7 KiB
Generic functions: Legacy syntax
Typevar must be used at least twice
If you're only using a typevar for a single parameter, you don't need the typevar — just use
object (or the typevar's upper bound):
from typing import TypeVar
T = TypeVar("T")
# TODO: error, should be (x: object)
def typevar_not_needed(x: T) -> None:
pass
BoundedT = TypeVar("BoundedT", bound=int)
# TODO: error, should be (x: int)
def bounded_typevar_not_needed(x: BoundedT) -> None:
pass
Typevars are only needed if you use them more than once. For instance, to specify that two parameters must both have the same type:
def two_params(x: T, y: T) -> T:
return x
or to specify that a return value is the same as a parameter:
def return_value(x: T) -> T:
return x
Each typevar must also appear somewhere in the parameter list:
def absurd() -> T:
# There's no way to construct a T!
raise ValueError("absurd")
Inferring generic function parameter types
If the type of a generic function parameter is a typevar, then we can infer what type that typevar is bound to at each call site.
from typing import TypeVar
T = TypeVar("T")
def f(x: T) -> T:
return x
reveal_type(f(1)) # revealed: Literal[1]
reveal_type(f(1.0)) # revealed: float
reveal_type(f(True)) # revealed: Literal[True]
reveal_type(f("string")) # revealed: Literal["string"]
Inferring “deep” generic parameter types
The matching up of call arguments and discovery of constraints on typevars can be a recursive process for arbitrarily-nested generic types in parameters.
from typing import TypeVar
T = TypeVar("T")
def f(x: list[T]) -> T:
return x[0]
# TODO: revealed: float
reveal_type(f([1.0, 2.0])) # revealed: Unknown
Inferring a bound typevar
from typing import TypeVar
from typing_extensions import reveal_type
T = TypeVar("T", bound=int)
def f(x: T) -> T:
return x
reveal_type(f(1)) # revealed: Literal[1]
reveal_type(f(True)) # revealed: Literal[True]
# error: [invalid-argument-type]
reveal_type(f("string")) # revealed: Unknown
Inferring a constrained typevar
from typing import TypeVar
from typing_extensions import reveal_type
T = TypeVar("T", int, None)
def f(x: T) -> T:
return x
reveal_type(f(1)) # revealed: int
reveal_type(f(True)) # revealed: int
reveal_type(f(None)) # revealed: None
# error: [invalid-argument-type]
reveal_type(f("string")) # revealed: Unknown
Typevar constraints
If a type parameter has an upper bound, that upper bound constrains which types can be used for that typevar. This effectively adds the upper bound as an intersection to every appearance of the typevar in the function.
from typing import TypeVar
T = TypeVar("T", bound=int)
def good_param(x: T) -> None:
reveal_type(x) # revealed: T
If the function is annotated as returning the typevar, this means that the upper bound is not
assignable to that typevar, since return types are contravariant. In bad, we can infer that
x + 1 has type int. But T might be instantiated with a narrower type than int, and so the
return value is not guaranteed to be compatible for all T: int.
def good_return(x: T) -> T:
return x
def bad_return(x: T) -> T:
# error: [invalid-return-type] "Return type does not match returned value: Expected `T`, found `int`"
return x + 1
All occurrences of the same typevar have the same type
If a typevar appears multiple times in a function signature, all occurrences have the same type.
from typing import TypeVar
T = TypeVar("T")
S = TypeVar("S")
def different_types(cond: bool, t: T, s: S) -> T:
if cond:
return t
else:
# error: [invalid-return-type] "Return type does not match returned value: Expected `T`, found `S`"
return s
def same_types(cond: bool, t1: T, t2: T) -> T:
if cond:
return t1
else:
return t2
All occurrences of the same constrained typevar have the same type
The above is true even when the typevars are constrained. Here, both int and str have __add__
methods that are compatible with the return type, so the return expression is always well-typed:
from typing import TypeVar
T = TypeVar("T", int, str)
def same_constrained_types(t1: T, t2: T) -> T:
# TODO: no error
# error: [unsupported-operator] "Operator `+` is unsupported between objects of type `T` and `T`"
return t1 + t2
This is not the same as a union type, because of this additional constraint that the two
occurrences have the same type. In unions_are_different, t1 and t2 might have different types,
and an int and a str cannot be added together:
def unions_are_different(t1: int | str, t2: int | str) -> int | str:
# error: [unsupported-operator] "Operator `+` is unsupported between objects of type `int | str` and `int | str`"
return t1 + t2
Typevar inference is a unification problem
When inferring typevar assignments in a generic function call, we cannot simply solve constraints eagerly for each parameter in turn. We must solve a unification problem involving all of the parameters simultaneously.
from typing import TypeVar
T = TypeVar("T")
def two_params(x: T, y: T) -> T:
return x
reveal_type(two_params("a", "b")) # revealed: Literal["a", "b"]
reveal_type(two_params("a", 1)) # revealed: Literal["a", 1]
When one of the parameters is a union, we attempt to find the smallest specialization that satisfies all of the constraints.
def union_param(x: T | None) -> T:
if x is None:
raise ValueError
return x
reveal_type(union_param("a")) # revealed: Literal["a"]
reveal_type(union_param(1)) # revealed: Literal[1]
reveal_type(union_param(None)) # revealed: Unknown
def union_and_nonunion_params(x: T | int, y: T) -> T:
return y
reveal_type(union_and_nonunion_params(1, "a")) # revealed: Literal["a"]
reveal_type(union_and_nonunion_params("a", "a")) # revealed: Literal["a"]
reveal_type(union_and_nonunion_params(1, 1)) # revealed: Literal[1]
reveal_type(union_and_nonunion_params(3, 1)) # revealed: Literal[1]
reveal_type(union_and_nonunion_params("a", 1)) # revealed: Literal["a", 1]
S = TypeVar("S")
def tuple_param(x: T | S, y: tuple[T, S]) -> tuple[T, S]:
return y
reveal_type(tuple_param("a", ("a", 1))) # revealed: tuple[Literal["a"], Literal[1]]
reveal_type(tuple_param(1, ("a", 1))) # revealed: tuple[Literal["a"], Literal[1]]
Inferring nested generic function calls
We can infer type assignments in nested calls to multiple generic functions. If they use the same
type variable, we do not confuse the two; T@f and T@g have separate types in each example below.
from typing import TypeVar
T = TypeVar("T")
def f(x: T) -> tuple[T, int]:
return (x, 1)
def g(x: T) -> T | None:
return x
reveal_type(f(g("a"))) # revealed: tuple[Literal["a"] | None, int]
reveal_type(g(f("a"))) # revealed: tuple[Literal["a"], int] | None