Type conversion during function call in python

Question

In the example below I'm passing float value to a function accepting and int argument (using type hints). Looks like the value read into the function arg is a float nonetheless (was expecting int(11.2) * 10 = 110 instead of 112)

Why is this the case?

def f(n:int):
    return n*10
l = [11.2, 2.2, 3.3, 4.4]
mfunc = map(f,l)
print(list(mfunc))

Result: [112.0, 22.0, 33.0, 44.0]

** Process exited - Return Code: 0 ** Press Enter to exit terminal

As you said, it's just a type hint. It tells the user that it expects to get an integer, but no enforcement action will be taken. — Mechanic Pig
– Mechanic Pig, Commented Jan 7, 2023 at 12:49
As @MechanicPig rightly said no implicit casting goes on when you used type hint. So you will have to do your casting explicitly. return int(n) * 10 — Maxwell D. Dorliea
– Maxwell D. Dorliea, Commented Jan 7, 2023 at 12:59

Ronin · Accepted Answer · 2023-01-07 13:31:33Z

2

As @Mechanic Pig said n:int in function signature is just a hint for developer and its not converts to int.

So you cast to int

def foo(n: int):
  if type(n) is float:
      n = int(n)
  return n * 10

Or you can use assert to raise error if n is not int

def foo(n: int):
    assert type(n) == int, "must be int"
    return n * 10

or

def foo(n: int):
    if type(n) is not int:
        raise Exception(f"Must by int instead of {type(n).__name__}")
    return n * 10

Hints more useful when you use IDE that support it.

Here -> int: describes what type function returns

def foo(n: int) -> int:
    return n * 10

edited Jan 7, 2023 at 13:31

answered Jan 7, 2023 at 13:02

Ronin

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