## Summary Use the type annotation of function parameters as bidirectional type context when inferring the argument expression. For example, the following example now type-checks: ```py class TD(TypedDict): x: int def f(_: TD): ... f({ "x": 1 }) ``` Part of https://github.com/astral-sh/ty/issues/168.
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Overloads
When ty evaluates the call of an overloaded function, it attempts to "match" the supplied arguments with one or more overloads. This document describes the algorithm that it uses for overload matching, which is the same as the one mentioned in the spec.
Note that all of the examples that involve positional parameters are tested multiple times: once with the parameters matched with individual positional arguments, and once with the parameters matched with a single positional argument that is splatted into the argument list. Overload resolution is performed after splatted arguments have been expanded, and so both approaches (TODO: should) produce the same results.
Arity check
The first step is to perform arity check. The non-overloaded cases are described in the function document.
overloaded.pyi
:
from typing import overload
@overload
def f() -> None: ...
@overload
def f(x: int) -> int: ...
from overloaded import f
# These match a single overload
reveal_type(f()) # revealed: None
reveal_type(f(*())) # revealed: None
reveal_type(f(1)) # revealed: int
reveal_type(f(*(1,))) # revealed: int
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f("a", "b")) # revealed: Unknown
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f(*("a", "b"))) # revealed: Unknown
Type checking
The second step is to perform type checking. This is done for all the overloads that passed the arity check.
Single match
overloaded.pyi
:
from typing import overload
@overload
def f(x: int) -> int: ...
@overload
def f(x: str) -> str: ...
@overload
def f(x: bytes) -> bytes: ...
Here, all of the calls below pass the arity check for all overloads, so we proceed to type checking which filters out all but the matching overload:
from overloaded import f
reveal_type(f(1)) # revealed: int
reveal_type(f(*(1,))) # revealed: int
reveal_type(f("a")) # revealed: str
reveal_type(f(*("a",))) # revealed: str
reveal_type(f(b"b")) # revealed: bytes
reveal_type(f(*(b"b",))) # revealed: bytes
Single match error
overloaded.pyi
:
from typing import overload
@overload
def f() -> None: ...
@overload
def f(x: int) -> int: ...
@overload
def f(x: int, y: int) -> int: ...
If the arity check only matches a single overload, it should be evaluated as a regular
(non-overloaded) function call. This means that any diagnostics resulted during type checking that
call should be reported directly and not as a no-matching-overload
error.
from typing_extensions import reveal_type
from overloaded import f
reveal_type(f()) # revealed: None
reveal_type(f(*())) # revealed: None
# error: [invalid-argument-type] "Argument to function `f` is incorrect: Expected `int`, found `Literal["a"]`"
reveal_type(f("a")) # revealed: Unknown
# error: [invalid-argument-type] "Argument to function `f` is incorrect: Expected `int`, found `Literal["a"]`"
reveal_type(f(*("a",))) # revealed: Unknown
More examples of this diagnostic can be found in the single_matching_overload.md document.
Multiple matches
overloaded.pyi
:
from typing import overload
class A: ...
class B(A): ...
@overload
def f(x: A) -> A: ...
@overload
def f(x: B, y: int = 0) -> B: ...
from overloaded import A, B, f
# These calls pass the arity check, and type checking matches both overloads:
reveal_type(f(A())) # revealed: A
reveal_type(f(*(A(),))) # revealed: A
reveal_type(f(B())) # revealed: A
reveal_type(f(*(B(),))) # revealed: A
# But, in this case, the arity check filters out the first overload, so we only have one match:
reveal_type(f(B(), 1)) # revealed: B
reveal_type(f(*(B(), 1))) # revealed: B
Argument type expansion
This step is performed only if the previous steps resulted in no matches.
In this case, the algorithm would perform argument type expansion and loops over from the type checking step, evaluating the argument lists.
Expanding the only argument
overloaded.pyi
:
from typing import overload
class A: ...
class B: ...
class C: ...
@overload
def f(x: A) -> A: ...
@overload
def f(x: B) -> B: ...
@overload
def f(x: C) -> C: ...
from overloaded import A, B, C, f
def _(ab: A | B, ac: A | C, bc: B | C):
reveal_type(f(ab)) # revealed: A | B
reveal_type(f(*(ab,))) # revealed: A | B
reveal_type(f(bc)) # revealed: B | C
reveal_type(f(*(bc,))) # revealed: B | C
reveal_type(f(ac)) # revealed: A | C
reveal_type(f(*(ac,))) # revealed: A | C
Expanding first argument
If the set of argument lists created by expanding the first argument evaluates successfully, the algorithm shouldn't expand the second argument.
overloaded.pyi
:
from typing import Literal, overload
class A: ...
class B: ...
class C: ...
class D: ...
@overload
def f(x: A, y: C) -> A: ...
@overload
def f(x: A, y: D) -> B: ...
@overload
def f(x: B, y: C) -> C: ...
