How to force `condition-case` to activate the debugger

Question

Often times I need to debug code wrapped in the BODYFORM of condition-case; however, the debugger is suppressed even though I have debug-on-error set to t. The doc-string states that we can add debug to the list of handlers, but I didn't find a good example by Googling of how this might be accomplished. Modifying the code of something that I am trying to debug to add a handler seems rather inefficient.

Is there an approach to force condition-case to always generate a debugging message without modifying the section of code that I am trying to debug? If not, then an explanation why along with a sample of how to use debug as a handler would be an acceptable answer.

Have you tried using `condition-case-unless-debug` instead of `condition-case`? — Drew, Feb 05 '17 at 17:31
@Drew -- thank you for the suggestion -- I'll give that a try this morning. — lawlist, Feb 05 '17 at 17:32
FYI, setting `debug-on-signal` can be useful for this too (although it's sometimes a bit too much). — npostavs, Feb 06 '17 at 13:46

score 4 · Accepted Answer · answered Feb 05 '17 at 17:38

Try using condition-case-unless-debug instead of condition-case.

You should be able to M-x debug-on-entry either foo or bar in this context:

(defun foo ()
  (condition-case-unless-debug nil
      (bar)
    (error nil)))

(defun bar ()
  (message "Cuckoo!")
  (pp-eval-expression '(cons 42 49)))

Note that condition-case-unless-debug just does this:

(macroexpand '(condition-case-unless-debug nil (bar) (error nil)))

gives

(condition-case nil
    (bar)
  ((debug error)
   nil))

How to force `condition-case` to activate the debugger

1 Answers1