Le 06/01/2011 09:48, Vidar Holen a écrit :
Finding the meaning of $? and $! in the man page is quite hard for people
not familiar with the layout and bash terminology (this frequently comes
up in Freenode #bash). It would be very helpful for them if you could
simply search for $! to find the
On 1/7/11 5:09 AM, Jan Schampera wrote:
Alexander Tiurin wrote:
~$ time for i in `seq 0 1` ; do echo /o/23/4 | cut -d'/' -f2 ; done
/dev/null
To track this a bit, I ran the exact command several times in a Bash 3.2,
seeing increasing execution times (40s up to ~2min), as reported.
On 1/6/11 8:17 PM, Alexander Tiurin wrote:
Hi!
I ran the command
~$ time for i in `seq 0 1` ; do echo /o/23/4 | cut -d'/' -f2 ; done
/dev/null
6 times in a row, and noticed to the increase in execution time:
[...]
how to interpret the results?
It's hard to say
On 1/7/11 10:03 AM, Chet Ramey wrote:
On 1/6/11 8:17 PM, Alexander Tiurin wrote:
Hi!
I ran the command
~$ time for i in `seq 0 1` ; do echo /o/23/4 | cut -d'/' -f2 ; done
/dev/null
6 times in a row, and noticed to the increase in execution time:
[...]
how to interpret the
Chet Ramey wrote:
I can't imagine this is just some debugging code still active (it's a beta).
Imagine. Anything that doesn't have a version tag of `release' has DEBUG
enabled for the preprocessor, which enables MALLOC_DEBUG. If you're using
the bash malloc, MALLOC_DEBUG turns on extensive
On 01/07/2011 08:39 AM, Chet Ramey wrote:
On 1/7/11 10:03 AM, Chet Ramey wrote:
On 1/6/11 8:17 PM, Alexander Tiurin wrote:
Hi!
I ran the command
~$ time for i in `seq 0 1` ; do echo /o/23/4 | cut -d'/' -f2 ; done
/dev/null
6 times in a row, and noticed to the increase in execution