@overload
def f(x: B, y: D) -> D: ...
from overloaded import A, B, C, D, f
def _(a_b: A | B):
reveal_type(f(a_b, C())) # revealed: A | C
reveal_type(f(*(a_b, C()))) # revealed: A | C
reveal_type(f(a_b, D())) # revealed: B | D
reveal_type(f(*(a_b, D()))) # revealed: B | D
# But, if it doesn't, it should expand the second argument and try again:
def _(a_b: A | B, c_d: C | D):
reveal_type(f(a_b, c_d)) # revealed: A | B | C | D
reveal_type(f(*(a_b, c_d))) # revealed: A | B | C | D
Expanding second argument
If the first argument cannot be expanded, the algorithm should move on to the second argument, keeping the first argument as is.
overloaded.pyi
:
from typing import overload
class A: ...
class B: ...
class C: ...
class D: ...
@overload
def f(x: A, y: B) -> B: ...
@overload
def f(x: A, y: C) -> C: ...
@overload
def f(x: B, y: D) -> D: ...
from overloaded import A, B, C, D, f
def _(a: A, bc: B | C, cd: C | D):
# This also tests that partial matching works correctly as the argument type expansion results
# in matching the first and second overloads, but not the third one.
reveal_type(f(a, bc)) # revealed: B | C
reveal_type(f(*(a, bc))) # revealed: B | C
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f(a, cd)) # revealed: Unknown
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f(*(a, cd))) # revealed: Unknown
Generics (legacy)
overloaded.pyi
:
from typing import TypeVar, overload
_T = TypeVar("_T")
class A: ...
class B: ...
@overload
def f(x: A) -> A: ...
@overload
def f(x: _T) -> _T: ...
from overloaded import A, f
def _(x: int, y: A | int):
reveal_type(f(x)) # revealed: int
reveal_type(f(*(x,))) # revealed: int
reveal_type(f(y)) # revealed: A | int
reveal_type(f(*(y,))) # revealed: A | int
Generics (PEP 695)
[environment]
python-version = "3.12"
overloaded.pyi
:
from typing import overload
class A: ...
class B: ...
@overload
def f(x: B) -> B: ...
@overload
def f[T](x: T) -> T: ...
from overloaded import B, f
def _(x: int, y: B | int):
reveal_type(f(x)) # revealed: int
reveal_type(f(*(x,))) # revealed: int
reveal_type(f(y)) # revealed: B | int
reveal_type(f(*(y,))) # revealed: B | int
Expanding bool
overloaded.pyi
:
from typing import Literal, overload
class T: ...
class F: ...
@overload
def f(x: Literal[True]) -> T: ...
@overload
def f(x: Literal[False]) -> F: ...
from overloaded import f
def _(flag: bool):
reveal_type(f(True)) # revealed: T
reveal_type(f(*(True,))) # revealed: T
reveal_type(f(False)) # revealed: F
reveal_type(f(*(False,))) # revealed: F
reveal_type(f(flag)) # revealed: T | F
reveal_type(f(*(flag,))) # revealed: T | F
Expanding tuple
overloaded.pyi
:
from typing import Literal, overload
class A: ...
class B: ...
class C: ...
class D: ...
@overload
def f(x: tuple[A, int], y: tuple[int, Literal[True]]) -> A: ...
@overload
def f(x: tuple[A, int], y: tuple[int, Literal[False]]) -> B: ...
@overload
def f(x: tuple[B, int], y: tuple[int, Literal[True]]) -> C: ...
@overload
def f(x: tuple[B, int], y: tuple[int, Literal[False]]) -> D: ...
from overloaded import A, B, f
def _(x: tuple[A | B, int], y: tuple[int, bool]):
reveal_type(f(x, y)) # revealed: A | B | C | D
reveal_type(f(*(x, y))) # revealed: A | B | C | D
Expanding type
There's no special handling for expanding type[A | B]
type because ty stores this type in it's
distributed form, which is type[A] | type[B]
.
overloaded.pyi
:
from typing import overload
class A: ...
class B: ...
@overload
def f(x: type[A]) -> A: ...
@overload
def f(x: type[B]) -> B: ...
from overloaded import A, B, f
def _(x: type[A | B]):
reveal_type(x) # revealed: type[A] | type[B]
reveal_type(f(x)) # revealed: A | B
reveal_type(f(*(x,))) # revealed: A | B
Expanding enums
Basic
overloaded.pyi
:
from enum import Enum
from typing import Literal, overload
class SomeEnum(Enum):
A = 1
B = 2
C = 3
class A: ...
class B: ...
class C: ...
@overload
def f(x: Literal[SomeEnum.A]) -> A: ...
@overload
def f(x: Literal[SomeEnum.B]) -> B: ...
@overload
def f(x: Literal[SomeEnum.C]) -> C: ...
from typing import Literal
from overloaded import SomeEnum, A, B, C, f
def _(x: SomeEnum, y: Literal[SomeEnum.A, SomeEnum.C]):
reveal_type(f(SomeEnum.A)) # revealed: A
reveal_type(f(*(SomeEnum.A,))) # revealed: A
reveal_type(f(SomeEnum.B)) # revealed: B
reveal_type(f(*(SomeEnum.B,))) # revealed: B
reveal_type(f(SomeEnum.C)) # revealed: C
reveal_type(f(*(SomeEnum.C,))) # revealed: C
reveal_type(f(x)) # revealed: A | B | C
reveal_type(f(*(x,))) # revealed: A | B | C
reveal_type(f(y)) # revealed: A | C
reveal_type(f(*(y,))) # revealed: A | C
Enum with single member
This pattern appears in typeshed. Here, it is used to represent two optional, mutually exclusive keyword parameters:
overloaded.pyi
:
from enum import Enum, auto
from typing import overload, Literal
class Missing(Enum):
Value = auto()
class OnlyASpecified: ...
class OnlyBSpecified: ...
class BothMissing: ...
