msh bug - backtick syntax doesn't set $?

[John S. Skogtvedt]

Test case:
dlginput=`dialog --stdout --inputbox Foo 0 0`
echo $?

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:
http://home.chello.no/~jskogtv/msh.c.diff

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 
grave_exitstatus.

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.)