What happens if you use raise without specifying an exception inside an except block?

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BlackTom AI · Accepted Answer

Consider the Python behavior described in the question about using raise inside an except block without specifying an exception.
Option 1: 'Prints an error message' — This is not what happens. Merely re-raising an exception does not automatically print a generic error message; printing would require an explicit print call. Therefore, this is not accurate.
Option 2: 'Raises TypeError' — Raising a TypeError would require that a TypeError instance (or type) is explicitly raised. The context here is about re-raising the current exception, not introducing a new, specific exception type. Hence, this is incorrect.
Option 3: 'Does nothing' — If you use raise by itself inside an except block, you don’t swallow the error; Python propagates the currently handled exception up the call stack. So it does something: it re-raises. This option is false.
Option 4: 'Exits the program silently' — Re-raising the current exception will propagate it, which may terminate the program if unhandled, but it is not silent behavior by itself. The typical outcome is an exception propagating, potentially crashing if unhandled, not a silent exit. Thus, this is misleading and incorrect.
Option 5: 'Raises ValueError' — Similar to the TypeError case, this would require explicitly raising a ValueError. Re-raising the existing exception does not switch to a different exception type. So this is not correct.
Option 6: 'Raises the same exception again' — This matches the standard Python behavior: inside an except block, using raise with no argument re-raises the current exception, effectively propagating the same error again. This is the correct understanding.

IS 4485-001 Fall 2025 Final Exam Practice 1

What happens if you use raise without specifying an exception inside an except block?

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