@overload
def f(*, a: int, b: Literal[Missing.Value] = ...) -> OnlyASpecified: ...
@overload
def f(*, a: Literal[Missing.Value] = ..., b: int) -> OnlyBSpecified: ...
@overload
def f(*, a: Literal[Missing.Value] = ..., b: Literal[Missing.Value] = ...) -> BothMissing: ...
from typing import Literal
from overloaded import f, Missing
reveal_type(f()) # revealed: BothMissing
reveal_type(f(a=0)) # revealed: OnlyASpecified
reveal_type(f(b=0)) # revealed: OnlyBSpecified
f(a=0, b=0) # error: [no-matching-overload]
def _(missing: Literal[Missing.Value], missing_or_present: Literal[Missing.Value] | int):
reveal_type(f(a=missing, b=missing)) # revealed: BothMissing
reveal_type(f(a=missing)) # revealed: BothMissing
reveal_type(f(b=missing)) # revealed: BothMissing
reveal_type(f(a=0, b=missing)) # revealed: OnlyASpecified
reveal_type(f(a=missing, b=0)) # revealed: OnlyBSpecified
reveal_type(f(a=missing_or_present)) # revealed: BothMissing | OnlyASpecified
reveal_type(f(b=missing_or_present)) # revealed: BothMissing | OnlyBSpecified
# Here, both could be present, so this should be an error
f(a=missing_or_present, b=missing_or_present) # error: [no-matching-overload]
Enum subclass without members
An Enum
subclass without members should not be expanded:
overloaded.pyi
:
from enum import Enum
from typing import overload, Literal
class MyEnumSubclass(Enum):
pass
class ActualEnum(MyEnumSubclass):
A = 1
B = 2
class OnlyA: ...
class OnlyB: ...
class Both: ...
@overload
def f(x: Literal[ActualEnum.A]) -> OnlyA: ...
@overload
def f(x: Literal[ActualEnum.B]) -> OnlyB: ...
@overload
def f(x: ActualEnum) -> Both: ...
@overload
def f(x: MyEnumSubclass) -> MyEnumSubclass: ...
from overloaded import MyEnumSubclass, ActualEnum, f
def _(actual_enum: ActualEnum, my_enum_instance: MyEnumSubclass):
reveal_type(f(actual_enum)) # revealed: Both
reveal_type(f(*(actual_enum,))) # revealed: Both
reveal_type(f(ActualEnum.A)) # revealed: OnlyA
reveal_type(f(*(ActualEnum.A,))) # revealed: OnlyA
reveal_type(f(ActualEnum.B)) # revealed: OnlyB
reveal_type(f(*(ActualEnum.B,))) # revealed: OnlyB
reveal_type(f(my_enum_instance)) # revealed: MyEnumSubclass
reveal_type(f(*(my_enum_instance,))) # revealed: MyEnumSubclass
No matching overloads
If argument expansion has been applied to all arguments and one or more of the expanded argument lists cannot be evaluated successfully, generate an error and stop.
overloaded.pyi
:
from typing import overload
class A: ...
class B: ...
class C: ...
class D: ...
@overload
def f(x: A) -> A: ...
@overload
def f(x: B) -> B: ...
from overloaded import A, B, C, D, f
def _(ab: A | B, ac: A | C, cd: C | D):
reveal_type(f(ab)) # revealed: A | B
reveal_type(f(*(ab,))) # revealed: A | B
# The `[A | C]` argument list is expanded to `[A], [C]` where the first list matches the first
# overload while the second list doesn't match any of the overloads, so we generate an
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f(ac)) # revealed: Unknown
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f(*(ac,))) # revealed: Unknown
# None of the expanded argument lists (`[C], [D]`) match any of the overloads, so we generate an
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f(cd)) # revealed: Unknown
# error: [no-matching-overload] "No overload of function `f` matches arguments"
reveal_type(f(*(cd,))) # revealed: Unknown
Optimization: Avoid argument type expansion
Argument type expansion could lead to exponential growth of the number of argument lists that needs to be evaluated, so ty deploys some heuristics to prevent this from happening.
Heuristic: If an argument type that cannot be expanded and cannot be assighned to any of the remaining overloads before argument type expansion, then even with argument type expansion, it won't lead to a successful evaluation of the call.
overloaded.pyi
:
from typing import overload
class A: ...
class B: ...
class C: ...
@overload
def f() -> None: ...
@overload
def f(**kwargs: int) -> C: ...
@overload
def f(x: A, /, **kwargs: int) -> A: ...
@overload
def f(x: B, /, **kwargs: int) -> B: ...
class Foo:
@overload
def f(self) -> None: ...
@overload
def f(self, **kwargs: int) -> C: ...
@overload
def f(self, x: A, /, **kwargs: int) -> A: ...
