python - Exit program without triggering exception handling -


i wasn't sure how title question, issue this; have program set guess number , handle exceptions. program loops until number guessed, when program exits, exception message shows @ same time. how can fix this?

num = none while num != 31:     try:         num = int(input("what age of creator? \n"))         if num < 31:             print("higher! guess again! \n")         elif num > 31:             print("lower! guess again! \n")         elif num == 31:             print("good guess!")             exit()     except:         print("numbers only! \n")

this output:

what age of creator? 31 guess! numbers only!

process finished exit code 0

if want keep exception message, recommend break instead of exit, more natural exit program. try this:

try:     ...     elif num == 31:         print("good guess!")         break except valueerror:     print("numbers only! \n")

also, should catch specific error rather catch-all bare except.

if want silence error messages, should use pass instead. docs:

the pass statement nothing. can used when statement required syntactically program requires no action.

try:     ... except valueerror:     pass

pass innocuous placeholder statement nothing.


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