如何限制python函数的参数必须是字符串或类似lambda表达式的函数

Question

For example, the following, the first parameter should be restricted to a string, and the second parameter should be a function. However, this is wrong syntax for both. Anyone can help suggest the correct syntax to impose the type restriction?

def GetArg(msg:String, converter:lambda, default):
    print("{}, the default is '{}':".format(msg, default))
    return converter(stdin.readline().strip())

It gives error

Traceback (most recent call last):
  File "E:/stdin_ext.py", line 4, in <module>
    def GetArg(msg:String, converter, default:String):
NameError: name 'String' is not defined

and

  File "E:/stdin_ext.py", line 4
    def GetArg(msg:String, converter:lambda, default:String):
                                       ^
SyntaxError: invalid syntax

Answer 1

You can use the typing module.

from typing import Callable

def get_arg(msg: str, converter: Callable[[str], str], default) -> str:
    print(f"{msg}, the default is '{default}':")
    return converter(stdin.readline().strip())

assuming your function converter takes a string a returns a string , and that get_arg returns a string . Note the code has also been modified to follow python's naming convention and uses f strings over older style format strings.

Also note that these type hints are just that, hints . They are not checked statically or at runtime. Although, certain IDE's will help you ensure they are correct.

Answer 2

You should use str instead of String and Callable instead of lambda:

from typing import Callable    

def GetArg(msg:str, converter:Callable, default):
    print("{}, the default is '{}':".format(msg, default))
    return converter(stdin.readline().strip())

That is, unless you have a specific class called String and you are expecting an argument of its type.

When you write lambda you are defining a new lambda, not specifying a type for a function, whence the error.

I believe it is also important to point out that types in python are only a useful notation but it does not raise any error by itself, for example I could still call GetArg(1,2,3) and don't get any error just by the type hinting (of course I would get an error trying to pass an argument to an int afterwards).