@overload
def f(self, x: B, /, **kwargs: int) -> B: ...
from overloaded import A, B, C, Foo, f
from typing_extensions import Any, reveal_type
def _(ab: A | B, a: int | Any):
reveal_type(f(a1=a, a2=a, a3=a)) # revealed: C
reveal_type(f(A(), a1=a, a2=a, a3=a)) # revealed: A
reveal_type(f(B(), a1=a, a2=a, a3=a)) # revealed: B
# Here, the arity check filters out the first and second overload, type checking fails on the
# remaining overloads, so ty moves on to argument type expansion. But, the first argument (`C`)
# isn't assignable to any of the remaining overloads (3 and 4), so there's no point in expanding
# the other 30 arguments of type `Unknown | Literal[1]` which would result in allocating a
# vector containing 2**30 argument lists after expanding all of the arguments.
reveal_type(
# error: [no-matching-overload]
# revealed: Unknown
f(
C(),
a1=a,
a2=a,
a3=a,
a4=a,
a5=a,
a6=a,
a7=a,
a8=a,
a9=a,
a10=a,
a11=a,
a12=a,
a13=a,
a14=a,
a15=a,
a16=a,
a17=a,
a18=a,
a19=a,
a20=a,
a21=a,
a22=a,
a23=a,
a24=a,
a25=a,
a26=a,
a27=a,
a28=a,
a29=a,
a30=a,
)
)
# Here, the heuristics won't come into play because all arguments can be expanded but expanding
# the first argument resutls in a successful evaluation of the call, so there's no exponential
# growth of the number of argument lists.
reveal_type(
# revealed: A | B
f(
ab,
a1=a,
a2=a,
a3=a,
a4=a,
a5=a,
a6=a,
a7=a,
a8=a,
a9=a,
a10=a,
a11=a,
a12=a,
a13=a,
a14=a,
a15=a,
a16=a,
a17=a,
a18=a,
a19=a,
a20=a,
a21=a,
a22=a,
a23=a,
a24=a,
a25=a,
a26=a,
a27=a,
a28=a,
a29=a,
a30=a,
)
)
def _(foo: Foo, ab: A | B, a: int | Any):
reveal_type(foo.f(a1=a, a2=a, a3=a)) # revealed: C
reveal_type(foo.f(A(), a1=a, a2=a, a3=a)) # revealed: A
reveal_type(foo.f(B(), a1=a, a2=a, a3=a)) # revealed: B
reveal_type(
# error: [no-matching-overload]
# revealed: Unknown
foo.f(
C(),
a1=a,
a2=a,
a3=a,
a4=a,
a5=a,
a6=a,
a7=a,
a8=a,
a9=a,
a10=a,
a11=a,
a12=a,
a13=a,
a14=a,
a15=a,
a16=a,
a17=a,
a18=a,
a19=a,
a20=a,
a21=a,
a22=a,
a23=a,
a24=a,
a25=a,
a26=a,
a27=a,
a28=a,
a29=a,
a30=a,
)
)
reveal_type(
# revealed: A | B
foo.f(
ab,
a1=a,
a2=a,
a3=a,
a4=a,
a5=a,
a6=a,
a7=a,
a8=a,
a9=a,
a10=a,
a11=a,
a12=a,
a13=a,
a14=a,
a15=a,
a16=a,
a17=a,
a18=a,
a19=a,
a20=a,
a21=a,
a22=a,
a23=a,
a24=a,
a25=a,
a26=a,
a27=a,
a28=a,
a29=a,
a30=a,
)
)
Optimization: Limit expansion size
To prevent combinatorial explosion, ty limits the number of argument lists created by expanding a single argument.
overloaded.pyi
:
from typing import overload
class A: ...
class B: ...
class C: ...
@overload
def f() -> None: ...
@overload
def f(**kwargs: int) -> C: ...
@overload
def f(x: A, /, **kwargs: int) -> A: ...
@overload
def f(x: B, /, **kwargs: int) -> B: ...
from overloaded import A, B, f
from typing_extensions import reveal_type
def _(a: int | None):
reveal_type(
# error: [no-matching-overload]
# revealed: Unknown
f(
A(),
a1=a,
a2=a,
a3=a,
a4=a,
a5=a,
a6=a,
a7=a,
a8=a,
a9=a,
a10=a,
a11=a,
a12=a,
a13=a,
a14=a,
a15=a,
a16=a,
a17=a,
a18=a,
a19=a,
a20=a,
a21=a,
a22=a,
a23=a,
a24=a,
a25=a,
a26=a,
a27=a,
a28=a,
a29=a,
a30=a,
)
)
Retry from parameter matching
As per the spec, the argument type expansion should retry evaluating the expanded argument list from the type checking step. However, that creates an issue when variadic arguments are involved because if a variadic argument is a union type, it could be expanded to have different arities. So, ty retries it from the start which includes parameter matching as well.
overloaded.pyi
:
from typing import overload
@overload
def f(x: int, y: int) -> None: ...
