msh bug - backtick syntax doesn't set $?
John S. Skogtvedt
jss2k2 at chello.no
Mon Sep 4 18:54:41 CEST 2006
dlginput=`dialog --stdout --inputbox Foo 0 0`
Should echo 0 if OK was selected, 1 if Cancel was selected or 255 if the
escape key was pressed. In msh it always echoes 0.
I've made a small patch that tries to solve this:
The patch is against user/busybox/shell/msh.c .
From my testing (shell script with much dialog usage) it works OK.
I have to admit I don't fully understand how msh works, and there may be
a better way of solving it.
Here's what I know:
When the shell encounters `...` grave() is run.
Currently it doesn't capture the exit status.
After each command ("ls", "var=foo", etc) forkexec() sets $? to 0
unless a external command was run, in which case $? is set to the exit
status of the command.
What the patch does:
Captures the exit status in grave() and assigns it to a global variable
Checks for grave_exitstatus != -1 in forkexec(), and if so sets $? to
the exit status and sets grave_exitstatus to -1.
Can you take a look at the patch, and if it looks OK, commit it?
Will make a bug report with busybox eventually (that is, if no one here
sees something wrong with the patch).
(I'm now subscribed to the list, no need to CC me.)
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