> #!/usr/local/plan9/bin/9 rc
> echo first at path: $path(1)
> for (i in `{seq 1 5}){
> if (test $i -eq 3) {
> echo $i equals 3
> }
> if not echo $i is different from 3
> }
NB: in general, you don't need 'test'. use "if (~ $i 3) {}".
that said, now I'm convinced there's a bug somewhere. I got your code
to fail under the condition that rc is started via the '9' environment
(9 rc). the type of 'test' binary doesn't matter -- both /bin/test and
$PLAN9/bin/test break under '9', but don't break without it. the '9'
script is pretty simple, but there appear do be some side-effects
(probably due to \r being transmitted over the network?). here's the
simplest version of your script:
$ cat t.rc
#!/usr/bin/env rc
exec rc
$
dial and test:
$ dial -e 'tcp!localhost!8080'
######## no breakage
/bin/test 1 -eq 2 ; echo $status
1
/bin/test 1 -eq 1; echo $status
/Users/andrey/plan9/bin/test 1 -eq 2; echo $status
1
/Users/andrey/plan9/bin/test 1 -eq 1; echo $status
######## now see it break
9 rc
/bin/test 1 -eq 2 ; echo $status
/bin/test 1 -eq 1; echo $status
/Users/andrey/plan9/bin/test 1 -eq 2; echo $status
/Users/andrey/plan9/bin/test 1 -eq 1; echo $status
ls | wc > /dev/null; echo $status
|
ls | wc > /dev/null; echo $status
|||
ls | wc > /dev/null; echo $status
|||||||
ls | wc > /dev/null; echo $status
|||||||||||||||
ls | wc > /dev/null; echo $status
|||||||||||||||||||||||||||||||
#####back to non-breaking rc:
exit
echo $status
ls | wc > /dev/null; echo $status
|
ls | wc > /dev/null; echo $status
|
ls | wc > /dev/null; echo $status
|
ls | wc > /dev/null; echo $status
|
ls | wc > /dev/null; echo $status
|
