[Python] How to match an empty dictionary?

Python supports Structural Pattern Matching since version 3.10. I came to notice that matching an empty dict doesn't work by simply matching {} as it does for lists. According to my naive approach, non-empty dicts are also matched (Python 3.10.4):

def match_empty(m):
match m:
case []:
print("empty list")
case {}:
print("empty dict")
case _:
print("not empty")


match_empty([]) # empty list
match_empty([1, 2]) # not empty
match_empty({}) # empty dict
match_empty({'a': 1}) # empty dict


Matching the constructors even breaks the empty list matching:

def match_empty(m):
match m:
case list():
print("empty list")
case dict():
print("empty dict")
case _:
print("not empty")


match_empty([]) # empty list
match_empty([1, 2]) # empty list
match_empty({}) # empty dict
match_empty({'a': 1}) # empty dict


Here is a solution, that works as I expect:

def match_empty(m):
match m:
case []:
print("empty list")
case d:
if isinstance(d, dict) and len(d) == 0:
print("empty dict")
return
print("not empty")


match_empty([]) # empty list
match_empty([1, 2]) # not empty
match_empty({}) # empty dict
match_empty({'a': 1}) # not empty


Now my questions are:

• Why do my first 2 approaches not work (as expected)?
• Is there a way to use structural pattern matching to match only an empty dict (without checking the dict length explicitly)?

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