overload in Python

Several signatures for one function, so the checker knows what comes back.

from typing import overload


@overload
def parse(text: str) -> int:
    ...


@overload
def parse(text: str, default: str) -> int | str:
    ...


def parse(text, default=None):
    try:
        return int(text)
    except ValueError:
        if default is None:
            raise
        return default


print(parse("42"), parse("x", "fallback"))
print(parse.__name__)

How it works

1. Each @overload stub declares one calling shape.
2. The real implementation follows, untyped by overload.
3. At run time only the implementation exists.

Keywords and builtins used here

• def
• except
• if
• int
• parse
• print
• raise
• return
• str
• try

The run, in numbers

Lines

24

Characters to type

319

Tokens

94

Three-star pace

110 tpm

At the three-star pace of 110 tokens a minute, this run takes about 51 seconds.

Type this snippet

Step 4 of 4 in Typing in depth, step 44 of 53 in Pythonic Python.

← Previous Next →