[red-knot] Port type inference tests to new test framework (#13719)

## Summary

Porting infer tests to new markdown tests framework.

Link to the corresponding issue: #13696

---------

Co-authored-by: Carl Meyer <carl@astral.sh>
This commit is contained in:
Alex
2024-10-15 21:23:46 +03:00
committed by GitHub
parent 5fa82fb0cd
commit d77480768d
56 changed files with 2025 additions and 3355 deletions

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# If expressions
## Simple if-expression
```py
x = 1 if flag else 2
reveal_type(x) # revealed: Literal[1, 2]
```
## If-expression with walrus operator
```py
y = 0
z = 0
x = (y := 1) if flag else (z := 2)
a = y
b = z
reveal_type(x) # revealed: Literal[1, 2]
reveal_type(a) # revealed: Literal[0, 1]
reveal_type(b) # revealed: Literal[0, 2]
```
## Nested if-expression
```py
x = 1 if flag else 2 if flag2 else 3
reveal_type(x) # revealed: Literal[1, 2, 3]
```
## None
```py
x = 1 if flag else None
reveal_type(x) # revealed: Literal[1] | None
```

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# If statements
## Simple if
```py
y = 1
y = 2
if flag:
y = 3
x = y
reveal_type(x) # revealed: Literal[2, 3]
```
## Simple if-elif-else
```py
y = 1
y = 2
if flag:
y = 3
elif flag2:
y = 4
else:
r = y
y = 5
s = y
x = y
reveal_type(x) # revealed: Literal[3, 4, 5]
reveal_type(r) # revealed: Unbound | Literal[2]
reveal_type(s) # revealed: Unbound | Literal[5]
```
## Single symbol across if-elif-else
```py
if flag:
y = 1
elif flag2:
y = 2
else:
y = 3
reveal_type(y) # revealed: Literal[1, 2, 3]
```
## if-elif-else without else assignment
```py
y = 0
if flag:
y = 1
elif flag2:
y = 2
else:
pass
reveal_type(y) # revealed: Literal[0, 1, 2]
```
## if-elif-else with intervening assignment
```py
y = 0
if flag:
y = 1
z = 3
elif flag2:
y = 2
else:
pass
reveal_type(y) # revealed: Literal[0, 1, 2]
```
## Nested if statement
```py
y = 0
if flag:
if flag2:
y = 1
reveal_type(y) # revealed: Literal[0, 1]
```
## if-elif without else
```py
y = 1
y = 2
if flag:
y = 3
elif flag2:
y = 4
x = y
reveal_type(x) # revealed: Literal[2, 3, 4]
```

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# Pattern matching
## With wildcard
```py
match 0:
case 1:
y = 2
case _:
y = 3
reveal_type(y) # revealed: Literal[2, 3]
```
## Without wildcard
```py
match 0:
case 1:
y = 2
case 2:
y = 3
reveal_type(y) # revealed: Unbound | Literal[2, 3]
```
## Basic match
```py
y = 1
y = 2
match 0:
case 1:
y = 3
case 2:
y = 4
reveal_type(y) # revealed: Literal[2, 3, 4]
```