Re: How come math/arithmetic cannot work by set -x
On Sat, Aug 13, 2022 at 06:51:04AM +0700, Budi wrote: > It doesn't work means no use on set -x, no value is shown set -x shows the command being executed, with arguments expanded. unicorn:~$ cat foo #!/bin/bash set -x a=5 b=7 c=$((a * b)) echo "$c" unicorn:~$ ./foo + a=5 + b=7 + c=35 + echo 35 35 In this case, $((a * b)) was expanded to 35 before executing the variable assignment, so that's what set -x shows. Similarly, "$c" was expanded to 35, so set -x shows that as well. What *exactly* were you expecting? We're all being forced to guess, and it's obnoxious.
Re: How come math/arithmetic cannot work by set -x
On Fri, Aug 12, 2022, at 7:40 PM, Dennis Williamson wrote: > It works for me. What are you expecting? > > It would help if you show what you're doing, the result you're getting and > what you expect instead. I'm guessing that instead of, for example % bash -xc 'a="(x=1)" b="2*3"; ((a+b))' + a='(x=1)' + b='2*3' + (( a+b )) they want something like % bash -xc 'a="(x=1)" b="2*3"; ((a+b))' + a='(x=1)' + b='2*3' + (( a+b )) + (( (x=1)+2*3 )) + (( 1+6 )) + (( 7 )) or whatever. Who knows. -- vq
Re: How come math/arithmetic cannot work by set -x
On Fri, Aug 12, 2022 at 6:51 PM Budi wrote: > It doesn't work means no use on set -x, no value is shown > > On 8/13/22, Dennis Williamson wrote: > > On Fri, Aug 12, 2022, 6:28 PM Budi wrote: > > > >> How come math/arithmetic ((i=k+l)) cannot make use of set -x > >> > >> Please help.. > >> (so annoying). > >> > > > > > > It works for me. What are you expecting? > > > > It would help if you show what you're doing, the result you're getting > and > > what you expect instead. > > > > "It doesn't work" conveys no information whatsoever. > > > >> > > > Hmmm... interesting. $ set -x; unset a; b=2; c=7; ((a = $b + $c)); echo "$a $b $c"; set +x + unset a + b=2 + c=7 + (( a = 2 + 7 )) + echo '9 2 7' 9 2 7 + set +x shows the values in the arithmetic expression, but set -x; unset a; b=2; c=7; ((a = b + c)); echo "$a $b $c"; set +x + unset a + b=2 + c=7 + (( a = b + c )) + echo '9 2 7' 9 2 7 + set +x without the dollar signs doesn't. Also, note that this message includes a thorough example of doing, result, expecting. -- Visit serverfault.com to get your system administration questions answered.
Re: How come math/arithmetic cannot work by set -x
It doesn't work means no use on set -x, no value is shown On 8/13/22, Dennis Williamson wrote: > On Fri, Aug 12, 2022, 6:28 PM Budi wrote: > >> How come math/arithmetic ((i=k+l)) cannot make use of set -x >> >> Please help.. >> (so annoying). >> > > > It works for me. What are you expecting? > > It would help if you show what you're doing, the result you're getting and > what you expect instead. > > "It doesn't work" conveys no information whatsoever. > >> >
Re: How come math/arithmetic cannot work by set -x
On Fri, Aug 12, 2022, 6:28 PM Budi wrote: > How come math/arithmetic ((i=k+l)) cannot make use of set -x > > Please help.. > (so annoying). > It works for me. What are you expecting? It would help if you show what you're doing, the result you're getting and what you expect instead. "It doesn't work" conveys no information whatsoever. >
How come math/arithmetic cannot work by set -x
How come math/arithmetic ((i=k+l)) cannot make use of set -x Please help.. (so annoying).
Re: Should the readline *-meta flags reset when $LANG changes?
On 8/11/22 5:56 PM, Koichi Murase wrote: I agree with option 4. Thank you for all your explanations. -- Can we also change the behavior of TERM in a similar way with option 4? I'll look at that for the next version. -- ``The lyf so short, the craft so long to lerne.'' - Chaucer ``Ars longa, vita brevis'' - Hippocrates Chet Ramey, UTech, CWRUc...@case.eduhttp://tiswww.cwru.edu/~chet/
