Describe how to annotate a function with return type depending on a flag #13884
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…tional argument I could not find how to do this in the documentation, but eventually saw the example in python#6580 (comment)
sobolevn
reviewed
Oct 21, 2022
docs/source/more_types.rst
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| # This overload variant applies when "rounded" is omitted or set to False. | ||
| @overload | ||
| def pi(rounded: Literal[False] = ...) -> float: ... |
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This is not correct. This function will always be matched. Without a default should be the first one.
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I’m happy to change this, but at least in this case, both overloads are apparently matched. I tested with this file:
from typing import overload, Literal
@overload
def pi(rounded: Literal[False] = ...) -> float: ...
@overload
def pi(rounded: Literal[True]) -> int: ...
def pi(rounded=False):
return 3 if rounded else 3.14159
reveal_type(pi(rounded=False))
reveal_type(pi())
reveal_type(pi(rounded=True))mypy 0.982 tells me:
default.py:12: note: Revealed type is "builtins.float"
default.py:13: note: Revealed type is "builtins.float"
default.py:14: note: Revealed type is "builtins.int"
Swapping the order of the overloads doesn’t change the output. There’s also no warning about shadowing, wouldn’t that be printed otherwise?
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I could not find how to do this in the documentation, but eventually saw the example in
#6580 (comment)
Reading this helped me improve my mental model of how overloads work, and perhaps others would benefit as well.