@overload
def f(x: int, y: str, z: int) -> None: ...
from overloaded import f
# Test all of the above with a number of different splatted argument types
def _(t: tuple[int, str]) -> None:
# This correctly produces an error because the first element of the union has a precise arity of
# 2, which matches the first overload, but the second element of the tuple doesn't match the
# second parameter type, yielding an `invalid-argument-type` error.
f(*t) # error: [invalid-argument-type]
def _(t: tuple[int, str, int]) -> None:
# This correctly produces no error because the first element of the union has a precise arity of
# 3, which matches the second overload.
f(*t)
def _(t: tuple[int, str] | tuple[int, str, int]) -> None:
# This produces an error because the expansion produces two argument lists: `[*tuple[int, str]]`
# and `[*tuple[int, str, int]]`. The first list produces produces a type checking error as
# described in the first example, while the second list matches the second overload. And,
# because not all of the expanded argument list evaluates successfully, we produce an error.
f(*t) # error: [no-matching-overload]
Filtering based on variaidic arguments
This is step 4 of the overload call evaluation algorithm which specifies that:
If the argument list is compatible with two or more overloads, determine whether one or more of the overloads has a variadic parameter (either
*args
or**kwargs
) that maps to a corresponding argument that supplies an indeterminate number of positional or keyword arguments. If so, eliminate overloads that do not have a variadic parameter.
This is only performed if the previous step resulted in more than one matching overload.
Simple *args
overloaded.pyi
:
from typing import overload
@overload
def f(x1: int) -> tuple[int]: ...
@overload
def f(x1: int, x2: int) -> tuple[int, int]: ...
@overload
def f(*args: int) -> int: ...
from overloaded import f
def _(x1: int, x2: int, args: list[int]):
reveal_type(f(x1)) # revealed: tuple[int]
reveal_type(f(x1, x2)) # revealed: tuple[int, int]
reveal_type(f(*(x1, x2))) # revealed: tuple[int, int]
# Step 4 should filter out all but the last overload.
reveal_type(f(*args)) # revealed: int
Variable *args
[environment]
python-version = "3.11"
overloaded.pyi
:
from typing import overload
@overload
def f(x1: int) -> tuple[int]: ...
@overload
def f(x1: int, x2: int) -> tuple[int, int]: ...
@overload
def f(x1: int, *args: int) -> tuple[int, ...]: ...
from overloaded import f
def _(x1: int, x2: int, args1: list[int], args2: tuple[int, *tuple[int, ...]]):
reveal_type(f(x1, x2)) # revealed: tuple[int, int]
reveal_type(f(*(x1, x2))) # revealed: tuple[int, int]
# Step 4 should filter out all but the last overload.
reveal_type(f(x1, *args1)) # revealed: tuple[int, ...]
reveal_type(f(*args2)) # revealed: tuple[int, ...]
Simple **kwargs
overloaded.pyi
:
from typing import overload
@overload
def f(*, x1: int) -> int: ...
@overload
def f(*, x1: int, x2: int) -> tuple[int, int]: ...
@overload
def f(**kwargs: int) -> int: ...
from overloaded import f
def _(x1: int, x2: int, kwargs: dict[str, int]):
reveal_type(f(x1=x1)) # revealed: int
reveal_type(f(x1=x1, x2=x2)) # revealed: tuple[int, int]
# Step 4 should filter out all but the last overload.
reveal_type(f(**{"x1": x1, "x2": x2})) # revealed: int
reveal_type(f(**kwargs)) # revealed: int
TypedDict
The keys in a TypedDict
are static so there's no variable part to it, so step 4 shouldn't filter
out any overloads.
overloaded.pyi
:
from typing import TypedDict, overload
@overload
def f(*, x: int) -> int: ...
@overload
def f(*, x: int, y: int) -> tuple[int, int]: ...
@overload
def f(**kwargs: int) -> tuple[int, ...]: ...
from typing import TypedDict
from overloaded import f
class Foo(TypedDict):
x: int
y: int
def _(foo: Foo, kwargs: dict[str, int]):
reveal_type(f(**foo)) # revealed: tuple[int, int]
reveal_type(f(**kwargs)) # revealed: tuple[int, ...]
Filtering based on Any
/ Unknown
This is the step 5 of the overload call evaluation algorithm which specifies that:
For all arguments, determine whether all possible materializations of the argument’s type are assignable to the corresponding parameter type for each of the remaining overloads. If so, eliminate all of the subsequent remaining overloads.
This is only performed if the previous step resulted in more than one matching overload.
Single list argument
overloaded.pyi
:
from typing import Any, overload
@overload
def f(x: list[int]) -> int: ...
@overload
def f(x: list[Any]) -> int: ...
@overload
def f(x: Any) -> str: ...
For the above definition, anything other than list
should match the last overload:
from typing import Any
from overloaded import f
# Anything other than `list` should match the last overload
reveal_type(f(1)) # revealed: str
reveal_type(f(*(1,))) # revealed: str
def _(list_int: list[int], list_any: list[Any]):
reveal_type(f(list_int)) # revealed: int
reveal_type(f(*(list_int,))) # revealed: int
reveal_type(f(list_any)) # revealed: int
reveal_type(f(*(list_any,))) # revealed: int
Single list argument (ambiguous)
The overload definition is the same as above, but the return type of the second overload is changed
to str
to make the overload matching ambiguous if the argument is a list[Any]
.
overloaded.pyi
:
from typing import Any, overload
@overload
def f(x: list[int]) -> int: ...
@overload
def f(x: list[Any]) -> str: ...
