How to write an empty indentation block in Python?

Question

The runtime keeps telling me:

expected an indented block

But I don’t want write nothing inside my except block, I just want it to catch and swallow the exception.

Asked By: Jader Dias

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Source

Answer 1

Just write

pass

as in

try:
    # Do something illegal.
    ...
except:
    # Pretend nothing happened.
    pass

EDIT: @swillden brings up a good point, viz., this is a terrible idea in general. You should, at the least, say

except TypeError, DivideByZeroError:

or whatever kinds of errors you want to handle. Otherwise you can mask bigger problems.

Answered By: Peter

Answer 2

I’ve never done this in more permanent code, but I frequently do it as a placeholder

if some_expression:
  True
else:
  do_something(blah)

Just sticking a True in there will stop the error. Not sure if there’s anything bad about this.

Answered By: Josh

Answer 3

For those who are very unclear as to why you would want to do this. Here is an example where I initially thought that an empty block would be a good idea:

def set_debug_dir(self, debug_dir=None):
    if debug_dir is None:
        debug_dir = self.__debug_dir
    elif isinstance(debug_dir, (Path, str)):
        debug_dir = debug_dir # this is my null operation
    elif isinstance(debug_dir, list):
        debug_dir = functools.reduce(os.path.join, debug_dir)
    else:
        raise TypeError('Unexpected type for debug_dir: {}'.format(type(debug_dir).__name__))

But it would be more clear to reorganize the statement:

def set_debug_dir(self, debug_dir=None):
    if debug_dir is None:
        debug_dir = self.__debug_dir
    elif isinstance(debug_dir, list):
        debug_dir = functools.reduce(os.path.join, debug_dir)
    elif not isinstance(debug_dir, (Path, str)):
        raise TypeError('Unexpected type for debug_dir: {}'.format(type(debug_dir).__name__))

Answered By: Jason Harrison

How to write an empty indentation block in Python?

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