Python match syntax for evaluating typing objects

Question

I'm trying to use a match block to evaluate the types being returned from typing.get_type_hints. I want to use the type hinting to determine what code path to take in order to instantiate the appropriate value.

class Foo:
  x: int = 0
  y: float = 0.0
  z: str = ''

name = 'x'  # comes from an external source
value = '42'
type_hints = get_type_hints(Foo)
foo = Foo()

match type_hints[name]:
  case int:
    foo[name] = int(value)
  case float:
    foo[name] = int(value)
  case str:
    foo[name] = value
  # TODO: other complex type parsing (e.g., Enums)

This is the error I'm getting:

C:\dev\bdk-git\bdk\venv\Scripts\python.exe C:/dev/bdk-git/bdk/material/reader.py
  File "C:\dev\bdk-git\bdk\material\reader.py", line 22
    case int:
         ^^^
SyntaxError: name capture 'int' makes remaining patterns unreachable

What is the correct syntax to accomplish what I want to do?

Chris · Accepted Answer · 2022-12-13 03:44:10Z

2

I'm not sure what you're trying to do with the type hints here, but to structurally match based on class you need to provide a variable to which the value will be assigned, e.g.

foo = some_func()

match foo:
    case int(value):
        print(f"{value} is an integer")
    case float(value):
        print(f"{value} is a float")

This checks whether foo's type is already an int or a float (or a subtype thereof) and binds it to value, so no need to do int(value) inside the match body.

edited Dec 13, 2022 at 3:44

answered Dec 13, 2022 at 2:50

Chris

139k139 gold badges316 silver badges293 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Barmar · Accepted Answer · 2022-12-13 02:36:11Z

1

match/case is used for pattern matching, not for simple equality testing. A simple name like int is a pattern that matches anything, and sets int to the value.

You should use ordinary if conditions.

h = type_hints[name]
if h == int:
    foo[name] = int(value)
elif h == float:
    foo[name] = float(value)
elif h == str:
    foo[name] = value
else:
    print(f"Unknown type {h}")

answered Dec 13, 2022 at 2:36

Barmar

789k57 gold badges554 silver badges669 bronze badges

3 Comments

xtofl Over a year ago

Thanks! How does that differ from match n: 5: return 'five'? h is a subject, int may be its value, right?

xtofl Over a year ago

Ah - I see: peps.python.org/pep-0634/#literal-patterns are limited to numbers, strings and booleans. No types are allowed. That's arbitrary.

Barmar Over a year ago

Types aren't literals, they're variables whose values are type objects.

Collectives™ on Stack Overflow

Python match syntax for evaluating typing objects

2 Answers 2

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related