@overload
def f(x: Any) -> str: ...
from typing import Any
from overloaded import f
# Anything other than `list` should match the last overload
reveal_type(f(1)) # revealed: str
reveal_type(f(*(1,))) # revealed: str
def _(list_int: list[int], list_any: list[Any]):
# All materializations of `list[int]` are assignable to `list[int]`, so it matches the first
# overload.
reveal_type(f(list_int)) # revealed: int
reveal_type(f(*(list_int,))) # revealed: int
# All materializations of `list[Any]` are assignable to `list[int]` and `list[Any]`, but the
# return type of first and second overloads are not equivalent, so the overload matching
# is ambiguous.
reveal_type(f(list_any)) # revealed: Unknown
reveal_type(f(*(list_any,))) # revealed: Unknown
Single tuple argument
overloaded.pyi
:
from typing import Any, overload
@overload
def f(x: tuple[int, str]) -> int: ...
@overload
def f(x: tuple[int, Any]) -> int: ...
@overload
def f(x: Any) -> str: ...
from typing import Any
from overloaded import f
reveal_type(f("a")) # revealed: str
reveal_type(f(*("a",))) # revealed: str
reveal_type(f((1, "b"))) # revealed: int
reveal_type(f(*((1, "b"),))) # revealed: int
reveal_type(f((1, 2))) # revealed: int
reveal_type(f(*((1, 2),))) # revealed: int
def _(int_str: tuple[int, str], int_any: tuple[int, Any], any_any: tuple[Any, Any]):
# All materializations are assignable to first overload, so second and third overloads are
# eliminated
reveal_type(f(int_str)) # revealed: int
reveal_type(f(*(int_str,))) # revealed: int
# All materializations are assignable to second overload, so the third overload is eliminated;
# the return type of first and second overload is equivalent
reveal_type(f(int_any)) # revealed: int
reveal_type(f(*(int_any,))) # revealed: int
# All materializations of `tuple[Any, Any]` are assignable to the parameters of all the
# overloads, but the return types aren't equivalent, so the overload matching is ambiguous
reveal_type(f(any_any)) # revealed: Unknown
reveal_type(f(*(any_any,))) # revealed: Unknown
Unknown
passed into an overloaded function annotated with protocols
Foo.join()
here has similar annotations to str.join()
in typeshed:
module.pyi
:
from typing_extensions import Iterable, overload, LiteralString, Protocol
from ty_extensions import Unknown, is_assignable_to
class Foo:
@overload
def join(self, iterable: Iterable[LiteralString], /) -> LiteralString: ...
@overload
def join(self, iterable: Iterable[str], /) -> str: ...
main.py
:
from module import Foo
from typing_extensions import LiteralString
def f(a: Foo, b: list[str], c: list[LiteralString], e):
reveal_type(e) # revealed: Unknown
# TODO: we should select the second overload here and reveal `str`
# (the incorrect result is due to missing logic in protocol subtyping/assignability)
reveal_type(a.join(b)) # revealed: LiteralString
reveal_type(a.join(c)) # revealed: LiteralString
# since both overloads match and they have return types that are not equivalent,
# step (5) of the overload evaluation algorithm says we must evaluate the result of the
# call as `Unknown`.
#
# Note: although the spec does not state as such (since intersections in general are not
# specified currently), `(str | LiteralString) & Unknown` might also be a reasonable type
# here (the union of all overload returns, intersected with `Unknown`) -- here that would
# simplify to `str & Unknown`.
reveal_type(a.join(e)) # revealed: Unknown
Multiple arguments
overloaded.pyi
:
from typing import Any, overload
class A: ...
class B: ...
@overload
def f(x: list[int], y: tuple[int, str]) -> A: ...
@overload
def f(x: list[Any], y: tuple[int, Any]) -> A: ...
@overload
def f(x: list[Any], y: tuple[Any, Any]) -> B: ...
from typing import Any
from overloaded import A, f
def _(list_int: list[int], list_any: list[Any], int_str: tuple[int, str], int_any: tuple[int, Any], any_any: tuple[Any, Any]):
# All materializations of both argument types are assignable to the first overload, so the
# second and third overloads are filtered out
reveal_type(f(list_int, int_str)) # revealed: A
reveal_type(f(*(list_int, int_str))) # revealed: A
# All materialization of first argument is assignable to first overload and for the second
# argument, they're assignable to the second overload, so the third overload is filtered out
reveal_type(f(list_int, int_any)) # revealed: A
reveal_type(f(*(list_int, int_any))) # revealed: A
# All materialization of first argument is assignable to second overload and for the second
# argument, they're assignable to the first overload, so the third overload is filtered out
reveal_type(f(list_any, int_str)) # revealed: A
reveal_type(f(*(list_any, int_str))) # revealed: A
# All materializations of both arguments are assignable to the second overload, so the third
# overload is filtered out
reveal_type(f(list_any, int_any)) # revealed: A
reveal_type(f(*(list_any, int_any))) # revealed: A
# All materializations of first argument is assignable to the second overload and for the second
# argument, they're assignable to the third overload, so no overloads are filtered out; the
# return types of the remaining overloads are not equivalent, so overload matching is ambiguous
reveal_type(f(list_int, any_any)) # revealed: Unknown
reveal_type(f(*(list_int, any_any))) # revealed: Unknown
LiteralString
and str
overloaded.pyi
:
from typing import overload
from typing_extensions import LiteralString
@overload
def f(x: LiteralString) -> LiteralString: ...
