ruff/crates/ty_python_semantic/resources/mdtest/del.md
Shunsuke Shibayama ef564094a9
[ty] support del statement and deletion of except handler names (#18593)
## Summary

This PR closes https://github.com/astral-sh/ty/issues/238.

Since `DefinitionState::Deleted` was introduced in #18041, support for
the `del` statement (and deletion of except handler names) is
straightforward.

However, it is difficult to determine whether references to attributes
or subscripts are unresolved after they are deleted. This PR only
invalidates narrowing by assignment if the attribute or subscript is
deleted.

## Test Plan

`mdtest/del.md` is added.

---------

Co-authored-by: Alex Waygood <Alex.Waygood@Gmail.com>
2025-06-12 07:44:42 -07:00

2.3 KiB

del statement

Basic

a = 1
del a
# error: [unresolved-reference]
reveal_type(a)  # revealed: Unknown

# error: [invalid-syntax] "Invalid delete target"
del 1

# error: [unresolved-reference]
del a

x, y = 1, 2
del x, y
# error: [unresolved-reference]
reveal_type(x)  # revealed: Unknown
# error: [unresolved-reference]
reveal_type(y)  # revealed: Unknown

def cond() -> bool:
    return True

b = 1
if cond():
    del b

# error: [possibly-unresolved-reference]
reveal_type(b)  # revealed: Literal[1]

c = 1
if cond():
    c = 2
else:
    del c

# error: [possibly-unresolved-reference]
reveal_type(c)  # revealed: Literal[2]

d = 1

def delete():
    # TODO: this results in `UnboundLocalError`; we should emit `unresolved-reference`
    del d

delete()
reveal_type(d)  # revealed: Literal[1]

def delete_global():
    global d
    del d

delete_global()
# The variable should have been removed, but we won't check it for now.
reveal_type(d)  # revealed: Literal[1]

Delete attributes

If an attribute is referenced after being deleted, it will be an error at runtime. But we don't treat this as an error (because there may have been a redefinition by a method between the del statement and the reference). However, deleting an attribute invalidates type narrowing by assignment, and the attribute type will be the originally declared type.

Invalidate narrowing

class C:
    x: int = 1

c = C()
del c.x
reveal_type(c.x)  # revealed: int

# error: [unresolved-attribute]
del c.non_existent

c.x = 1
reveal_type(c.x)  # revealed: Literal[1]
del c.x
reveal_type(c.x)  # revealed: int

Delete an instance attribute definition

class C:
    x: int = 1

c = C()
reveal_type(c.x)  # revealed: int

del C.x
c = C()
# This attribute is unresolved, but we won't check it for now.
reveal_type(c.x)  # revealed: int

Delete items

Deleting an item also invalidates the narrowing by the assignment, but accessing the item itself is still valid.

def f(l: list[int]):
    del l[0]
    # If the length of `l` was 1, this will be a runtime error,
    # but if it was greater than that, it will not be an error.
    reveal_type(l[0])  # revealed: int

    # error: [call-non-callable]
    del l["string"]

    l[0] = 1
    reveal_type(l[0])  # revealed: Literal[1]
    del l[0]
    reveal_type(l[0])  # revealed: int