Antonio Macchi wrote:
> using ^V to pass an ascii character to read, the behavior is incoherent
> beetween simple read, and read -nx
When using the -n option, bash puts the terminal into character-at-a-time
non-canonical mode, which disables the tty special characters. When
using read without tha
Greg Wooledge wrote:
> The other two messages I sent today were just things I encountered while
> bringing my bash 4.0 up to the current patch level. This is the real
> problem I've been chasing.
>
> imadev:/var/tmp/bash-4.0$ bash-3.1.17 -c 'printf -v foo %s bar'
> imadev:/var/tmp/bash-4.0$ bash-
Ken Irving wrote:
>> $ printf "%d %s\n" 1 ok -
>> 1 ok
>> -bash: printf: -: invalid number
>> 0
>>
>> why getting error here, and not in the previous?
>> why "invalid number" ?
>> what is that zero?
>
> Again, you have more arguments than operators, so it makes another pass,
> and on the second p
There's nothing to fix. It might help if you provide some markers
sorry, and thanks for your patience...
in your test patterns so you can see where each argument begins and
ends, e.g.,
$ printf "(%d) {%s}\n" 1 ok -
(1) {ok}
-bash: printf: -: invalid number
(0) {}
... and f
On Thu, Nov 12, 2009 at 01:47:52AM +0100, Antonio Macchi wrote:
> $ printf "%s\n" ok -
> ok
> -
>
> why that score in the newline?
Your pattern has one format operator, but there are two arguments,
so the pattern is applied again.
>
>
> $ printf "%d %s\n" 1 ok -
> 1 ok
> -bash: printf: -
$ printf "%s\n" ok -
ok
-
why that score in the newline?
$ printf "%d %s\n" 1 ok -
1 ok
-bash: printf: -: invalid number
0
why getting error here, and not in the previous?
why "invalid number" ?
what is that zero?
$ printf "%2s\n" qwerty
qwerty
strings larger than fixed-wi
using ^V to pass an ascii character to read, the behavior is incoherent
beetween simple read, and read -nx
$ read
^V^A
$ hd <(echo -n $REPLY)
01|.|
0001
$ read -n1
^V
$ hd <(echo -n $REPLY)
16
The other two messages I sent today were just things I encountered while
bringing my bash 4.0 up to the current patch level. This is the real
problem I've been chasing.
imadev:/var/tmp/bash-4.0$ bash-3.1.17 -c 'printf -v foo %s bar'
imadev:/var/tmp/bash-4.0$ bash-4.0.10 -c 'printf -v foo bar'
ima
/net/appl/gcc-3.3/bin/gcc -L./builtins -L./lib/readline -L./lib/readline
-L./lib/glob -L./lib/tilde -L./lib/malloc -L./lib/sh-g -O2 -o bash shell.o
eval.o y.tab.o general.o make_cmd.o print_cmd.o dispose_cmd.o execute_cmd.o
variables.o copy_cmd.o error.o expr.o flags.o jobs.o subst.o hashcm
make[1]: Entering directory `/var/tmp/bash-4.0/lib/glob'
rm -f glob.o
/net/appl/gcc-3.3/bin/gcc -c -DHAVE_CONFIG_H -DSHELL -I. -I../.. -I../..
-I../../include -I../../lib -DHPUX -g -O2 glob.c
glob.c:1026:69: missing terminating ' character
make[1]: *** [glob.o] Error 1
make[1]: Leaving directory
> oups sorry you want to see the other errors from kill ...
Then invoke kill before redirecting stderr.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRUc...@case.eduhttp://cnswww.cns.cwru.edu/~c
> > Nothing wrong, but can be made more efficient because "| grep" means
> > another subprocess which can be eliminated if the shell silents the
> > Terminate command in the first place.
Then why not try the approach I outlined in the previous message, instead
of sticking with something that you d
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