@overload
def f(x: str) -> str: ...
from typing import Any
from typing_extensions import LiteralString
from overloaded import f
def _(literal: LiteralString, string: str, any: Any):
reveal_type(f(literal)) # revealed: LiteralString
reveal_type(f(*(literal,))) # revealed: LiteralString
reveal_type(f(string)) # revealed: str
reveal_type(f(*(string,))) # revealed: str
# `Any` matches both overloads, but the return types are not equivalent.
# Pyright and mypy both reveal `str` here, contrary to the spec.
reveal_type(f(any)) # revealed: Unknown
reveal_type(f(*(any,))) # revealed: Unknown
Generics
overloaded.pyi
:
from typing import Any, TypeVar, overload
_T = TypeVar("_T")
class A: ...
class B: ...
@overload
def f(x: list[int]) -> A: ...
@overload
def f(x: list[_T]) -> _T: ...
@overload
def f(x: Any) -> B: ...
from typing import Any
from overloaded import f
def _(list_int: list[int], list_str: list[str], list_any: list[Any], any: Any):
reveal_type(f(list_int)) # revealed: A
reveal_type(f(*(list_int,))) # revealed: A
reveal_type(f(list_str)) # revealed: str
reveal_type(f(*(list_str,))) # revealed: str
reveal_type(f(list_any)) # revealed: Unknown
reveal_type(f(*(list_any,))) # revealed: Unknown
reveal_type(f(any)) # revealed: Unknown
reveal_type(f(*(any,))) # revealed: Unknown
Generics (multiple arguments)
overloaded.pyi
:
from typing import Any, TypeVar, overload
_T = TypeVar("_T")
@overload
def f(x: int, y: Any) -> int: ...
@overload
def f(x: str, y: _T) -> _T: ...
from typing import Any
from overloaded import f
def _(integer: int, string: str, any: Any, list_any: list[Any]):
reveal_type(f(integer, string)) # revealed: int
reveal_type(f(*(integer, string))) # revealed: int
reveal_type(f(string, integer)) # revealed: int
reveal_type(f(*(string, integer))) # revealed: int
# This matches the second overload and is _not_ the case of ambiguous overload matching.
reveal_type(f(string, any)) # revealed: Any
reveal_type(f(*(string, any))) # revealed: Any
reveal_type(f(string, list_any)) # revealed: list[Any]
reveal_type(f(*(string, list_any))) # revealed: list[Any]
Generic self
overloaded.pyi
:
from typing import Any, overload, TypeVar, Generic
_T = TypeVar("_T")
class A(Generic[_T]):
@overload
def method(self: "A[int]") -> int: ...
@overload
def method(self: "A[Any]") -> int: ...
class B(Generic[_T]):
@overload
def method(self: "B[int]") -> int: ...
@overload
def method(self: "B[Any]") -> str: ...
from typing import Any
from overloaded import A, B
def _(a_int: A[int], a_str: A[str], a_any: A[Any]):
reveal_type(a_int.method()) # revealed: int
reveal_type(a_str.method()) # revealed: int
reveal_type(a_any.method()) # revealed: int
def _(b_int: B[int], b_str: B[str], b_any: B[Any]):
reveal_type(b_int.method()) # revealed: int
reveal_type(b_str.method()) # revealed: str
reveal_type(b_any.method()) # revealed: Unknown
Variadic argument
overloaded.pyi
:
from typing import Any, overload
class A: ...
class B: ...
@overload
def f1(x: int) -> A: ...
@overload
def f1(x: Any, y: Any) -> A: ...
@overload
def f2(x: int) -> A: ...
@overload
def f2(x: Any, y: Any) -> B: ...
@overload
def f3(x: int) -> A: ...
@overload
def f3(x: Any, y: Any) -> A: ...
@overload
def f3(x: Any, y: Any, *, z: str) -> B: ...
@overload
def f4(x: int) -> A: ...
@overload
def f4(x: Any, y: Any) -> B: ...
@overload
def f4(x: Any, y: Any, *, z: str) -> B: ...
from typing import Any
from overloaded import f1, f2, f3, f4
def _(arg: list[Any]):
# Matches both overload and the return types are equivalent
reveal_type(f1(*arg)) # revealed: A
# Matches both overload but the return types aren't equivalent
reveal_type(f2(*arg)) # revealed: Unknown
# Filters out the final overload and the return types are equivalent
reveal_type(f3(*arg)) # revealed: A
# Filters out the final overload but the return types aren't equivalent
reveal_type(f4(*arg)) # revealed: Unknown
Non-participating fully-static parameter
Ref: https://github.com/astral-sh/ty/issues/552#issuecomment-2969052173
A non-participating parameter would be the one where the set of materializations of the argument type, that are assignable to the parameter type at the same index, is same for the overloads for which step 5 needs to be performed.
overloaded.pyi
:
from typing import Literal, overload
@overload
def f(x: str, *, flag: Literal[True]) -> int: ...
@overload
def f(x: str, *, flag: Literal[False] = ...) -> str: ...
@overload
def f(x: str, *, flag: bool = ...) -> int | str: ...
