Thanks for giving me a heads up here. I have updated the bashdb documentation for the skip command make clear that $? is no changed in skipping an instruction.
More generally bash saves and restores some state on entry and exit to extdebug, like the variable $? (the extdebug function can issue a command which inside sets $? but on return $? I think is restored) However there is other global state that isn't saved and restored. I think this includes the set flags. I'm not sure that what gets saved and restored is fully documented, but it probably should be. On Fri, Apr 7, 2017 at 9:06 PM, Chet Ramey <chet.ra...@case.edu> wrote: > On 4/4/17 5:08 PM, nesro wrote: > > > > > To have both commands similar, let's assume the while as this: > > > > while (( 1 > 2 )); do :; done > > > > Thanks for explaining me why it happen, but now I don't know why for > cycle > > works and while does not. > When extdebug is enabled, the command is skipped if the DEBUG trap returns > non-zero. This doesn't change $? (traps never do) and it doesn't cause the > command that would have been run to return a non-zero exit status. If it > did, the combination of DEBUG and `set -e' would exit the shell, for > example, and I'm sure it would not behave as the bash debugger would > expect. For that matter, why should a command that's not run fail? > > Since that command returns successfully, when it's used as the `while' > command test, it causes the while command to execute forever. > > -- > ``The lyf so short, the craft so long to lerne.'' - Chaucer > ``Ars longa, vita brevis'' - Hippocrates > Chet Ramey, UTech, CWRU c...@case.edu http://cnswww.cns.cwru.edu/~ > chet/ >
