[red-knot] - Flow-control for boolean operations (#13940)

## Summary

As python uses short-circuiting boolean operations in runtime, we should
mimic that logic in redknot as well.
For example, we should detect that in the following code `x` might be
undefined inside the block:

```py
if flag or (x := 1):
    print(x) 
```

## Test Plan

Added mdtest suit for boolean expressions.

---------

Co-authored-by: Carl Meyer <carl@astral.sh>
This commit is contained in:
TomerBin 2024-10-27 05:33:01 +02:00 committed by GitHub
parent b6ffa51c16
commit 66c3aaa307
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3 changed files with 125 additions and 1 deletions

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@ -0,0 +1,78 @@
# Short-Circuit Evaluation
## Not all boolean expressions must be evaluated
In `or` expressions, if the left-hand side is truthy, the right-hand side is not evaluated.
Similarly, in `and` expressions, if the left-hand side is falsy, the right-hand side is not
evaluated.
```py
def bool_instance() -> bool:
return True
if bool_instance() or (x := 1):
# error: [possibly-unresolved-reference]
reveal_type(x) # revealed: Unbound | Literal[1]
if bool_instance() and (x := 1):
# error: [possibly-unresolved-reference]
reveal_type(x) # revealed: Unbound | Literal[1]
```
## First expression is always evaluated
```py
def bool_instance() -> bool:
return True
if (x := 1) or bool_instance():
reveal_type(x) # revealed: Literal[1]
if (x := 1) and bool_instance():
reveal_type(x) # revealed: Literal[1]
```
## Statically known truthiness
```py
if True or (x := 1):
# TODO: infer that the second arm is never executed so type should be just "Unbound".
# error: [possibly-unresolved-reference]
reveal_type(x) # revealed: Unbound | Literal[1]
if True and (x := 1):
# TODO: infer that the second arm is always executed so type should be just "Literal[1]".
# error: [possibly-unresolved-reference]
reveal_type(x) # revealed: Unbound | Literal[1]
```
## Later expressions can always use variables from earlier expressions
```py
def bool_instance() -> bool:
return True
bool_instance() or (x := 1) or reveal_type(x) # revealed: Literal[1]
# error: [unresolved-reference]
bool_instance() or reveal_type(y) or (y := 1) # revealed: Unbound
```
## Nested expressions
```py
def bool_instance() -> bool:
return True
if bool_instance() or ((x := 1) and bool_instance()):
# error: "Name `x` used when possibly not defined"
reveal_type(x) # revealed: Unbound | Literal[1]
if ((y := 1) and bool_instance()) or bool_instance():
reveal_type(y) # revealed: Literal[1]
# error: [possibly-unresolved-reference]
if (bool_instance() and (z := 1)) or reveal_type(z): # revealed: Unbound | Literal[1]
# error: [possibly-unresolved-reference]
reveal_type(z) # revealed: Unbound | Literal[1]
```

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@ -1106,6 +1106,27 @@ where
},
);
}
ast::Expr::BoolOp(ast::ExprBoolOp {
values,
range: _,
op: _,
}) => {
// TODO detect statically known truthy or falsy values (via type inference, not naive
// AST inspection, so we can't simplify here, need to record test expression for
// later checking)
let mut snapshots = vec![];
for (index, value) in values.iter().enumerate() {
// The first item of BoolOp is always evaluated
if index > 0 {
snapshots.push(self.flow_snapshot());
}
self.visit_expr(value);
}
for snapshot in snapshots {
self.flow_merge(snapshot);
}
}
_ => {
walk_expr(self, expr);
}