In the following example, for the f(any, flag=True)
call, the materializations of first argument
type Any
that are assignable to str
is same for overloads 1 and 3 (at the time of step 5), so
for the purposes of overload matching that parameter can be ignored. If Any
materializes to
anything that's not assignable to str
, all of the overloads would already be filtered out which
will raise a no-matching-overload
error.
from typing import Any
from overloaded import f
def _(any: Any):
reveal_type(f(any, flag=True)) # revealed: int
reveal_type(f(*(any,), flag=True)) # revealed: int
reveal_type(f(any, flag=False)) # revealed: str
reveal_type(f(*(any,), flag=False)) # revealed: str
Non-participating gradual parameter
overloaded.pyi
:
from typing import Any, Literal, overload
@overload
def f(x: tuple[str, Any], flag: Literal[True]) -> int: ...
@overload
def f(x: tuple[str, Any], flag: Literal[False] = ...) -> str: ...
@overload
def f(x: tuple[str, Any], flag: bool = ...) -> int | str: ...
from typing import Any, Literal
from overloaded import f
def _(any: Any):
reveal_type(f(any, flag=True)) # revealed: int
reveal_type(f(*(any,), flag=True)) # revealed: int
reveal_type(f(any, flag=False)) # revealed: str
reveal_type(f(*(any,), flag=False)) # revealed: str
def _(args: tuple[Any, Literal[True]]):
reveal_type(f(*args)) # revealed: int
def _(args: tuple[Any, Literal[False]]):
reveal_type(f(*args)) # revealed: str
Argument type expansion
This filtering can also happen for each of the expanded argument lists.
No ambiguity
overloaded.pyi
:
from typing import Any, overload
class A: ...
class B: ...
@overload
def f(x: tuple[A, B]) -> A: ...
@overload
def f(x: tuple[B, A]) -> B: ...
@overload
def f(x: tuple[A, Any]) -> A: ...
@overload
def f(x: tuple[B, Any]) -> B: ...
Here, the argument tuple[A | B, Any]
doesn't match any of the overloads, so we perform argument
type expansion which results in two argument lists:
tuple[A, Any]
tuple[B, Any]
The first argument list matches overload 1 and 3 via Any
materialization for which the return
types are equivalent (A
). Similarly, the second argument list matches overload 2 and 4 via Any
materialization for which the return types are equivalent (B
). The final return type for the call
will be the union of the return types.
from typing import Any
from overloaded import A, B, f
def _(arg: tuple[A | B, Any]):
reveal_type(f(arg)) # revealed: A | B
reveal_type(f(*(arg,))) # revealed: A | B
One argument list ambiguous
The example used here is same as the previous one, but the return type of the last overload is changed so that it's not equivalent to the return type of the second overload, creating an ambiguous matching for the second argument list.
overloaded.pyi
:
from typing import Any, overload
class A: ...
class B: ...
class C: ...
@overload
def f(x: tuple[A, B]) -> A: ...
@overload
def f(x: tuple[B, A]) -> B: ...
@overload
def f(x: tuple[A, Any]) -> A: ...
@overload
def f(x: tuple[B, Any]) -> C: ...
from typing import Any
from overloaded import A, B, C, f
def _(arg: tuple[A | B, Any]):
reveal_type(f(arg)) # revealed: A | Unknown
reveal_type(f(*(arg,))) # revealed: A | Unknown
Both argument lists ambiguous
Here, both argument lists created by expanding the argument type are ambiguous, so the final return
type is Any
.
overloaded.pyi
:
from typing import Any, overload
class A: ...
class B: ...
class C: ...
@overload
def f(x: tuple[A, B]) -> A: ...
@overload
def f(x: tuple[B, A]) -> B: ...
@overload
def f(x: tuple[A, Any]) -> C: ...
@overload
def f(x: tuple[B, Any]) -> C: ...
from typing import Any
from overloaded import A, B, C, f
def _(arg: tuple[A | B, Any]):
reveal_type(f(arg)) # revealed: Unknown
reveal_type(f(*(arg,))) # revealed: Unknown
Bidirectional Type Inference
[environment]
python-version = "3.12"
Type inference accounts for parameter type annotations across all overloads.
from typing import TypedDict, overload
class T(TypedDict):
x: int
@overload
def f(a: list[T], b: int) -> int: ...
@overload
def f(a: list[dict[str, int]], b: str) -> str: ...
def f(a: list[dict[str, int]] | list[T], b: int | str) -> int | str:
return 1
def int_or_str() -> int | str:
return 1
x = f([{"x": 1}], int_or_str())
reveal_type(x) # revealed: int | str
# TODO: error: [no-matching-overload] "No overload of function `f` matches arguments"
# we currently incorrectly consider `list[dict[str, int]]` a subtype of `list[T]`
f([{"y": 1}], int_or_str())
Non-matching overloads do not produce diagnostics:
from typing import TypedDict, overload
class T(TypedDict):
x: int
@overload
def f(a: T, b: int) -> int: ...
@overload
def f(a: dict[str, int], b: str) -> str: ...
def f(a: T | dict[str, int], b: int | str) -> int | str:
return 1
x = f({"y": 1}, "a")
reveal_type(x) # revealed: str
from typing import SupportsRound, overload
@overload
def takes_str_or_float(x: str): ...
@overload
def takes_str_or_float(x: float): ...
def takes_str_or_float(x: float | str): ...
takes_str_or_float(round(1.0))