_H   -I.  -I../bash -I../bash/include
> -I../bash/lib   -g -O2 -Wall
> uname output: Linux mmarkk-desktop 2.6.27-11-generic #1 SMP Thu Jan 29
> 19:24:39 UTC 2009 i686 GNU/Linux
> Machine Type: i486-pc-linux-gnu
> 
> Bash Version: 3.2
> Patch Level: 39
> Release Status: release
> 
> Description:
>       TIMEFORMAT variable improperly handled
> 
> Repeat-By:
>       $ TIMEFORMAT='test'; time echo aaa
>       aaa
>       test
>       (works OK, but:)
> 
>       $ TIMEFORMAT='test' time echo aaa
>       outputs something like this:
>       aaa
>       0.00user 0.00system 0:00.00elapsed 200%CPU (0avgtext+0avgdata 
> 0maxresident)k
>       0inputs+0outputs (0major+198minor)pagefaults 0swaps
>       ( it seems memory leak or undocumented features of the bash...)
> 
>       BASH 2.05 said:
>       $ TIMEFORMAT=1 time echo q
>       -bash: time: command not found

`time' is a bash reserved word.  It's essentially a pipeline modifier
that allows pipelines, shell builtins, shell functions, and shell
programming constructs to be timed.  The bash built-in command timing
is what understands and interprets $TIMEFORMAT.

Preceding a reserved word with an assignment statement prevents it from
being recognized as such.  When you use the assignment statement, you
end up running /usr/bin/time, which does not interpret $TIMEFORMAT and
has a different output format.

Chet

-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer

Chet Ramey, ITS, CWRU    chet@case.edu    http://cnswww.cns.cwru.edu/~chet/